Finding a set and an injection

[Bashyboy]

Homework Statement

Let $G$ be a group. I need to find a set $X$ and an injective function from $G$ into $Sym(X)$

Homework Equations

The Attempt at a Solution

I am having difficulty with this problem, and I want to make sure I understand exactly what it is asking.
If I understand the question correctly, we have some group $G$, and we want to find some set $X$ and create some mapping (specifically, an injection) which associates the elements in $G$ with the elements in $Sym(X)$, so that we can clearly see that the elements of $G$, along with its binary operator $\star$, form functions?

Here are some ideas: I am pretty certain that I need to choose an $X$ such that $Sym(X)$ is at least as large as, if not greater than, $G$. I know that the binary operator $\star$ is actually a function, so could I base my injection off of this? I have this suspicion that $G=X$ would work, but I can't quite figure it out.

EDIT: I also know that $(Sym(X), \circ)$ forms a group, so I am basically trying to find an injection between two different groups.

Does anyone have any suggestions?

 

[Dick]

Homework Statement

Let $G$ be a group. I need to find a set $X$ and an injective function from $G$ into $Sym(X)$

Homework Equations

The Attempt at a Solution

I am having difficulty with this problem, and I want to make sure I understand exactly what it is asking.
If I understand the question correctly, we have some group $G$, and we want to find some set $X$ and create some mapping (specifically, an injection) which associates the elements in $G$ with the elements in $Sym(X)$, so that we can clearly see that the elements of $G$, along with its binary operator $\star$, form functions?

Here are some ideas: I am pretty certain that I need to choose an $X$ such that $Sym(X)$ is at least as large as, if not greater than, $G$. I know that the binary operator $\star$ is actually a function, so could I base my injection off of this? I have this suspicion that $G=X$ would work, but I can't quite figure it out.

EDIT: I also know that $(Sym(X), \circ)$ forms a group, so I am basically trying to find an injection between two different groups.

Does anyone have any suggestions?

Yes, take G=X. Associate an element of the group g with the mapping g:X->X defined by g(x)=gx. Is that an element of Sym(G)?

 

[Bashyboy]

Wait. From which set does $g$ come from? Are you saying that it is in $G$, or in $Sym(G)$?

 

[Dick]

Wait. From which set does $g$ come from? Are you saying that it is in $G$, or in $Sym(G)$?

g is in G.

 

[Bashyboy]

Okay. So, let me reiterate things, to see if I understand fully. We want to find a set $X$ and $\pi$ such that $\pi : G \mapsto Sym(X)$ is an injective map. We let $X=G$, and so $Sym(G)$ consists of all the bijective functions acting over the elements of $G$; that is, $Sym(G) = \{\theta_1 , \theta_2, \theta_3,... \}$, where $\theta_1 : G \mapsto G$, $\theta_2 : G \mapsto G$, etc.

Let $g \in G$ be arbitrary. I need to find an element in $Sym(G)$ to which $\pi$ can map it. If $g \in G$, then there exists a mapping $\theta \in Sym(G)$ such that

$\theta : g \mapsto a$ or $\theta(g) = a$, where $a \in G$ is unique, because $\theta$ is a bijection. So,

$\pi : g \mapsto \theta (g)$

$\pi : g \mapsto a$.

So, $\pi$ will actually map $g$ to a unique element in $G$.

This is where I am not certain of what I can conclude. In fact, I am not certain if any of what I said is valid. Could you pick apart my work?

 

[Dick]

Okay. So, let me reiterate things, to see if I understand fully. We want to find a set $X$ and $\pi$ such that $\pi : G \mapsto Sym(X)$ is an injective map. We let $X=G$, and so $Sym(G)$ consists of all the bijective functions acting over the elements of $G$; that is, $Sym(G) = \{\theta_1 , \theta_2, \theta_3,... \}$, where $\theta_1 : G \mapsto G$, $\theta_2 : G \mapsto G$, etc.

Let $g \in G$ be arbitrary. I need to find an element in $Sym(G)$ to which $\pi$ can map it. If $g \in G$, then there exists a mapping $\theta \in Sym(G)$ such that

$\theta : g \mapsto a$ or $\theta(g) = a$, where $a \in G$ is unique, because $\theta$ is a bijection. So,

$\pi : g \mapsto \theta (g)$

$\pi : g \mapsto a$.

So, $\pi$ will actually map $g$ to a unique element in $G$.

This is where I am not certain of what I can conclude. In fact, I am not certain if any of what I said is valid. Could you pick apart my work?

You are thinking of it way too abstractly. My hint is to point out that for any g in G the map f from G->G defined by f(x)=gx is a bijection. Prove that first. Then think about the overall problem.

 

[Bashyboy]

Are you saying that I need to demonstrate that $f(x) = g \star x$ is a bijection, where $x \in G$, I presume? So, the function/map $f$ is defined in terms of the operator $\star$? Well, the binary operator $\star$ is well-defined, so it maps uniquely; and because $\star$ has closure, the binary operator, and therefore the function $f$, will map every element $G$ to some element in $G$.

 

[Dick]

Are you saying that I need to demonstrate that $f(x) = g \star x$ is a bijection, where $x \in G$, I presume? So, the function/map $f$ is defined in terms of the operator $\star$? Well, the binary operator $\star$ is well-defined, so it maps uniquely; and because $\star$ has closure, the binary operator, and therefore the function $f$, will map every element $G$ to some element in $G$.

You also have to show it's one-to-one and onto to show it's a bijection.

 

[Bashyboy]

Don't closure and being well-defined demonstrate that it's one-to-one and onto? Closure means that every element in G gets mapped to G, so f maps the elements of G onto G; and well-defined means that it takes $g \star x$ to a unique element in G.

 

[Dick]

Don't closure and being well-defined demonstrate that it's one-to-one and onto? Closure means that every element in G gets mapped to G, so f maps the elements of G onto G; and well-defined means that it takes $g \star x$ to a unique element in G.

That just tells you f is a function. How is that supposed to imply that it's one to one or onto? Say how those are defined and spell it out in detail. You need to use that G is a group, not just that * is a binary operator. Saying f maps G TO G is not the same thing as saying f maps G ONTO G. What's the difference?

 

[Bashyboy]

Oh, okay. Let $g \in G$ be some arbitrary element. The function we have defined is $f(x) = g \star x$, which maps elements from $G$ to the same set.

If the function is injective, then $f(x) = f(y) \implies x = y ~\forall x,y \in G$. Let $x,y \in G$ be arbitrary; and let us suppose that $f(x) = f(y)$, and see whether this leads to $x=y$.

$f(x) = f(y) \iff g \star x = g \star y$. Using the cancellation law, we get $x = y$. Therefore, it is injective.

If the function is surjective, then $f$ should take every element $b \in G$ in the range and associate it with some element in the domain. Let $b \in G$ be some arbitrary element.

$b = f(x) \implies b = g \star x$. Because $g \in G$, then $g^{-1} \in G$. Multiplying both sides by this, we get

$g^{-1} \star b = x$. By closure of $\star$ over $G$, $g^{-1} \star b$ is some element in $G$, which is the domain of $f$; because $x$ is equal to this element, we have found that there does in fact exist an $x$ in the domain which is associated with $b$ in the range.

How does this appear?

 

[Dick]

Oh, okay. Let $g \in G$ be some arbitrary element. The function we have defined is $f(x) = g \star x$, which maps elements from $G$ to the same set.

If the function is injective, then $f(x) = f(y) \implies x = y ~\forall x,y \in G$. Let $x,y \in G$ be arbitrary; and let us suppose that $f(x) = f(y)$, and see whether this leads to $x=y$.

$f(x) = f(y) \iff g \star x = g \star y$. Using the cancellation law, we get $x = y$. Therefore, it is injective.

If the function is surjective, then $f$ should take every element $b \in G$ in the range and associate it with some element in the domain. Let $b \in G$ be some arbitrary element.

$b = f(x) \implies b = g \star x$. Because $g \in G$, then $g^{-1} \in G$. Multiplying both sides by this, we get

$g^{-1} \star b = x$. By closure of $\star$ over $G$, $g^{-1} \star b$ is some element in $G$, which is the domain of $f$; because $x$ is equal to this element, we have found that there does in fact exist an $x$ in the domain which is associated with $b$ in the range.

How does this appear?

Much better. Now you have a correspondence between every element g of G and a bijection of G->G. That's an element of Sym(G), yes? Now you just want to prove that correspondence is an injection.

 

[Bashyboy]

So, I need to prove that $\pi : g \mapsto f(x)$ is an injection; that is, $\pi (g) = f(x)$?

Let $a,b \in G$ be arbitrary. Let us suppose that $\pi(a) = \pi(b)$.

$\pi(a) = \pi(b) \iff$

$a \star x = b \star x$ Using the cancellation law,

$a = b$.

Is this right?

Would it be less ambiguous if we wrote $f(x)$ as $f_g(x)$?

 

[Dick]

So, I need to prove that $\pi : g \mapsto f(x)$ is an injection; that is, $\pi (g) = f(x)$?

Let $a,b \in G$ be arbitrary. Let us suppose that $\pi(a) = \pi(b)$.

$\pi(a) = \pi(b) \iff$

$a \star x = b \star x$ Using the cancellation law,

$a = b$.

Is this right?

Would it be less ambiguous if we wrote $f(x)$ as $f_g(x)$?

The idea is right. The notation is pretty messy. Sure, make it $f_g(x)=g \star x$. So $\pi(g)=f_g$.