I Question about "A Group Epimorphism is Surjective"

[elias001]

TL;DR Summary

I have questions about the paper: "A Group Epimorphism is Surjective".

The following is from the paper "A Group Epimorphism is Surjective" by C.E. Linderholm



Background

Let $\phi: F\rightarrow G$ to mean that $F$ and $G$ are groups and $\phi$ is a homomorphism defined on $F$ and taking values in $G$. If $\psi:G\rightarrow H$ the composite homomorphism from $F$ to $H$ is written $\phi\psi.$ If $x\in F$ the image under $\phi$ is written $(x)\phi.$ A homomorphism $\psi$ that cancels on the right in every equation $\phi_1\psi=\phi_2\psi$ is a monomorphism. If $\psi:G\rightarrow H$ then $\psi$ is a monomorphism if for each group $F$ and for each pair $\phi_1, \phi_2$ of homomorphism from $F$ to $G$ such that $\phi_1\psi=\phi_2\psi,$ the equation $\phi_1=\phi_2$ holds. A homomorphism $\phi$ that cancels on the left in every equation $\phi_1\psi=\phi_2\psi$ is an epimorphism. If $\phi:F\rightarrow G$ and if an inverse homomorphism $\psi$ from $G$ to $F$ exists - i.e., one such that $(x)\phi\psi=x$ for each element $x$ of $F$ and $(y)\psi\phi=y$ for each element $y$ of $G$- then $\phi$ is an isomorphism. We prove that "A Group Epimorphism is Surjective".

Proof: Let $\phi$ be an epimorphism from a group $G$ to a group $H,$ and let $A$ be the image subgroup. We must show that $A=H.$ To do this we construct two homomorphism $\psi_1,\psi_2$ from $H$ to a group $K$ such that $\phi\psi_1=\phi\psi_2,$ and we use the resulting equation $\psi_1=\psi_2$ to prove that $A=H.$

Let $H/A$ be the set of all right cosets of $A;$ let $A'$ be something that is not a right coset of $A,$ and let $S$ be the set $H/A\cup\{A'\}.$ Let $K$ be the group of all permutations (bijections) of $S.$ If $h,h_1,h_2\in H$ and if $Ah_1$ and $Ah_2$ are the same coset, then $A(h_1h)$ and $A(h_2h)$ are the same coset, and hence the function from $A/H$ to $A/H$ that sends $Ah'$ to $A(h'h)$ is well defined. It is easily seen to be bijective. (in fact its inverse is given by $Ah'\rightarrow Ah'h^{-1}.$) If $A'$ is sent to itself, this defines a bijection of $S,$ which we write $(h)\psi_1.$ The function $\psi_1$ from $H$ to $K$ so defined is easily seen to be a homomorphism. Let $\sigma$ be the permutation of $S$ that interchanges $A$ and $A'$ and leaves everything else fixed, and if $h\in H$ write $(h)\psi_2$ for the composite permutation $\sigma^{-1}((h)\psi_1)\sigma.$ Then the function $\psi_2$ is the composition of $\psi_1$ with an inner automorphism of $K,$ and hence is a homomorphism from $H$ to $K$.

If $a\in A,$ then $(a)\psi_1$ leaves both $A$ and $A'$ fixed, and $\sigma$ leaves every other element of $S$ fixed Hence $\sigma$ and $(a)\psi_1$ commute, and $(a)\psi_1=\sigma^{-1}((a)\psi_1)\sigma.$ Since $\psi_1$ and $\psi_2$ agree on the range of $\phi,$ the equation $\phi\psi_1=\phi\psi_2$ holds. Since $\phi$ is an epimorphism, $\psi_1=\psi_2.$ Hence $(h)\psi_1$ commutes with $\sigma$ for each element $h$ of $H.$ Since $(h)\psi_1$ leaves $A'$ fixed and $\sigma$ interchanges $A'$ and $A.$ It follows that $(h)\psi_1$ leave $A$ fixed. Since $(h)\psi_1$ sends $A$ to $Ah,$ it follows that $h\in A.$ Since $h$ was an arbitrary element of the group $H,$ it follows that $A=H$ and $\phi$ is surjective.

Note: where it says the function from $A/H$ to $A/H$, should it not be $H/A$ to $H/A?$


Questions

There are some points about the above proof I am not clear on.

I understand that $A=\phi(G)$, but do the set $K$ consist of both the permutations $\sigma$ and $(h)\psi_1.$

For the function $(h)\psi_1$, it is defined as $(h) \psi_1 : H\to K$ but what about at various places where it says it sends $A\to Ah,$ $A'$ to itself. In terms of elements evaluated by $\psi_1$, what is the difference between when it sends $A'$ to itself versus $(a)\psi_1.$ Earlier in the proof, it already states that $(h)\psi_1$ sends $A'$ to itself, but $(a)\psi_1$ leaves both $A$ and $A'$ fixed. The notation for the elements of $\psi_1$ are confusing. $h, a$ are elements of $H, A$ respectively, then when it says $(h)\psi_1$ leaves $A'$ to $A'$, what about $(h)\psi_1$ where it sends $A$ to $Ah.$ Also in the case of $A$ to $Ah$, does it mean $a\mapsto ah?$ Other points of confusion for me is how do both $\sigma, (a)\psi_1$ and $\sigma, (h)\psi_1$ commute, lastly, how does $(a)\psi_2=\sigma^{-1}((a)\psi_1)\sigma=(a)\sigma_1.$


Thank you in advance.

 

[fresh_42]

Where do you find those texts? The entire world, ok, most of it, writes consecutive mappings from right to left, and the variable on the right, but you found a text that does it the other way around. Moreover, it deals with left and right cosets. That's a bit confusing. And the title is strange, too, like it was "A circle is round". Epimorphism is simply another word for surjective morphism. Correction: epimorphisms and monomorphisms are here defined by their cancellation property, so that this equivalence has to be shown. I would leave all that categorical stuff until you understood the basics. It does not really provide insights. And it certainly does not help to use both kinds of notations as long as you are in the learning process.

 

[elias001]

@fresh_42 I post the original question here. I will let you read the exchange of comments. Arturo wrote out an explanation of the same article in a different post which he linked.

The reason I made the MSE post was because it was a contrapositive to a statement in Arbib and Manes' exercise which I asked here

 

[elias001]

@fresh_42 I just want to say I don't know what Arturo in that MSE reply meant when he said I don't understand what group action is. I understand group actions in terms of left multiplication, which was from Hungerford's abstract algebra text. Am I missing something like the full blown definition of what group action is. Definitely. But, but would I have even understood it even after having seen it. My answer is: NO!!!. Here is why. In the book A Course on Group Theory by; John S. Rose, Chapter 4 titled Group Actions on Sets. In the first paragraph of that chapter, somewhere on the second line, he says:

Let $X$ be some system, that is to say a set with a certain distinguished structure, it may be algebraic or geometric, and let $G$ be a subgroup of the symmetry group of $X$ (the group of all structure-preserving permutations of $X$). Then each $g\in G$ moves each $x\in X$ to some elements of $X$ (where in using the word 'moves' we allow the possibility that $f$ fixes $x$).

[Does that mean when I am stirring spaghetti in a pot of boiling water when making pasta can be consider as "group action", the stirring is the moving part, and the spaghetti being moved are the $x$s.]

Every other books in abstract algebra or group theory just present group actions with the standard definition consisting of the two axioms. I have not managed to find any other source that would give such poetic description of the definition. Of course in Karlheinz Spindler's two volumes of Abstract Algebra with Applications volume 1 where in the Group Actions chapter, he present examples of group actions from differential equations. But does that example fit Rose's poetic description of what a group action does to an element or how it fits. I don't know if anyone would care, since I am not even sure if that is what it means to understand the concept of Group Actions in algebra.

 

[elias001]

@fresh_42 I asked few LLMs with a list of questions together and five of them separately. Here is the result:

Questions to LLM:$\\$
In the $\textit{Proof}$ of the stated $\textbf{Theorem}$ within the following quoted passage, answer using mathematical notations $\textbf{Questions:}$ (1) to (11). $\\$

In mathematical notations list and define all the group actions in the form of $\textbf{Assumed definition 1}$, $\textbf{Assumed definition 2}$ and $\textbf{Assumed definition 3}.\\$

For each group actions listed, using mathematical notations, show how each of the following map: $(a)\psi_1, (h)\psi_1, (h)\psi_2, \sigma, \psi_1,\psi_2$ is each conceptually connected to each of the listed respective group actions and at the same time explain each connection making in the context and making reference to $\textbf{Assumed definition 1, Assumed definition 2}$ and $\textbf{Assumed definition 3}.\\$

Rewrite out each group actions homomorphism maps within the $\textit{Proof}$ in accordance to each of the three $\textbf{Assumed definitions}.\\$

$\textbf{Questions:}\\$

(1) Where it says the function from $A/H$ to $A/H$, should it not be $H/A$ to $H/A?\\$

(2) In the set $A=\phi(G)$, does the set $K$ just consists of both the permutations $\sigma$ and $(h)\psi_1?\\$

(3) For the function $(h)\psi_1$, it is defined as $(h)\psi_1: H\to K$ but what about at various places where it says it sends $A\to Ah,$ and $A'$ to itself?$\\$

(4) In terms of elements evaluated by $\psi_1$, what is the difference between when it sends $A'$ to itself versus $(a)\psi_1?\\$

(5) Where it stated that $(h)\psi_1$ sends $A'$ to itself, does $(a)\psi_1$ also leaves both $A$ and $A'$ fixed?$\\$

(6) In terms of elements, how does $\sigma$ as the permutation of $S$ interchanges $A$ and $A'$ and leaves everything else fixed?$\\$

(7) How is the function from $A/H$ to $A/H$ that sends $Ah'$ to $A(h'h)$ well defined?$\\$

(8) The notations for the elements of $\psi_1$ are confusing. $h, a$ are elements of $H, A$ respectively, then when it says $(h)\psi_1$ leaves $A'$ to $A'$, what about $(h)\psi_1$ where it sends $A$ to $Ah?\\$

(9) For the map that sends $A$ to $Ah$, does it mean $a\mapsto ah?\\$

(10) How does $\sigma$ commute with (a)$\psi_1,$ and also $\sigma$ commute with (h)$\psi_1?\\$

(11) How does $(a)\psi_2=\sigma^{-1}((a)\psi_1)\sigma=(a)\psi_1?\\$

$\textbf{Assumed definition 1:}$ A $\textit{group action }$ of a group $G$ on a set $X$ is a function from $G\times X\to X,$ with outputs written as $\cdot x,$ satisfying $\\$

(a) (Compatibility) $g_1\cdot(g_2\cdot x)=(g_1g_2)\cdot x,$ or all $g_1,g_2\in G$ and $x\in X;\\$

(b) (Identity) $1\cdot x=x,$ for all $x\in X.\\$

If there exists a group action of $G$ on $X,$ we say that $G$ $\textit{acts }$ on $X.$$\\$


$\textbf{Assumed definition 2:}$ Let $G$ be a group acting on a set $X.\\$

(a') for all $g\in G,$ the function $\sigma_g$ is a permutation of $X.\\$

(b') The map $\rho:G\to G_X$ defined by $\rho(g)=\sigma_g$ is a homomorphism.$\\$

$\textbf{Assumed definition 3:}$ A group action is a triple $(G,X,\rho)$ where $G$ is a group, $X$ is a set, and $\rho:G\to S_X$ is a homomorphism. The image element $\rho(g)(x)$ is simply often denoted by gx.$\\$

Analysis of Group Actions in the Proof of the Theorem: A Group Epimorphism is Surjective

Responses from LLM:$\\$

Introduction$\\$

The proof of the theorem ``A Group Epimorphism is Surjective'' involves a group $H$ acting on a set $S,$ defined as $S = H/A \cup \{A'\},$ where $A = \phi(G)$ is the image subgroup of an epimorphism $\phi: G \to H, H/A$ is the set of right cosets of $A,$ and $A'$ is an element not in $H/A.$ The group $K$ is the group of all permutations (bijections) of $S$. We analyze the group actions involved, define them according to the provided definitions, connect the maps $(a)\psi_1, (h)\psi_1, (h)\psi_2, \sigma, \psi_1,$ and $\psi_2$ to these actions, and rewrite the homomorphism maps.$\\$

Assumed Definitions$\\$

$\textbf{Assumed Definition 1:}$ A group action of a group $G$ on a set $X$ is a function $G \times X \to X,$ denoted $g \cdot x$, satisfying:$\\$

(Compatibility) $g_1 \cdot (g_2 \cdot x) = (g_1 g_2) \cdot x$, for all $g_1, g_2 \in G, x \in X.\\$

(Identity) $1 \cdot x = x$, for all $x \in X.\\$

If such a function exists, $G \textit{ acts }$ on $X.\\$

$\textbf{Assumed Definition 2:}$ For a group $G$ acting on a set $X:\\$

[(a')] For all $g \in G$, the function $\sigma_g: X \to X$, defined by $\sigma_g(x) = g \cdot x$, is a permutation of $X.\\$

[(b')] The map $\rho: G \to S_X$, defined by $\rho(g) = \sigma_g$, is a homomorphism, where $S_X$ is the symmetric group on $X.\\$

$\textbf{Assumed Definition 3:}$ A group action is a triple $(G, X, \rho)$, where $G$ is a group, $X$ is a set, and $\rho: G \to S_X$ is a homomorphism. The action is denoted $\rho(g)(x) = g x.\\$

Group Actions in the Proof$\\$

The proof involves a single group action: the group $H$ acting on the set $S = H/A \cup \{A'\}$, where $K = S_S$ is the symmetric group on $S$. We define this action according to each assumed definition.$\\$

Group Action According to Assumed Definition 1$\\$

Define the action of $H$ on $S$ by:$\\$

$h \cdot s = \begin{cases} A(h'h) & \text{if } s = Ah' \in H/A, \\ A' & \text{if } s = A', \end{cases}$ for $h \in H, s \in S.\\$

$\textbf{Verification:}\\$

$\textbf{Compatibility:}$ For $h_1, h_2 \in H, s \in S:\\$

If $s = Ah' \in H/A$, then $h_2 \cdot s = A(h'h_2)$, and $h_1 \cdot (h_2 \cdot s) = h_1 \cdot A(h'h_2) = A(h'h_2 h_1)$. Also, $(h_1 h_2) \cdot s = (h_1 h_2) \cdot Ah' = A(h'h_1 h_2)$, which is equal since group multiplication is associative.$\\$

If $s = A'$, then $h_2 \cdot A' = A'$, and $h_1 \cdot (h_2 \cdot A') = h_1 \cdot A' = A'$. Also, $(h_1 h_2) \cdot A' = A'$, so compatibility holds.$\\$

$\textbf{Identity:}$ For the identity $1 \in H, 1 \cdot s = s:\\$

If $s = Ah'$, then $1 \cdot Ah' = A(h' \cdot 1) = Ah'.\\$

If $s = A'$, then $1 \cdot A' = A'.\\$

Thus, this defines a group action of $H$ on $S.\\$

Group Action According to Assumed Definition 2$\\$

For each $h \in H$, define $\sigma_h: S \to S$ by:$\\$

$\sigma_h(s) = h \cdot s = \begin{cases} A(h'h) & \text{if } s = Ah', \\ A' & \text{if } s = A'. \end{cases}\\$

[(a')] $\textbf{Permutation:}$ The map $\sigma_h$ is bijective:$\\$

$\textbf{Injectivity:}$ Suppose $\sigma_h(s_1) = \sigma_h(s_2).\\$

If $s_1 = Ah_1, s_2 = Ah_2$, then $A(h_1 h) = A(h_2 h)$, so $Ah_1 = Ah_2$, since right multiplication by $h$ is a bijection on cosets (as shown in the proof).$\\$

If $s_1 = Ah_1, s_2 = A'$, then $A(h_1 h) = A'$, which is impossible since $A(h_1 h) \in H/A, A' \notin H/A.\\$

If $s_1 = s_2 = A'$, then $A' = A'.\\$

Thus, $s_1 = s_2.\\$

$\textbf{Surjectivity:}$ For any $s' \in S:\\$

If $s' = A k \in H/A$, let $s = A(k h^{-1})$. Then $\sigma_h(s) = A(k h^{-1} h) = A k$.$\\$

If $s' = A'$, let $s = A'$. Then $\sigma_h(A') = A'.\\$

[(b')] $\textbf{Homomorphism:}$ Define $\rho: H \to S_S$ by $\rho(h) = \sigma_h$. For $h_1, h_2 \in H, \rho(h_1 h_2) = \sigma_{h_1 h_2}$, and $\sigma_{h_1} \sigma_{h_2}(s) = \sigma_{h_1}(h_2 \cdot s)$. If $s = Ah'$, then $\sigma_{h_1} \sigma_{h_2}(Ah') = \sigma_{h_1}(A(h'h_2)) = A(h'h_2 h_1) = \sigma_{h_1 h_2}(Ah')$. If $s = A'$, then $\sigma_{h_1} \sigma_{h_2}(A') = \sigma_{h_1}(A') = A' = \sigma_{h_1 h_2}(A')$. Thus, $\rho(h_1 h_2) = \rho(h_1) \rho(h_2).\\$

Group Action According to Assumed Definition 3$\\$

The action is the triple $(H, S, \rho)$, where $\rho: H \to S_S, \rho(h) = \sigma_h$, and $\sigma_h(s) = h s$ as above. The action is denoted $h s$, matching the definition in Assumed Definition 2, and $\rho$ is a homomorphism as verified.$\\$

Connections of Maps to the Group Action$\\$

We connect the maps $(a)\psi_1, (h)\psi_1, (h)\psi_2, \sigma, \psi_1,$ and $\psi_2$ to the group action.$\\$

$\mathbf{\psi_1: H \to K:}$ The map $\psi_1$ is defined by $\psi_1(h) = (h)\psi_1$, where $(h)\psi_1 = \sigma_h$, the permutation induced by the action of $h \in H$ on $S$. In terms of Assumed Definition 2, $\psi_1 = \rho$, since $\rho(h) = \sigma_h$. In Assumed Definition 3, $\psi_1$ is the homomorphism $\rho: H \to S_S$. For $s = Ah', (h)\psi_1(Ah') = A(h'h)$, and $(h)\psi_1(A') = A'$, matching the action $h \cdot s.\\$

$\mathbf{(h)\psi_1:}$ For $h \in H, (h)\psi_1 = \sigma_h$, the permutation in $K = S_S$ defined by the action. In Assumed Definition 1, $(h)\psi_1(s) = h \cdot s$. In Assumed Definition 2, $(h)\psi_1 = \sigma_h$, and in Assumed Definition 3, $(h)\psi_1(s) = h s$. This map sends $A h' \to A(h'h)$ and $A' \to A'.\\$

$\mathbf{(a)\psi_1:}$ For $a \in A \subseteq H, (a)\psi_1 = \sigma_a$, where $A = \phi(G)$. Since $a \in A, (a)\psi_1(A h') = A(h' a)$. But $h' a = h' a'$ for some $a' \in A$, since $A$ is a subgroup, so $A(h' a) = A h'$. Thus, $(a)\psi_1$ fixes $H/A$ pointwise and fixes $A'$, i.e., $(a)\psi_1 = \text{id}_S$. This is consistent with Assumed Definitions 1, 2, and 3, as it is the action of $a \in H$ on $S.\\$

$\mathbf{(h)\psi_2:}$ Defined as $(h)\psi_2 = \sigma^{-1} (h)\psi_1 \sigma$, where $\sigma \in K$ interchanges $A$ (the coset $A1$) and $A'$, fixing all other elements of $S$. In Assumed Definition 3, $(h)\psi_2$ is a permutation in $S_S$, obtained by conjugating the action permutation $\sigma_h$. For $s \in S:\\$

$(h)\psi_2(s) = \sigma^{-1} (\sigma_h (\sigma(s))).\\$

This is connected to the action via conjugation in $S_S.\\$

$\mathbf{\psi_2: H \to K:}$ Defined by $\psi_2(h) = (h)\psi_2$. Since $(h)\psi_2 = \sigma^{-1} \sigma_h \sigma, \psi_2(h) = \sigma^{-1} \rho(h) \sigma$, an inner automorphism of $K$. This is a homomorphism (Assumed Definition 2(b')), and relates to the action by modifying the permutation $\sigma_h.\\$

$\mathbf{\sigma:}$ The permutation $\sigma \in K$interchanges $A$ and $A'$, fixing all other elements of $S$. It is not directly the action of an element of $H$, but an element of $K = S_S$. In Assumed Definition 3, it is a fixed permutation used to define $\psi_2$, not part of the action $\rho$, but interacts with action permutations via conjugation.$\\$

Rewriting Homomorphism Maps$\\$

The homomorphisms in the proof are $\psi_1, \psi_2: H \to K.\\$

$\mathbf{\psi_1:}$ In Assumed Definition 3, $\psi_1 = \rho$, where:$\\$

$\psi_1(h)(s) = h s = \begin{cases} A(h'h) & \text{if } s = Ah', \\ A' & \text{if } s = A'. \end{cases}$

In Assumed Definition 1, this is the action $h \cdot s.$

$\mathbf{\psi_2:}$ In Assumed Definition 3:

$\psi_2(h) = \sigma^{-1} \rho(h) \sigma, \quad \psi_2(h)(s) = \sigma^{-1} (\rho(h) (\sigma(s))).\\$

Compute explicitly:$\\$

If $s = A k, \sigma(A k) = A k$ since $A k \neq A, A'$, so $\rho(h) (\sigma(A k)) = A(k h)$, and $\sigma^{-1}(A(k h)) = A(k h).\\$

If $s = A, \sigma(A) = A'$, so $\rho(h)(A') = A'$, and $\sigma^{-1}(A') = A.\\$

If $s = A', \sigma(A') = A$, so $\rho(h)(A) = A h$, and $\sigma^{-1}(A h) = A h.\\$

Thus:$\\$

$\psi_2(h)(s) = \begin{cases} A(k h) & \text{if } s = A k, k \neq 1, \\ A & \text{if } s = A', \\ A h & \text{if } s = A. \end{cases}\\$

Answers to Questions

1 $\textbf{Where it says the function from$\mathbf{A/H}$to$\mathbf{A/H}$, should it not be$\mathbf{H/A}$to$\mathbf{H/A}?}$Yes, it should be$H/A$to$H/A$. The set$H/A$denotes right cosets$\{ A h \mid h \in H \}$. The function$A h' \to A(h' h)$is defined on right cosets, so$H/A$is correct, as$A/H$typically denotes left cosets, which are not used here. 2$\textbf{In the set $\mathbf{A = \phi(G)}$, does the set $\mathbf{K}$ just consist of both the permutations $\mathbf{\sigma}$ and $\mathbf{(h)\psi_1}?}$

No, $K = S_S$, the symmetric group on $S$, contains all permutations of $S$. The permutations $\sigma$ and $(h)\psi_1$ (for all $h \in H$) are elements of $K$, but $K$ is much larger, including all bijections $S \to S.$

3 $\textbf{For the function$\mathbf{(h)\psi_1: H \to K}$, what about where it says it sends$\mathbf{(h)\psi_1: H \to KA}$, and$\mathbf{A' \to A'}?}$The map$(h)\psi_1 = \sigma_h \in K$is defined by the action:$(h)\psi_1(A h') = A(h' h)$, and$(h)\psi_1(A') = A'$. Specifically, for the coset$A = A 1, (h)\psi_1(A) = A(1 \cdot h) = A h$. This is consistent with Assumed Definition 1$h \cdot A = A h$and Assumed Definition 3$h A = A h.$4$\textbf{In terms of elements evaluated by $\mathbf{\psi_1}$, what is the difference between when it sends $\mathbf{A' \to A'}$ versus $\mathbf{(a)\psi_1}?}$

For $h \in H, (h)\psi_1(A') = A'$, meaning the permutation fixes $A'$. For $a \in A, (a)\psi_1 = \sigma_a$, which fixes all elements of $S$: for $s = A h', (a)\psi_1(A h') = A(h' a) = A h'$, and $(a)\psi_1(A') = A'$. Thus, $(a)\psi_1$ is the identity permutation, while $(h)\psi_1$ generally is not, unless $h \in A.$

5 $\textbf{Does$(a)\psi_1$also leave both$A$and$A'$fixed?}$

Yes, $(a)\psi_1 = \sigma_a$ is the identity: $(a)\psi_1(A h') = A(h' a) = A h'$, and $(a)\psi_1(A') = A'$. Thus, it fixes $A = A 1, A'$, and all of $S.$

6 $\textbf{How does$\sigma$interchange$A$and$A'$, leaving everything else fixed?}$

The permutation $\sigma \in K$ is defined by:#

$\sigma(s) = \begin{cases} A' & \text{if } s = A, \\ A & \text{if } s = A', \\ s & \text{if } s \in H/A, s \neq A. \end{cases}$

For elements $s = A h, h \neq 1, \sigma(A h) = A h; \sigma(A) = A'; \sigma(A') = A.$

7 $\textbf{How is the function from$\mathbf{H/A}$to$\mathbf{H/A}$that sends$\mathbf{A h' \to A(h' h)}$well defined?}$

If $A h_1 = A h_2$, then $h_1 h_2^{-1} \in A$, so $h_1 h_2^{-1} a = a' \in A$. Thus, $A(h_1 h) = A(h_2 (h_2^{-1} h_1 h)) = A(h_2 a' h) = A(h_2 h)$, since $a' \in A$. The function is bijective, with inverse $A h' \to A(h' h^{-1}).$

8 $\textbf{When$\mathbf{(h)\psi_1}$leaves$\mathbf{A' \to A'}$, what about sending$\mathbf{A \to A h}?}$For$h \in H, (h)\psi_1(A h') = A(h' h)$, so$(h)\psi_1(A) = A h$, and$(h)\psi_1(A') = A'$. Both are part of the action definition in Assumed Definition 1 and 3. 9$\textbf{Does $\mathbf{A \to A h}$ mean $\mathbf{a \mapsto a h}?}$

No, $A \to A h$ refers to the coset $A = A 1 \in H/A$, so $(h)\psi_1(A) = A h$. This is a coset action, not an element-wise map $a \mapsto a h.$

10 $\textbf{How does$\sigma$commute with$(a)\psi_1$and$(h)\psi_1$?}$

For $(a)\psi_1$, since $(a)\psi_1 = \text{id}_S, \sigma (a)\psi_1 = (a)\psi_1 \sigma$. For $(h)\psi_1$, the proof shows $\psi_1 = \psi_2$, so $(h)\psi_1 = \sigma^{-1} (h)\psi_1 \sigma$, implying $\sigma (h)\psi_1 = (h)\psi_1 \sigma.$

11 $\textbf{How does$\mathbf{(a)\psi_2 = \sigma^{-1} ((a)\psi_1) \sigma = (a)\psi_1}?}$Since$(a)\psi_1 = \text{id}_S,$compute:$(a)\psi_2 = \sigma^{-1} (a)\psi_1 \sigma = \sigma^{-1} \text{id}_S \sigma = \text{id}_S = (a)\psi_1##

 

[elias001]

@fresh_42 Here is what the response was from MSE


I know this is an old question, but this came up recently in an answer to another question, so I figure we might as well have a standard answer on the site.$\\$

A classic proof of this fact is due Schreier; he used free products with amalgamation. He then also deduced the nontrivial fact that epimorphisms are surjective in the category of \textit{finite } groups.$\\$

Carl Linderholm (of \textit{ Mathematics Made Difficult } fame) published a 1 page elementary proof in "A Group Epimorphism is Surjective", in the American Mathematical Monthly \textbf{ 77}, pp. 176-177. I gave the proof on `sci.math` back in 2000(This is this proof with slightly different notation.$\\$


Let $f\colon H\to K$ be an epimorphism. We seek to find a group $S$ and two morphisms $g,h\colon K\to S$ such that $gf=hf$, and then use the conclusion that $g=h$ to conclude that $f(H)=K.\\$


Let $X=K/f(H)$ be the set of right cosets of $f(H)$ in $K$. Let $\infty$ be something which is not an element of $X$, and let $Y=X\cup\{\infty\}$. Let $S$ be the group of permutations on $Y.\\$


The right action of $K$ on $X$ induces an embedding of $K$ into $S$ as permutations that fix $\infty$; call this map $g$. Now let $\sigma\in S$ be the permutation that exchanges the coset $f(H)$ with $\infty$ and fixes everything else. let $h\colon K\to S$ be the homomorphism we get by composing $g$ with conjugation by $\sigma$ in $K$.$\\$


Now, consider $h$ and $g$. If $x\in H$, then $f(x)$ fixes the coset $f(H)$ and fixes $\infty$. Thusm the support of $g(f(x))$ and $\sigma$ are disjoint hence they commute: $h(f(x))=\sigma^{-1}\circ g(f(x))\circ \sigma=g(f(x))$. That is, $h\circ f = g\circ f$.$\\$

Since $f$ is an epimorphism, we conclude that $h=g.\\$

But that means that for all $k$, $h(k)=g(k)$. In particular, $g(k)$ must commute with $\sigma$ for all $k$. This requires that $g(k)$ leave $f(H)$ fixed. But that requires $k\in f(H)$. That is, we must have $f(H)=K$, so $f$ is surjective, as claimed. $\\$

Note, along the way, that when $K$ is finite the group $S$ is also finite. So this also proves that epimorphisms are surjective in the category of all finite groups, something which is not immediately obvious from Schreier’s Theorem. However, Schreier’s Theorem is stronger than the assertion that epimorphisms are surjective: it proves that every subgroup is an equalizer subgroup; i.e., if $H\leq G$, then there exist a group $K$ and morphisms $f,g\colon G\to K$ such that the equalizer of $f$ and $g$ is exactly $H:\mathrm{Eq}(f,g)=\{x\in G\mid f(x)=g(x)\}=H$.

 

[renormalize]

@elias001 what are you trying to accomplish with this long dense post? It clearly violates one of the rules of Physics Forums:

• Answering a science or math question with AI-generated text, even with attribution, is not allowed. AI-generated text apps like ChatGPT are not valid sources.

 

[elias001]

@renormalize I have not finished my post since I am trying to edit the laTex. Yes I am asking LLM and I am being very transparent about it. I am asking an AI to clarify somethings in the first post which has to do with a paper i posted, and bringing here anything that I don't understand from the answers I got.

 

[renormalize]

I am asking an AI to clarify somethings in the first post which has to do with a paper i posted, and bringing here anything that I don't understand from the answers I got.

Then ask your questions as expressed in your own words without quoting the AI answers. Posting AI quotes is against the PF rules!

 

[elias001]

@renormalize I thought posting AI answer without attributing or acknowledging it is against forum rules. I am being transparent here and not attributing any of the material here as my own. I actually mentioned it to fresh about it and i informed him i will ask the AI and I will post the answer here and let him know what seems to be the troubling points. When I ask the questions, it will be based on what the AI has said, so i might as well write out precisely what the AI told me.

 

[renormalize]

When I ask the questions, it will be based on what the AI has said, so i might as well write out precisely what the AI told me.

I appreciate your transparency, but again I quote the specific rule:

• Answering a science or math question with AI-generated text, even with attribution, is not allowed. AI-generated text apps like ChatGPT are not valid sources.

So if you "write out precisely what the AI told me" you unambiguously break PF rules. You must find a different way to formulate your questions and their proposed answers.

 

[elias001]

@renormalize this is mathematics and the material are expressed in math notations. I specifically ask the AI to write the answers in mathematical notations. Also, i was being very specific, I asked the LLM to rephrase or express certain material within a paper (the one in the first post) in terms of quoted definitions. There was something I did not understand in the answer given to me. All of the LLMs even states with a disclaimer that it might contain mistakes. Basically you are saying that I can't ask anything on here that I have come across from asking an LLM, am I correct? Also, i have not finished writing out my posts. There are reasons why I am quoting the LLM's answer here. If you let me finish what I have to ask, you will see why.

 

[PeterDonis]

Thread closed for moderation.

 

[berkeman]

After a Mentor discussion this thread will remain closed. @elias001 -- please reread the PF rules, particularly the part about not trying to use AI as a reference in the technical forums.