I Dfns group action & group, written in terms of the other

[elias001]

Below in the quoted passage is expressing the definition of group action in terms of the definition of a group and vice versa. Can someone check if it there any mistakes please. I had helped with one of the LLMs. I always want to know for the two definitions, if one can be written in terms of the other. It also includes numerical examples for both cases. I want to know how I reconcile with the definition of a group with the definition for group action. I had my suspicion that the definition of a group is the definition of a group action on the set that a group is defined on, or is it defined via faithful action? I wasn't sure in either case. Another thing is that the examples I requested deliberately avoids any mention of permutation groups. I know about Cayley's theorem, but like I said, I am trying to feel translating between the two different two definitions and having permutation group sandwiched between the two definitions. Also I am not sure about my math English ability to express things precisely.

Definitions and Examples of Groups and Group Actions Definition 1: Group in Terms of a Group Action

A $\textit{group}$ can be defined via a faithful group action. Suppose a set $G$ acts on a set $X$ via a map $\phi: G \times X \to X$, satisfying the group action axioms:

(Identity) There exists an element $e \in G$ such that for all $x \in X$, $\phi(e, x) = x$.
(Compatibility) For all $g, h \in G$ and $x \in X$, $\phi(g, \phi(h, x)) = \phi(gh, x)$, where $gh$ denotes a binary operation on $G$.


The action is $\textit{faithful}$ if for all $g, h \in G$, if $\phi(g, x) = \phi(h, x)$ for all $x \in X$, then $g = h$. If $\phi$ is faithful, then $G$ forms a group under the binary operation induced by the action, with:

Binary operation: The product $gh$ is defined such that $\phi(gh, x) = \phi(g, \phi(h, x))$.
Identity: The element $e \in G$ satisfying $\phi(e, x) = x$ for all $x \in X$.
Inverses: For each $g \in G$, there exists $g^{-1}\in G$ such that $\phi(g^{-1}, \phi(g, x)) = x$ and $\phi(g, \phi(g^{-1}, x)) = x$ for all $x \in X$.
Associativity: For all $f, g, h \in G, (fg)h = f(gh)$, which follows from the compatibility condition.

Thus, $(G, \cdot)$ is a group if the action is faithful, with the group operation defined via the action.

Example for Definition 1

Consider the set $G = \{1, -1\}$ acting on the set $X = \mathbb{R}$ (the real numbers) via the action $\phi: G \times \mathbb{R}\to \mathbb{R}$ defined by $\phi(g, x) = g \cdot x,$ where $\cdot$ is the usual multiplication in $\mathbb{R}$. We verify that this defines a group structure on $G$.


Identity: Check if there exists $e \in G$ such that $\phi(e, x) = x$ for all $x \in \mathbb{R}$. For $e = 1$, we have $\phi(1, x) = 1 \cdot x = x$, which holds for all $x \in \mathbb{R}$. Thus, $e = 1$ is the identity element.
Compatibility: For $g, h \in G$ and $x \in \mathbb{R}$, compute $\phi(g, \phi(h, x)) = \phi(g, h \cdot x) = g \cdot (h \cdot x) = (gh) \cdot x = \phi(gh, x)$, where $gh$ is the product in $G$ under usual multiplication. This holds since multiplication in $\mathbb{R}$ is associative.
Faithfulness: If $\phi(g, x) = \phi(h, x)$ for all $x \in \mathbb{R}$, then $g \cdot x = h \cdot x$, implying $(g - h)x = 0$. Since this must hold for all $x \in \mathbb{R}$, and $x \neq 0$ exists, we have $g - h = 0$, so $g = h$. The action is faithful.
Inverses: For $g = 1, \phi(1, x) = x$, so $1$ is its own inverse. For $g = -1$, check if there exists $g^{-1} \in G$ such that $\phi(g^{-1}, \phi(-1, x)) = x$. Since $\phi(-1, x) = -x$, we need $\phi(g^{-1}, -x) = x$. Try $g^{-1} = -1: \phi(-1, -x) = (-1) \cdot (-x) = x$. Thus, $(-1)^{-1} = -1$.


Define the group operation on $G$ via the action: $g \cdot h = gh$ (usual multiplication). The group $(G, \cdot)$ is:


Elements: $G = \{1, -1\}$.
Operation: Multiplication, e.g., $1 \cdot 1 = 1, 1 \cdot (-1) = -1, (-1) \cdot 1 = -1, (-1) \cdot (-1) = 1$.
Identity: $e = 1$.
Inverses: $1^{-1}= 1, (-1)^{-1}= -1$.
Associativity: Follows from the associativity of multiplication in $\mathbb{R}$.


Computation for Example 1

Verify the group axioms numerically:


Associativity: For $f = -1, g = -1, h = 1$, compute $(f \cdot g) \cdot h = ((-1) \cdot (-1)) \cdot 1 = 1 \cdot 1 = 1$, and $f \cdot (g \cdot h) = (-1) \cdot ((-1) \cdot 1) = (-1) \cdot (-1) = 1$. Equal.
Identity: For $g = -1$, check $1 \cdot (-1) = -1$ and $(-1) \cdot 1 = -1$.
Inverses: For $g = -1, (-1) \cdot (-1) = 1$, the identity.


Thus, $G = \{1, -1\}$ with multiplication forms a group, isomorphic to the cyclic group of order 2.

Explanation

The action $\phi(g, x) = g \cdot x$ induces a group structure on $G$ because it is faithful and satisfies the action axioms. The operation defined by $\phi(gh, x) = \phi(g, \phi(h, x))$ corresponds to multiplication in $G$, and the faithfulness ensures distinct elements in $G$ produce distinct transformations, yielding a group structure.

Definition 2: Group Action in Terms of a Group

A $\textit{group action}$ of a group $(G, \cdot)$ on a set $X$ is a map $\phi: G \times X \to X$ such that:


(Identity) There exists an identity element $e \in G$ such that $\phi(e, x) = x$ for all $x \in X$.
(Compatibility) For all $g, h \in G$ and $x \in X$, $\phi(g, \phi(h, x)) = \phi(g \cdot h, x)$, where $\cdot$ is the group operation in $G$.

Here, $(G, \cdot)$ is a group with a binary operation $\cdot$, identity $e$, inverses, and associativity.

Example for Definition 2

Let $G = (\mathbb{Z}/4\mathbb{Z}, +)$ be the group of integers modulo 4 under addition, with elements $\{0, 1, 2, 3\}$. Let $X = \mathbb{R}^2$, the 2-dimensional real plane. Define the action $\phi: G \times \mathbb{R}^2 \to \mathbb{R}^2$ by
$\phi([n], (x, y)) = (x \cos(n\pi/2) - y \sin(n\pi/2), x \sin(n\pi/2) + y \cos(n\pi/2))$, where $[n]$ denotes the class of $n$ modulo 4. This represents rotations of $\mathbb{R}^2$ by multiples of $90^\circ$.

Verify the group action axioms:

Identity: For $[0] \in G$, compute
$\phi([0], (x, y)) = (x \cos(0 \cdot \pi/2) - y \sin(0 \cdot \pi/2), x \sin(0 \cdot \pi/2) + y \cos(0 \cdot \pi/2)) = (x \cdot 1 - y \cdot 0, x \cdot 0 + y \cdot 1) = (x, y)$. Thus, $[0]$ acts as the identity.
Compatibility: For $[m], [n] \in G$, compute $\phi([m], \phi([n], (x, y)))$. First, $\phi([n], (x, y)) = (x \cos(n\pi/2) - y \sin(n\pi/2), x \sin(n\pi/2) + y \cos(n\pi/2)) = (x', y')$. Then, $\phi([m], (x', y')) = (x' \cos(m\pi/2) - y' \sin(m\pi/2), x' \sin(m\pi/2) + y' \cos(m\pi/2))$. Substituting $x', y'$ and using trigonometric identities, $\cos((m+n)\pi/2) = \cos(m\pi/2)\cos(n\pi/2) - \sin(m\pi/2)\sin(n\pi/2)$, $\sin((m+n)\pi/2) = \sin(m\pi/2)\cos(n\pi/2) + \cos(m\pi/2)\sin(n\pi/2)$, we get $\phi([m], \phi([n], (x, y))) = \phi([m+n], (x, y))$, since $[m+n]$ is addition modulo 4.


Computation for Example 2

Compute the action for $[n] = [1]$ and $(x, y) = (1, 0):\phi([1], (1, 0)) = (1 \cos(\pi/2) - 0 \sin(\pi/2), 1 \sin(\pi/2) + 0 \cos(\pi/2)) = (1 \cdot 0 - 0 \cdot 1, 1 \cdot 1 + 0 \cdot 0) = (0, 1).$
For $[n] = [2]:\phi([2], (1, 0)) = (1 \cos(\pi) - 0 \sin(\pi), 1 \sin(\pi) + 0 \cos(\pi)) = (1 \cdot (-1) - 0 \cdot 0, 1 \cdot 0 + 0 \cdot (-1)) = (-1, 0).$
Verify compatibility for $[m] = [1], [n] = [1], (x, y) = (1, 0):\phi([1], \phi([1], (1, 0))) = \phi([1], (0, 1)) = (0 \cos(\pi/2) - 1 \sin(\pi/2), 0 \sin(\pi/2) + 1 \cos(\pi/2)) = (0 - 1, 0 + 0) = (-1, 0).$
Since $[1+1] = [2]$, compute $\phi([2], (1, 0)) = (-1, 0)$, which matches.

Explanation

The group $G = \mathbb{Z}/4\mathbb{Z}$ acts on $\mathbb{R}^2$ by rotating vectors by multiples of $90^\circ$. The action respects the group structure because the composition of rotations corresponds to the sum of angles modulo $360^\circ$ (or $2\pi$), and the identity element $[0]$ corresponds to a $0^\circ$ rotation, leaving points fixed.

I am not sure why when I place things within a quoted passage, the displayed LaTex is not respecting default paragraph formatting. Some of the math is running off the pages. I don't know how to fix that. If anyone can provide suggestions, that would be greatly appreciated it.




Thank you in advance.

 

[fresh_42]

There are three different wordings of what a group action is: representations, homomorphisms, or operations. They all can be used to define a group action. The group itself should be considered as something different. You should not confuse them. Let's start with a group $G$ that acts on a set $X.$

a) Representation:
The elements $g\in G$ are said to be represented by transformations of $X$ by
$$
g\longmapsto (x\longmapsto g.x)
$$
This means we look at an element $g\in G$ and investigate what it does with $X.$

b) Homomorphism:
A group action of $G$ on $X$ is a group homomorphism $G\longrightarrow \operatorname{Sym}(X).$
This wording focuses on the rules $(gh).x=g.(h.x))$ and $1.x=x$ that defines a group action. In case $X=V$ is a vector space, $\operatorname{Sym}(X)$ becomes $\operatorname{GL}(V)$ and we call the group action linear.

c) Operation:
This is only another word for group action. We can either say that $G$ acts on $X$ or that $G$ operates on $X.$

All these wordings are equivalent, which becomes obvious if we add the additional rules of what a representation / homomorphism / operation means, namely the two conditions I have listed.

It is also clear that we have two different sets: $G$ and $X.$ Yes, we can also consider $X=G,$ but you should not start with this specific case unless you understand the concept in general. The risk of confusion is simply high here. We already have three different languages for the group action itself, and now you complete confusion be considering $X=G$ and bringing in the group operation, which is a fourth terminology describing the binary operation that defines a group. It has at prior nothing to do with the action of $G$ on $X$ that's also called an operation.

The definition of a faithful representation in your post is right away nonsense.

A group action of $G$ on $X$ is called faithful if the group homomorphism $G \longrightarrow \operatorname{Sym}(X)$ is a monomorphism, i.e., injective. These are specific representations, since they allow us to consider $G$ as a subgroup of $\operatorname{Sym}(X)$ by that homomorphism.

-----------------

To answer your question about the specific case $X=G,$ we consider the group homomorphism
$$
\phi\, : \,G\longrightarrow \operatorname{Sym}(G)
$$
defined by $(\phi(g))(x)=g\cdot x.$ We must be sure that $\phi$ is a group homomorphism. This means that $(\phi(g\cdot h))(k)=(\phi(g)\circ \phi(h))(k)=\phi(g)((\phi(h))(k))$ which is the associativity in $G$ if we write it out. Of course, we also have $(\phi(1))(g)=1\cdot g=g.$

If we consider the kernel of $\phi$ then we have
\begin{align*}
\operatorname{ker}(\phi)&=\{g\in G\,|\,\phi(g)=\operatorname{id}_G\}\\&=\{g \in G\,|\,(\phi(g))(h)=h\text{ for all }h\in G\}\\&=\{g\in G\,|\,gh=h\text{ for all }h\in G\}\\&=\{1\}.
\end{align*}
This means that this operation is faithful. However, not every faithful operation looks like this. It's only a very special case. The "standard operation" of $G$ on itself is usually the conjugation, $g\longmapsto (h\longmapsto h^{-1}gh).$

-----------------

Do not use LLMs if you want to learn something. E.g., from Wikipedia:

A large language model (LLM) is a language model trained with self-supervised machine learning on a vast amount of text, designed for natural language processing tasks, especially language generation.

Do you see the contradiction? The definition of LLMs says that you train them, not the other way around! It is ultimately a self-training group without any control instance. They are the answer to the question: Who does not learn anything from whom?

LLMs only have a larger library. Unfortunately, this library contains the entire internet, including its vast amount of absolute nonsense.

My approach: I searched for "group operation + pdf" and my first hit was
https://kconrad.math.uconn.edu/blurbs/grouptheory/gpaction.pdf
on a server from the University of Connecticut. Way better than "anything" on the internet. My filter is simply better than yours! Take the time to study real university texts rather than relying on the arbitrariness of LLMs.

 

[elias001]

@fresh_42 i specifically am looking for two things, phrasing the one of the definition in terms of the other, and not making any reference to symmetric/permutation groups, including cayley's theorem. I want to purely work with the two definitions.

That all translates into, I have a set $X$ and some binary operation $\circ$ defined on it that satisfied the group axioms. How does that satisfied the definition of group action.

Conversely, i have a group $G$, and a set $X$, with some binary operation $\circ$, and the so called group $G$ "acting on" $X$, but the "acting on" part, is that the same operation as the the binary operation $\circ$?

I am sticking with the keep it simple and short principle and only stick to the two definitions and nothing else.

Also once permutstion group get introduce as an example of group actions, things get more complicated and muddy the water as far as if one wants to know if the two definitions are referring to the sane mathematical situation. Basically i have a single group $G$, a single set $X$ and a binary operation $\circ$ thar satisfies the group axioms. Those are the only cast of characters I am limiting to. I am not interested in paying for any extra concepts, or examples like permutation, symmetric, diherdral, or general linear group, vector spaces, lie groups, linear transformations, etc, unless any of them only uses those three characters. The situation in the definition for a group, how can that be rephrase using the definition of group action? The definition of group action, how can that be rephrase using purely the definition of a group? What you and Conrad has written are causing more confusion for me. All i am asking something very simple. Call it the equivalent of moving out into the rural side of mathematics, because all that extra stuff feels like living in the math equivalent of NYC.

 

[fresh_42]

@fresh_42 i specifically am looking for two things, phrasing the one of the definition in terms of the other, and not making any reference to symmetric/permutation groups, including cayley's theorem. I want to purely work with the two definitions.

I do not see two different definitions, only two different subjects: the action of $G$ on $X$ and the binary group multiplication $G\times G\longrightarrow G.$ You cannot compare what is already different by definition. The second can be viewed (but usually is not) as a special case of the first.

If we define $X=G$ and set $g.x=g\cdot x$ then $g\longmapsto (x\longmapsto g\cdot x),$ which is the group multiplication, is also a group action.

Proof: $(g\cdot h)(x)=(g\cdot h)\cdot x=g\cdot (h\cdot x) = g\cdot (h.x)=g.(h.x)$ and $1.x=1\cdot x=x.$

Conversely, there is no way to define a general group action by the group multiplication. We would need a group homomorphism $G\longrightarrow \operatorname{Sym}(X),$ and how would $X$ come into play here?

That all translates into, I have a set $X$ and some binary operation $\circ$ defined on it that satisfied the group axioms. How does that satisfied the definition of group action.

See my proof above. This question, reworded, means: If $X$ is a group, then how is its multiplication a group action?

Conversely, i have a group $G$, and a set $X$, with some binary operation $\circ$, and the so called group $G$ "acting on" $X$, but the "acting on" part, is that the same operation as the the binary operation $\circ$?

No. Consider a matrix group, e.g.,
$$
G=\left\{\begin{pmatrix}\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi)&\cos(\varphi)\end{pmatrix}\right\}
$$
and $X=\mathbb{R}^2$ with the binary operation $\circ=+.$ Then
$$
g.\vec{v}=\begin{pmatrix}\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi)&\cos(\varphi)\end{pmatrix}\begin{pmatrix}x_v\\y_v\end{pmatrix}=
\begin{pmatrix}x_v\cos(\varphi)-y_v\sin(\varphi)\\ x_v\sin(\varphi)+y_v\cos(\varphi)\end{pmatrix}
$$
which has nothing to do with
$$
g\cdot h=\begin{pmatrix}\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi)&\cos(\varphi)\end{pmatrix}\cdot \begin{pmatrix}\cos(\psi)&-\sin(\psi)\\ \sin(\psi)&\cos(\psi)\end{pmatrix}=
\begin{pmatrix}\cos(\varphi+\psi)&-\sin(\varphi+\psi)\\ \sin(\varphi+\psi)&\cos(\varphi+\psi)\end{pmatrix}
$$

I am sticking with the keep it simple and short principle and only stick to the two definitions and nothing else.

You want to compare two different concepts. At best, one is a special (but widely ignored) case of the other.

The idea behind a group action is that if we only look at the group as in my example, then we only see the addition theorems of the sine and cosine function. But if we consider the group action, here its natural action on the two-dimensional vector space, then we see rotations. And rotations preserve angles and lengths, which provides a larger field of applications than the group multiplication alone.

And there are countless other operations of $G$ on various examples of $X.$ The definition $X=G$ and $g.h=\phi(g)(h)=g\cdot h$ by the group multiplication is only one of them. Other, more important operations would be the conjugation within the group or the adjoint representation on its tangent space.

Also once permutstion group get introduce as an example of group actions, things get more complicated and muddy the water as far as if one wants to know if the two definitions are referring to the sane mathematical situation.

They don't, and that's what sets your question on false assumptions. You should distinguish between the two concepts.

 

[elias001]

@fresh_42 i explicitly said " I am not interested in paying for any extra concepts, or examples like permutation, symmetric, diherdral, or general linear group, vector spaces, lie groups, linear transformations, etc, unless any of them only uses those three characters." That means what ever binary operation gets introduced should be the same for both definitions. So yes, you can talk about conjugation, but for both definitions, we stick to conjugations. I am not asking if the two definitions are equivalent, I am asking how one definitions can he phrased using the other definition and vice versa. I don't care if one of them is a special case of the other, or if the other one is the brother in law to the other one's wife call the pernutarion call. Also, i am not comparing two different concepts. That is not what I am trying to do or understand. What you almost said If $X$ a set with a binary operation $\circ$ that satisfied the group axioms, then how is $X$'s $\circ$ operation a group action?

Also given the definition of group action, why can't it be rephrase purely using the definition of a group. The "acting on" part only means a single binary operation. Also, we are talking about groups, so one binary operation only.

 

[fresh_42]

@fresh_42 i explicitly said " I am not interested in paying for any extra concepts, or examples like permutation, symmetric, diherdral, or general linear group, vector spaces, lie groups, linear transformations, etc, unless any of them only uses those three characters." That means what ever binary operation gets introduced should be the same for both definitions. So yes, you can talk about conjugation, but for both definitions, we stick to conjugations. I am not asking if the two definitions are equivalent, I am asking how one definitions can he phrased using the other definition and vice versa. I don't care if one of them is a special case of the other, or if the other one is the brother in law to the other one's wife call the pernutarion call.

You are defining a fruit and an apple. An apple is a fruit, but not vice versa.

The two required conditions of a group action are $g.(h.x)=(g\cdot h)(x),$ which is the group's associativity in case the operation is the group operation, and $1.x=x,$ which is the definition of the group's neutral element. This specific operation is faithful,
$$
g.x=x \Longrightarrow g=1,
$$
by the uniqueness of the neutral element in the group.

That's it.

However, and that has to be noted here, we still have two different subjects. One is the group multiplication, and the other is an identification of group elements $g$ by what they do on $G.$ We identify
$$
g \leftrightarrow (h\longmapsto g\cdot h)
$$
where $g\neq (h\longmapsto g\cdot h).$ One lives in $G$ and the other one in $\operatorname{Sym}(G).$ The latter is automatically associative by the definition of a consecutive application of transformations. The identification is a monomorphism, which corresponds to the statement that $G$ operates faithfully on itself by left-multiplication.

The group action here is this specific identification. They are not the same, only a linguistic correspondence.

 

[elias001]

@freah_42 But in one of the example of group actions, it uses the example of a triangle with its three vertices which can be the set $X$ we are talking about and the binary operation is rotation by $0^\circ, 120^\circ, 360^\circ$ So we have one binary operation which is rotation. Group axioms that any binary operation has to satisfy also contains associativity. To me, rotation of vertices of a triangle satisfies the definition of a group, and according to the standard examples lore, it satisfie the definition for group action.


When we ask LLMs, it even say that the person doing the asking should consult with experts since the answers might contain mistakes. If we want AIs to be like humans, it better learn to make mistakes like humans and it should learn like humans, by making mistakes.

 

[fresh_42]

@freah_42 But in one of the example of group actions, it uses the example of a triangle with its three vertices which can be the set X we are talking about and the binary operation is rotation by 0∘,120∘,360∘ So we have one binary operation which is rotation. Group axioms that any binary operation has to satisfy also contains associativity.

Yes, that's basically the group $\mathbb{Z}_3$ represented by rotations of $120°.$ The associativity in $\mathbb{Z}_3$ corresponds to the associativity of the rotations. The homomorphism property of
$$
\bigl\langle a\,|\,a^3=1 \bigr\rangle \longrightarrow \begin{pmatrix}\cos(120°)&-\sin(120°)\\ \sin(120°)&\cos(120°)\end{pmatrix}^a
$$
corresponds to the action conditions.

I have particularly used $\mathbb{Z}_3$ to emphasize the difference between the two groups. One is simply the cyclic group with three elements, and the other one is the symmetry group of the equilateral triangle. They are formally not the same, although there is only one group with three elements - up to isomorphisms. And the one above is one of them. $a \longmapsto a+3\cdot \mathbb{Z}$ would be another one.

 

[elias001]

@freah_42 your matrix group example with $g\\.\bar{v}$ and $g\cdot h$, are you referring to associativity? Why is there two different binary operations? That is not allowed. I only want to stick to one binary operation. Also, i don't care what other group the group for rotating three vertices of a triangle is isomorphic to. We have one example, but two different definitions can apply to it.

So we have a group that acts on a set $X$, and that means with a single binary operation. How can that be reformulated using the definition of a group.

 

[fresh_42]

@freah_42 your matrix group example with $g\ldot \bar{v}$ and $g\cdot h$, are you referring to associativity? Why is there two different binary operations? That is not allowed. I only want to stick to one binary operation.

You can't. One binary operation is the group multiplication, the other one is the left-multiplication. They are formally not the same.

Given a (multiplicatively written) group $G$ then this multiplication is given by
$$
G\times G\longrightarrow G\quad,\quad (g,h)\longmapsto g\cdot h
$$
whereas the action of the group on itself is formally given by
$$
G \longrightarrow \operatorname{Sym}(G)\quad,\quad g\longmapsto L_g
$$
where $L_g$ denotes the left-multiplication by the element $g.$ That is $L_g(h)=g\cdot h.$

You can identify $g$ with $L_g,$ but they are formally not the same. The action here is the fact that $L_{g\cdot h}=L_g\circ L_h$ defines an injective group homomorphism. The action property is the homomorphism property. Associativity in $G$ and $\operatorname{Sym}(G)$ are given by the requirement that these are both groups. It is used to prove the homomorphism property, however. So in a way, the action is a consequence of group associativity. It is therefore sufficient to get this specific action.

Whether it is necessary is a logical problem. We need it to set up an action, but given the action, how could we prove its components were groups without already requiring that they are? We cannot phrase the conditions of a group action without already having a group multiplication.

So we have a group that acts on a set $X$, and that means with a single binary operation. How can that be reformulated using the definition of a group.

By saying that $G\longmapsto \operatorname{Sym}(X)$ is a group homomorphism.

The other way around is the problem. You cannot set up a group action without having groups.

Like I said, I know you are very good at being the capo dei capi of the math mafia by drawing in other stuff, but just for this post, let's try to conduct normal legitimate business. :)

It's not legitimate to compare fruits and apples. They belong to different categories.

 

[elias001]

@fresh_42 wait, why does left multiplication come in? Hungerford only talk about that and excited out the topic of group action.

Basically we are saying if we have a definition of a group, the definition of a group can be written in terms of the definition of group action.

But if we are given the definition of group action, then group.action definition can not be written in terms of the definition of a group.

But to go back to our example of rotation of triangle vertices, we have a group of rotation, a set consisting of vertices, and rotation operaton as the binary operation. It seems like we have the definition of a group? Why is left multiplication being involved? When i see the non mathematical notation explainations, it doesn't match for me what I see from the math notations.

Also, I asked LLMs because I still have questions in some of my past posts here. It might be better i ask it and post it in those posts and ask you to take a look at them to check if they are correct.

 

[fresh_42]

@fresh_42 wait, why does left multiplication come in? Hungerford only talk about that and excited out the topic of group action.

That's because it is the action you wanted to talk about: $G\times X\longrightarrow X$ defined as $X=G$ and $(g,x)\longmapsto g\cdot x = L_g(x).$ The principle of using actions instead of only regarding the group operation is investigating what $G$ does on $X.$ And this is shuffling the elements of $X$ even in the case of $X=G.$ We change from inspecting $g\in G$ to inspecting $L_g\in \operatorname{Sym}(G).$

The fact that they can be identified $g\leftrightarrow L_g$ only adds confusion to the topic. They are formally different things: $g$ is operating (acting) via left-multiplication on $G.$ Hence, there has a decision to be made: to talk about $g$ or its action on $G$ by multiplication? It is from the left because this is the associative law in $G.$ If we considered right-multiplication instead, then we get
$$
R_{gh}(x)=x(gh)=(xg)h=R_h(xg)=R_h(R_g(x))=(R_h\circ R_g)(x)
$$
instead, which means that the group homomorphism would become an anti-homomorphism, or - and that is the common way to resolve this issue - we would need to define the consecutive application of functions from left to right: $(R_h\circ R_g)(x)=R_g(R_h(x)).$

Left-multiplication doesn't get us into this notational inconvenience.

Basically we are saying if we have a definition of a group, the definition ofva group can be written in terms of the definition of group action.

Basically, yes, but nobody does this except for having a specific example of a group action. If we already have a group and a binary operation $G\times G \longrightarrow G,$ why use another terminology? You have to start with a group to even define group action, only to end up with what you already had from the beginning: a group.

That's a bit as if you wanted to talk about oaks and start talking about forests, requiring that there are oaks in the woods.

But if we are given the definition of group action, then group.action definition can not be written in terms of the definition of a group.

Let me analyze this:

But if we are given the definition of group action, ...

Means: we have a group homomorphism $G\longrightarrow \operatorname{Sym}(G).$

... then group.action definition ...

The homomorphism property.

... can not be written in terms of the definition of a group.

That would require defining an object of a category, here the category GRP of groups, by means of its morphisms. I don't know whether homological algebra does have or does not have a kind of duality to solve this logical problem. In regular terms: no, since we need the objects and their structures before we can talk about what it means to preserve these structures by morphisms. Maybe there is such a thing as a cocategory that exchanges objects and morphisms, but I'm not sure.

A group action is a homomorphism. Group axioms are what are preserved by homomorphisms.

But to go back to our example of rotation of triangle vertices, we have a group of rotation, a set consisting of vertices, and rotation operaton as the binary operation. It seems like we have the definition of a group?

Yes, you defined the term symmetry group (of an equilateral triangle). You can show in a next step that this group is $\mathbb{Z}_3$ and that $\mathbb{Z}_3$ operates on the vertices of the triangle by rotations. However, you did not define the multiplication in $\mathbb{Z}_3.$ You only used the given multiplication of the consecutive applications of rotations. It is quite obvious that this is identical to the multiplication in $\mathbb{Z}_3$ since there is only one group (up to isomorphisms) with three elements, so that the identification (the isomorphism) is obvious. But what if I gave you the symmetry group of Rubik's cube instead? How would you define its group operations such that we can identify the group and show the isomorphism?

Again, $\{g\,|\,g\in G\}=G$ and $\{L_g\,|\,g\in G\} \subseteq \operatorname{Sym}(G)$ can be treated as isomorphic groups, but they are not identical.

 

[elias001]

@fresh_42 So if we are given the definition of group action, then group.action definition can not be written in terms of the definition of a group. Can this statement be rewritten in more math notations, and can the claim be proven?

Also in the definition of group action, we have a binary operation $\circ,$ and $G$ is a group. Suppose $G=X$ or $G\neq X$, in either case, for $g\in G$ what is $g$? How is $g$ related to $\circ?$ For $L_g$, what is the notation mean? If $L_g(x)=g\cdot x,$ does $g\cdot x$ mean the same as $g\circ x?$

I should also mention, Michael Artin's Algebra texr does the same as Hungerford.

For the longest time, the $g$, and the two operations $\\., \cdot$ in $g\\.(h\cdot v)=(g\\.h)\cdot v$ drove me bannanas.

 

[fresh_42]

@fresh_42 So if we are given the definition of group action, then group.action definition can not be written in terms of the definition of a group.

It can. By saying that the group action defines a group homomorphism. This requires knowing what a group is beforehand. The other way around is the problem, namely, to retrieve the group axioms from such a homomorphism. If we invest all that we need, then it is possible. E.g., if $\varphi\, : \, G\longrightarrow \operatorname{Smy}(X)$ is an injective homomorphism of bijective transformations on $X,$ then $G\cong \operatorname{im}(\varphi)$ and we can use this image to define the group operation in $G.$ But what did we invest? We required a faithful group action of bijective transformations on a set $X$. And what did we gain? A group that is isomorphic to $\operatorname{im}(\varphi).$ That doesn't help us to understand $G.$ It only helps in understanding the subgroups of $\operatorname{Sym}(X).$

You would use $G$ to describe it by $\operatorname{im}(\varphi)$ but you wouldn't use $\operatorname{im}(\varphi)$ to describe $G,$ simply because you don't know anything about $G.$

Your example with the triangle shows that. You can define the symmetry group. But you wouldn't use rotations to describe the remainders of a division by three.

Can this statement be rewritten in more math notations, and can the claim be proven?

Which claim? You could try to formalize the equivalence of

1. $G\longrightarrow \operatorname{Sym}(X)$ is a group homomorphism.
2. The group $G$ acts on the set $X.$
3. The group $G$ is represented by bijective transformations of $X.$

which is the general case. Your specific statement would be:

A group $G$ operates on itself via left-multiplication.

If you try to show something in the opposite direction, then you end up with trivial statements. We would start with a subgroup of $\operatorname{Sym}(X)$ and would show that it is a group. That doesn't make sense. And if we considered any subset of transformations of $X$ instead, we weren't able to derive anything.

Also in the definition of group action, we have a binary operation $\circ,$ and $G$ is a group. Suppose $G=X$ or $G\neq X$, in either case, for $g\in G$ what is $g$? How is $g$ related to $\circ?$ For $L_g$, what is the notation mean? If $L_g(x)=g\cdot x,$ does $g\cdot x$ mean the same as $g\circ x?$

That depends on what you mean by $g\circ x.$ As there are different terminologies to describe a group action so there are different notations.

What's going on is that an element $g\in G$ of the groups "shuffles" the elements of a set $X$ and obeys certain conditions. Hence, we have a group element $g,$ we have shuffling the elements of $X,$ and we have the result of that process.
$$
g \stackrel{action}{\longrightarrow } \left( X \stackrel{shuffle}{\longrightarrow }X \right)
$$
or a bit more mathematical
\begin{align*}
G&\longrightarrow (X\longrightarrow X)\\
g&\longmapsto (x\longmapsto g.x)\\[6pt]
G\times X &\longrightarrow X\\
(g,x)&\longmapsto g.x
\end{align*}
We have a map from a group into a group of mappings. The first notation emphasizes the homomorphism from $G$, the second one with the direct product emphasizes the shuffle of $X$.

My favorite notation is that of a group homomorphism $\varphi\, : \,G\longrightarrow \operatorname{Sym}(X).$ This means $\varphi(g\cdot h)=\varphi(g)\varphi(h).$ The multiplication on the right is a consecutive application of first $\varphi(h)$ and then $\varphi(g).$ It is sometimes written as $\varphi(g)\varphi(h)=\varphi(g)\circ \varphi(h)$ but also as $\varphi(g)\varphi(h)=\varphi(g)\cdot\varphi(h).$

The elements $\varphi(g)$ and $\varphi(h)$ are bijective transformations of $X,$ i.e., we can apply them to an element $x\in X.$ This can be written as
$$
\varphi(g)(x)=g.x=gx=g\circ x
$$
all of them being correct, and depending on the author and which part of the process you want to emphasize. In case of $X=G$ and the left-multiplication $L_g$, we even have
$$
\varphi(g)(x)=g.x=gx=g\circ x=L_g(x).
$$
This is the result of the shuffling process, namely where $x\in X$ is located after we applied $g.$

I should also mention, Michael Artin's Algebra texr does the same as Hungerford.

I am not surprised. Both are reputable addresses. But again, it is only a matter of perspective and where you want to lay the emphasis. If I'm only scribbling or want to be fast, then I write $g.x$ instead of $g\circ x$ or $\varphi(g)(x).$ If I want to distinguish whether I am in $G,$ in $X,$ or in the orbit of $g,$ then I prefer the extensive notation $\varphi(g)(x)$ that allows me to "see" where I am.

When we are talking about one single group, one binary operation, one set, if we are not talking about legitimate business, it would feel like you are trying to cheat me metaphorically mathematically speaking. Is like you can force a rock to bleed out blood when it itself don't even contain any trace of water.

This is the case because of the variety of different languages and different notations when it comes to group action. Plus, the fact that you asked about something that is impossible in your very first sentence:

Below in the quoted passage is expressing the definition of group action in terms of the definition of a group and vice versa.

There is no vice versa.

 

[elias001]

@fresh_42 wait wait,

Which of the folllowing two situations is possible, state a yes or a no

Express the definition of a group action in terms of the definition of a group.

Express the definition of a group in terms of the definition of a group action.

Next,

Say we have the set $X$ consisting of the three vertices of a triangle and the rotation operation as its binary operation $\circ$. This binary operation satisfies the group axioms. So the binary operation along with the set $X$ form a group $G$.
Next we define a group action $G$ "acting on" $X$... What would be the specific elements of $g\in G$?

Still sticking with the same example, when i read the equality $g\cdot(h\cdot x)=gh\cdot x$, the operations $gh, g\cdot h$ are actually two distinct and different operations? If so, how does that relates to $\circ$?

What are the difference between "action" vs "shuffle"? Again are they referring to the two distinct and different operations $gh, g\cdot h$?

 

[fresh_42]

@fresh_42 wait wait,

Which of the folllowing two situations is possible, state a yes or a no

Express the definition of a group action in terms of the definition of a group.

Yes.

Express the definition of a group in terms of the definition of a group action.

No.

Next,

Say we have the set $X$ consisting of the three vertices of a triangle and the rotation operation as its binary operation $\circ$. This binary operation satisfies the group axioms. So the binary operation along with the set $X$ form a group $G$.

Yes.

Next we define a group action $G$ "acting on" $X$...

So we have a group homomorphism $\varphi\, : \,G\longrightarrow \operatorname{Sym}(X).$

I need this notation to answer your next question.

What would be the specific elements of $g\in G$?

You can define $G=\operatorname{im}(\varphi).$ But you need $G$ in the first place. Otherwise, this question doesn't make sense, since we already had to use $G$ in order to define the action $\varphi.$

Still sticking with the same example, when i read the equality $g\cdot(h\cdot x)=gh\cdot x$, the operations $gh, g\cdot h$ are actually two distinct and different operations? If so, how does that relates to $\circ$?

The notation $g \circ x$ is rare to unused, so better forget it. I only listed it as a possibility because you came up with it. The usual notation is $g.x$ or simply $gx$ if the context is clear.

The sign $\circ$ usually refers to the consecutive application of functions. Since $\varphi(G)$ are functions (of $X$), we can write $\varphi(g\cdot h)=\varphi(g)\circ \varphi(h).$ The dot is the multiplication in $G,$ and the circle is the multiplication of functions in $\operatorname{Sym}(X).$

If we apply this equation to a specific element $x\in X$ then we get
$$
(g\cdot h).x=\varphi(g\cdot h)(x)=(\varphi(g)\circ \varphi(h))(x)=\varphi(g)(\varphi(h)(x))= g.(h.x).
$$

What are the difference between "action" vs "shuffle"?

That was only a metaphore for what $g$ does with the elements of $X.$ $G$ acts on $X$ is the general picture, $g$ shuffles the elements of $X$ is what happens: $X\longrightarrow \{g.x\,|\,x\in X\}.$ If you number your vertices of the triangle, then any rotation will change these labels. The order will be different afterward.

Again are they referring to the two distinct and different operations $gh, g\cdot h$?

I don't know what you mean. $gh$ looks like the group multiplication, as does $g\cdot h,$ so there is no difference. If $gh$ stands for some kind of operation of $G$ on itself, then you should write $g.h$ because there are different possible actions of this kind. And the lower dot has to be explicitly defined. $g.h=g\cdot h$ is one possibility, $g.h=h^{-1}\cdot g\cdot h$ is another.

 

[elias001]

@fresh_42 For the case of $g\in G,$ let's try a concrete example. Say we have a set $X=\{x_1,x_2,x_3\},$ where each $x_i$ denotes one of the three vertex of a triangle, or what if the case for $X=G$? I am sure there is a technical difference, but not sure what specifically. Also let the rotation binary operation be $\circ$, consisting of the rotations $=\{R_{0°}R_{120°}R_{240°}\}=\{r_0,r_1,r_2\},$ and the three reflections operations be $\{t,u,v\}$ Then $G=\{r_0,r_1,r_2,t,u,v\}$. Then $g$ will be one of the 6 elements of $G$. So if we have $\phi_g(x)=g\circ x$, that means I specific $g$ to be some element of $G$, say $g=r_1$. Then since $G=X$ or $G\neq X$, $G$ acts on $X$ means I apply the binary operation to $g=r_1$ and every elements of $X$? I rinse and repeat by letting $g$ equal to each of the other elements in $G$, that is $g\circ x, \forall x\in G$. Is this correct interpretation?

If we have $\phi:G\times X\to S_X$ as a group action homomorphism map, and again $X$ might equal to $G$ or not. Let $A(G)$ denote the permutation group of $G$ and $S_X\subset A(G),$ is the symmetric group of $G$. The Cayley theorem states the map $f:G\to A(G)$ is an isomorphism along with the corollary $f:G\to S_X$ is also an isomorphism. The relationship between $\phi, f$ is what? i feel like there must be some sort of commutative diagram and there should be some sort of universal mapping property around the corner? Lastly what are the relationships between the left regular multiplication $L_g,$ $\phi_g(x), f?$

If expressing the definition of a group in terms of the definition of a group action is not possible. How can it be formally proven that is not possible?

 

[fresh_42]

@fresh_42 For the case of $g\in G,$ let's try a concrete example. Say we have a set $X=\{x_1,x_2,x_3\},$ where each $x_i$ denotes one of the three vertex of a triangle, or what if the case for $X=G$? I am sure there is a technical difference, but not sure what specifically.

The technical difference is the group structure on $G.$ If you set $G=X,$ then $X$ is an unstructured set, and you only defined the group elements, not how they should be multiplied. By writing $X=G,$ you implicitly assume that $X$ is carrying the group structure, too, without telling which it is.

If you only start with the set $X=\{x_1,x_2,x_3\}$ of vetices, then there are two possible groups acting on it: Even permutations that correspond to rotations of $0°, 120°,$ and $240°,$ and all permutations that also allow reflections along the heights of the equilateral triangle. The first group has three elements, the second has six. $G=X$ rules out the second possibility, but we still don't know what the group structure is. Ok, there is only one group with three elements, so we know it in this case (up to isomorphisms), but what if we choose a square instead? Then there are two possible group structures on $X= \{x_1,x_2,x_3,x_4\}.$ In this case, the notation $X=G$ only makes sense if we have already defined the group structure on $G.$ Otherwise, it is simply a set of four vertices.

Also let the rotation binary operation be $\circ$, consisting of the rotations $=\{R_{0°}R_{120°}R_{240°}\}=\{r_0,r_1,r_2\},$ and the three reflections operations be $\{t,u,v\}$ Then $G=\{r_0,r_1,r_2,t,u,v\}$. Then $g$ will be one of the 6 elements of $G$. So if we have $\phi_g(x)=g\circ x$, that means I specific $g$ to be some element of $G$, say $g=r_1$. Then since $G=X$ or $G\neq X$, $G$ acts on $X$ means I apply the binary operation to $g=r_1$ and every elements of $X$? I rinse and repeat by letting $g$ equal to each of the other elements in $G$, that is $g\circ x, \forall x\in G$. Is this correct interpretation?

Yes, up to some clumsy wordings. You had the same idea as I had in the previous section. Both groups $\mathbb{Z}_3=A_3$ and $\operatorname{Sym}(3)=S_3$ operate on the set $X=\{x_1,x_2,x_3\}.$

A group action associates a bijective mapping $X\longrightarrow X$ to every group element $g\in G.$ This makes the difference between an action and the group itself. The elements $g\in G$ aren't mappings, except that you defined them as such. That's why the situation is a bit different in your example, where you had those three rotations right from the beginning. In that case, you started the definition of $G$ by its action on $X=\{x_1,x_2,x_3\}.$ This isn't the case if we begin with an arbitrary group.

However, if you identify the group elements with the left-multiplication in $G,$ then they become mappings. That's where the entire discussion comes from. $G$ operates on $X=G$ by the definition $g.x=L_g(x)=g\cdot x.$ But that's only one possible action of $G$ on itself, and there are even more if we allow $X$ to be any set. And you made the identification $g \longleftrightarrow L_g.$ This changes the starting point of the discussion. If you do so, you have to say it.

If we have $\phi:G\times X\to S_X$ as a group action homomorphism map, and again $X$ might equal to $G$ or not. Let $A(G)$ denote the permutation group of $G$ and $S_X\subset A(G),$ is the symmetric group of $G$.

That is a very unfortunate notation. We usually write $S_n$ or $\operatorname{Sym}(n)$ for the entire permutation group of $n$ elements, and reserve $A_n$ for the even permutations, a normal subgroup of $S_n$ of index two.

The Cayley theorem states the map $f:G\to A(G)$ is an isomorphism along with the corollary $f:G\to S_X$ is also an isomorphism. The relationship between $\phi, f$ is what? i feel like there must be some sort of commutative diagram and there should be some sort of universal mapping property around the corner?

I have serious trouble understanding your question. It seems to me that you consider the action of a group with $n$ elements on itself, i.e., on a set with also $n$ elements. The difference is still the missing group structure on $X=\{x_1,x_2,x_3,\ldots,x_n\}.$

I do not see a difference between your $A(G)$ and $S(X).$ They both look as if they should be $\operatorname{Sym}(X)=S_n.$

If $f\, : \,G\longrightarrow S_n$ is an isomorphism, then you started with an isomorphic copy of the permutation group of $n$ elements.

Cayley's theorem says that every group is isomorphic to a subgroup of the permutation group.

So let's go back to a group $G$ with $n$ elements. Then we associate every group element $g\in G$ with its left-multiplication $L_g\, : \,G\longrightarrow G$ defined by $L_g(h)=g\cdot h.$ We already know that this is a faithful representation, i.e., $\varphi(g):=L_g$ is an injective homomorphism $\varphi\, : \,G\longrightarrow \operatorname{Sym}(G)=\operatorname{Sym}(n)=S_n.$ This already proves Cayley's theorem. The wording comes from the fact that $\operatorname{im}(\varphi)\subseteq S_n$ is a subgroup, and if we only consider the image points in the codomain, then the injective group homomorphism is also surjective, hence an isomorphism, so $G$ has an isomorphic subgroup in $S_n.$

The diagram is only a line:
$$
G \stackrel{1:1}{\longleftrightarrow } \{L_g\,|\,g\in G\} \subseteq \operatorname{Sym}(G)=S_n
$$
The identification $\varphi(g)=L_g$ is the key that defines an action, and the restriction to only those permutations that occur as an image point $L_g$ makes it isomorphic to a subgroup.

The equilateral triangle with the rotations is such an example of a proper subgroup. We have
$$
G=\mathbb{Z}_3\stackrel{1:1}{\longleftrightarrow }\{L_g\,|\,g\in \mathbb{Z}_3\}=A_3 \triangleleft S_3
$$
because the rotations correspond to only the even permutation in $A_3$ with $3$ elements, whereas the group of all permutations (if we allow reflections) $S_3$ has $6$ elements.

Lastly what are the relationships between the left regular multiplication $L_g,$ $\phi_g(x), f?$

I hope I answered this since I still don't know what $f$ and $\phi$ actually are.

If expressing the definition of a group in terms of the definition of a group action is not possible. How can it be formally proven that is not possible?

I'm not sure it is impossible. I don't know all the tricks in homological algebra. But in the realm of ordinary algebra, it is impossible because you cannot define a group action without a group. It is a logical impossibility. How would you define a group action without defining what $g\cdot h$ means?

 

[elias001]

@fresh_42 Hey I DM you. :) Also, since you don't know if it requires homological algebra, I was hoping I can see a simple proof by contradiction kind of argument. Should I post this on MSE? What should I say/how should i phrase the question so that I don't have to deal with certain egotistical crazy Karens on there.

Also, the two $f$s are the two isomorphic maps I listed related to cayley's theorem and $\phi_g$ is the group action homomorphism map. Is $L_g$ same as $\phi_g$? I thought there are relations between them. Also, when you use the term "identify", what do you mean by that in mathematical notations?

 

[fresh_42]

@fresh_42 Hey I DM you. :)

I know, I will answer later. I was / am a bit busy.

Also, since you don't know if it requires homological algebra, I was hoping I can see a simple proof by contradiction kind of argument. Should I post this on MSE? What should I say/how should i phrase the question so that I don't have to deal with certain egotistical crazy Karens on there.

I have no idea. I always run into them on MSE. It is as if you needed an answer before posting the question. If you don't have this answer, they call you silly, and if you have it, their only contribution is to tell you that there are mistakes in the wording.

I haven't yet figured out a single instance where MSE was actually helpful. It is from braggers for braggers.

The basic problem stands: How do you define a group action without using the group? Try it, it won't work.

Also, the two $f$s are the two isomorphic maps I listed related to cayley's theorem and $\phi_g$ is the group action homomorphism map. Is $L_g$ same as $\phi_g$?

In this case we have $f(g)=\phi(g)=\varphi(g)=L_g.$

The trick is that $g\in G$ are any elements of any groups. But in order to make them functions, we have somehow to identify those elements with functions. The map that is used in Cayley's theorem is $g\longmapsto L_g.$ It turns ordinary elements into functions, permutations of $G$ to be exact. And if $G$ has $n$ elements, then $\operatorname{Sym}(n)=S_n$ has $n!$ many elements. This is a significantly higher number, so $\varphi(G)$ can only be seen as a subgroup.

I thought there are relations between them. Also, when you use the term "identify", what do you mean by that in mathematical notations?

I mean that we consider the function $\varphi\, : \,G\longrightarrow \operatorname{Sym}(G)$ defined by $(\varphi(g))(h)=L_g(h)=g\cdot h.$

You would have to show:

a) $L_g$ maps $G$ into $G$.
b) $L_g$ is bijective.
c) $\varphi$ is a group homomorphism.
d) $\varphi$ is injective.

All these properties are necessary if we want to "identify" an element $g$ with a function $L_g.$ It means, since $\varphi$ is a monomorphism, we can happily switch between elements $g$ of the group and certain permutations $\varphi(g)$ of the group. That's what Cayley says.

 

[elias001]

@fresh_42 the map $L_g$ maps $G$ to $G$ or is it to $A(G)$? Also i think you used $\varphi_g$ where as I used $\phi_g$ to mean the same map?

 

[fresh_42]

@fresh_42 the map $L_g$ maps $G$ to $G$ or is it to $A(G)$? Also i think you used $\varphi_g$ where as I used $\phi_g$ to mean the same map?

$\varphi(g)=L_g$ maps $G$ to $G$.
$\varphi=L_{.}$ maps $G$ to $\operatorname{Sym}(G).$

I do not want to use the notation $A(G)$ because it is ambiguous, see post #18.

I used $\varphi\, : \,G\longrightarrow \operatorname{Sym}(G)$ and $\varphi(g)\, : \,G\longrightarrow G$ from the beginning for an arbitrary action, and $\varphi(g)=L_g$ for the specific action by left-multiplication.

Your usage of $f$ or $\phi$ came later and wasn't really defined, so I tried to ignore them. I guess, they are the same, but I can't tell since you haven't properly defined them.

It is important in mathematics to know where you are! $g\, , \,x\, , \,\varphi\, , \,\varphi(g)\, , \,(\varphi(g))(x)$ are in four different spaces. It is necessary to distinguish between them.

 

[elias001]

@fresh_42 I think my $f, \phi$ corresponds to your $\varphi$. I don't think neither of us used a letter for the map $G\times X\to X$. By the way, instead of a group acting on a set, we can have a group acting on another group? So we have two groups $G,H$, we can have the following situation: $G\times H\to H$?

 

[fresh_42]

@fresh_42 I think my $f, \phi$ corresponds to your $\varphi$. I don't think neither of us used a letter for the map $G\times X\to X$. By the way, instead of a group acting on a set, we can have a group acting on another group? So we have two groups $G,H$, we can have the following situation: $G\times H\to H$?

Sure. The first groups that come to mind are automorphism groups. An automorphism of a group $G$ is a bijective homomorphism $G\longrightarrow G.$ They carry a group structure because the homomorphisms are invertible. So the elements of an automorphism group are already functions in $\operatorname{Sym}(G).$ A special case are inner automorphisms. These are the conjugations $h \longmapsto g^{-1}hg$ by a certain element $g$. Every $g\in G$ defines therefore an inner automorphism, again building a subgroup of $\operatorname{Sym}(G).$

Automorphism groups play a central role in the concept of semidirect products. Say we have two different groups $H,N$ and a group homomorphism from $H$ into the automorphism group of $N,$ i.e.
$$
\varphi \, : \,H\longrightarrow \operatorname{Aut}(N).
$$
Then $\varphi$ defines an operation of $H$ on the group $N$ by $h.n=(\varphi(h))(n)$ which can be used to define a group structure on the set $G:=H\times N$ by setting
$$
(h_1\, , \,n_2)\cdot (h_2\, , \,n_2) = (h_1\cdot h_2\, , \,n_1 \cdot (\varphi(h_1))(n_2)).
$$
The resulting group is only a direct product of groups if $\varphi\equiv \operatorname{id}_{N}$ in which case we write $G=H\times N$ as groups, not only as sets. Both factors $H,N$ are normal subgroups of $G$ in this case.

If $\varphi\not\equiv \operatorname{id}_{N}$ then we still have $H,N$ as subgroups of $G$ but only $N$ is a normal subgroup. We write $G=H \ltimes_{\varphi} N$ in this case to note that fact.

Hence, whenever we see a semidirect product $H\ltimes N$, then we have an operation of the subgroup $H$ on the normal subgroup $N.$ A prominent example of semidirect products are the dihedral groups.

 

[elias001]

@fresh_42 a group acting on another group is not limited to semidirect product? But semidirect product is always an example of one group acting in another?

 

[fresh_42]

@fresh_42 a group acting on another group is not limited to semidirect product? But semidirect product is always an example of one group acting in another?

Yes.

Another example are groups of linear transformations. They act on the additive group of vector spaces.

 

[elias001]

@fresh_42 ah kk. I think i have a better handle of group actions. Thank you so much. I might post something about semidirect product, outer automorphism or why the direct sum is not the coproduct in the category of groups. Before that, if will deal with my other posts first.