What is the Meaning of Operation gx in Group Theory?

[mathmari]

Hey!

Let $G$ be a group and let $g\in G$. We define: \begin{align*}&\lambda_g:G\rightarrow G, \ x\mapsto gx \\ &\gamma_g:G\rightarrow G, \ x\mapsto gxg^{-1}\end{align*}

Show for all $g,h\in G$:

1. $\lambda_{1_G}=\text{id}_G$ and $\lambda_{gh}=\lambda_g\circ \lambda_h$
2. $\lambda_g$ is a permutation and it holds that $\lambda_g^{-1}=\lambda_{g^{-1}}$
3. The map $\lambda:G\rightarrow \text{Sym}(G), \ g\mapsto \lambda_g$ is a group monomorphism
4. $\gamma_{1_G}=\text{id}_G$ and $\gamma_{gh}=\gamma_g\circ \gamma_h$
5. $\gamma_g\in \text{Sym}(G)$ and it holds that $\gamma_g^{-1}=\gamma_{g^{-1}}$
6. For all $x\in G$ it holds that $\gamma_x(gh)=\gamma_x(g)\gamma_x(h)$
7. The map $\gamma:G\rightarrow \text{Sym}(G), \ g\mapsto \gamma_g$ is a group homomorphism
8. Determine $\ker (\gamma)$

First of all, to understand the operation... At the map $\lambda_g:G\rightarrow G, \ x\mapsto gx$ the $gx$ is a multiplication of functions or is it a composition or is it like $g(x)$ ?
Because at the first question we have $g=1_G$. Then we have for some $x\in G$ that $\lambda_{1_G}(x)=1_Gx$. What operation is this?

:unsure:

 

[HallsofIvy]

"g" is a given member of G. "x" is an arbitrary member of G. "gx" is whatever the operation is defined to be in the group G. If G were "the integers with multiplication modulo 3" then the members of G can be thought of as the equivalence classes [0]= {0, 3, 6, ...}, [1]= {1, 4, 7, …}, and [2]= {2, 5, 8, …}. With g= [2], gx would be given by g0= {0, 6, 12,...}= [0], g[1]= {2, 8, 14, ...}= [2], and g[2]= {4, 10, 16,..}= [1].

 

[mathmari]

Ok! But I still don't understand then $\lambda_{1_G}(x)=1_Gx$. We have that $1_G(y)=\begin{cases}1 & \text{ if } y\in G \\ 0 & \text{ if } y\notin G\end{cases}$. In this case we have just $1_Gx$. Which element do we check if it is in $G$, like at the definition the element $y$ ? Or do we say that in $G$ it holds that $1_G=1$ and so $1_Gx=x$ in $G$ ? :unsure:

 

[I like Serena]

Ok! But I still don't understand then $\lambda_{1_G}(x)=1_Gx$. We have that $1_G(y)=\begin{cases}1 & \text{ if } y\in G \\ 0 & \text{ if } y\notin G\end{cases}$. In this case we have just $1_Gx$. Which element do we check if it is in $G$, like at the definition the element $y$ ? Or do we say that in $G$ it holds that $1_G=1$ and so $1_Gx=x$ in $G$ ?

Hey mathmari!

$1_G$ is not the indicator function. Instead it is the identity element with respect to the operation that is defined in the group G. (Thinking)
That operation is not (necessarily) function composition or function multiplication - it is 'just' the abstract operation as it is defined.

 

[mathmari]

Ahhh! I was confused because I considered it to be the indicator function and so I got stuck! Thanks for clarifying!

Therefore for question 1 we have the following:
Let $x\in G$. Then we have that $\lambda_{1_G}(x)=1_Gx=x=\text{id}_G(x)$ and since $x$ is arbitrary we get that $\lambda_{1_G}=\text{id}_G$.
We also have that $\lambda_{gh}(x)=ghx$. Considering the right side of the desired equation we have that $(\lambda_g\circ\lambda_h)(x)=\lambda_g\left (\lambda_h(x)\right )=\lambda_g\left (hx\right )=ghx$. So both sides are equal and so the equality $\lambda_{gh}=\lambda_g\circ\lambda_h$ follows.

Is everything correct? :unsure: For question 2:
To show that $\lambda_g$ is a permutation we have to show that the map is bijective, right? :unsure: For question 3:
We have that $\lambda (gh)=\lambda_{gh}\ \overset{\text{ Question 1}}{=} \ \lambda_g\circ\lambda_h=\lambda(g)\circ \lambda(h)$ which means that $\lambda$ is a group homomorphism.
Now it is left to show that the map is injective. We have that $\lambda (g)=\lambda (h) \Rightarrow \lambda_g=\lambda_h$. Then for $x\in G$ we have that $\lambda_g(x)=\lambda_h(x) \Rightarrow gx=hx\Rightarrow gxx^{-1}=hxx^{-1} \Rightarrow g=h$ which means that $\lambda$ is injective and so it is a group monomorphism.
Is everything correct? :unsure: For question 4:
Let $x\in G$. Then we have that $\gamma_{1_G}(x)=1_Gx1_G^{-1}=x=\text{id}_G(x)$ and since $x$ is arbitrary we get that $\gamma_{1_G}=\text{id}_G$.

We also have that $\gamma_{gh}(x)=(gh)x(gh)^{-1}=ghxh^{-1}g^{-1}$. Considering the right side of the desired equation we have that $(\gamma_g\circ\gamma_h)(x)=\gamma_g\left (\gamma_h(x)\right )=\gamma_g\left (hxh^{-1}\right )=ghxh^{-1}g^{-1}$. So both sides are equal and so the equality $\gamma_{gh}=\gamma_g\circ\gamma_h$ follows.

Is everything correct? :unsure:For question 5:
To show that $\gamma_g$ is a permutation we have to show that the map is bijective, right? :unsure: For question 6:
We have that $\gamma_x(gh)=xghx^{-1}=xgx^{-1}xhx^{-1}=\left (xgx^{-1}\right )\left (xhx^{-1}\right )=\gamma_x(g)\gamma_x(h)$.

Is that correct? :unsure: For question 7:
We have that $\gamma (gh)=\gamma_{gh}\ \overset{\text{ Question 4}}{=} \ \gamma_g\circ\gamma_h=\gamma (g)\circ \gamma (h)$ which means that $\gamma$ is a group homomorphism.

Is that correct? :unsure: For question 8:
We have that $$\ker \gamma =\{g\in G\mid \gamma (g)=0\}=\{g\in G\mid \gamma_g=0\}$$ How can we continue? :unsure:

 

[I like Serena]

Is everything correct?

All good. (Happy)

For question 8:
We have that $$\ker \gamma =\{g\in G\mid \gamma (g)=0\}=\{g\in G\mid \gamma_g=0\}$$ How can we continue?

0 is not an element of $\operatorname{Sym}(G)$ is it? (Worried)
And can't we write it out a bit more?
There is actually a name for the result.

 

[mathmari]

0 is not an element of $\operatorname{Sym}(G)$ is it? (Worried)
And can't we write it out a bit more?
There is actually a name for the result.

Ahh I should have written $\text{id}_G$ instead of $0$, right? :unsure:

 

[I like Serena]

Ahh I should have written $\text{id}_G$ instead of $0$, right?

Yep.
And what does it mean that $\gamma_g=\operatorname{id}_G$? (Wondering)

 

[mathmari]

Yep.
And what does it mean that $\gamma_g=\operatorname{id}_G$? (Wondering)

According to question 4, do we get that $g=1_G$ and so $$\ker \gamma =\{g\in G\mid \gamma (g)=\text{id}_G\}=\{g\in G\mid \gamma_g=\text{id}_G\}=\{1_G\}$$ Or do we just know that $1_G$ is contained but we don't know if there are also other elements? :unsure:

 

[I like Serena]

According to question 4, do we get that $g=1_G$ and so $$\ker \gamma =\{g\in G\mid \gamma (g)=\text{id}_G\}=\{g\in G\mid \gamma_g=\text{id}_G\}=\{1_G\}$$ Or do we just know that $1_G$ is contained but we don't know if there are also other elements?

I don't get that from question 4.

Don't we get that for any $x\in G$ we must have that $\gamma_g(x)=\operatorname{id}_G(x) \iff gxg^{-1}=x$?

 

[mathmari]

Don't we get that for any $x\in G$ we must have that $\gamma_g(x)=\operatorname{id}_G(x) \iff gxg^{-1}=x$?

Ahh yes! So we have that $$\ker \gamma =\{g\in G\mid \gamma (g)=\text{id}_G\}=\{g\in G\mid \gamma_g=\text{id}_G\}=\{g\in G \mid gxg^{-1}=x\}=\{g\in G \mid gx=xg\}$$ Right? Can we simplify that further? :unsure:

 

[I like Serena]

Ahh yes! So we have that $$\ker \gamma =\{g\in G\mid \gamma (g)=\text{id}_G\}=\{g\in G\mid \gamma_g=\text{id}_G\}=\{g\in G \mid gxg^{-1}=x\}=\{g\in G \mid gx=xg\}$$ Right? Can we simplify that further?

Isn't $x$ hanging loose without quantification. (Worried)

 

[mathmari]

So you mean that we have to add that $x\in G$ ? So the following?
\begin{align*}\ker \gamma &=\{g\in G\mid \gamma (g)=\text{id}_G\}=\{g\in G\mid \gamma_g=\text{id}_G\}=\{g\in G\mid \gamma_g(x)=x \ , \ \forall x\in G\}\\ & =\{g\in G \mid gxg^{-1}=x \ , \ \forall x\in G\}=\{g\in G \mid gx=xg \ , \ \forall x\in G\}\end{align*} :unsure:

 

[I like Serena]

So you mean that we have to add that $x\in G$ ? So the following?
\begin{align*}\ker \gamma &=\{g\in G\mid \gamma (g)=\text{id}_G\}=\{g\in G\mid \gamma_g=\text{id}_G\}=\{g\in G\mid \gamma_g(x)=x \ , \ \forall x\in G\}\\ & =\{g\in G \mid gxg^{-1}=x \ , \ \forall x\in G\}=\{g\in G \mid gx=xg \ , \ \forall x\in G\}\end{align*}

Yep.
And it has a name.
From wiki:

the center of a group G is the set of elements that commute with every element of G. It is denoted Z(G), from German 'Zentrum' meaning center. In set-builder notation,
$$Z(G) = \{z \in G \mid \forall g \in G, zg = gz \}$$

:geek:

 

[mathmari]

Ahh yes!

As for 2:
I have shown that $\lambda_g$ is a permutation. It is left to show that $\lambda_g^{-1}=\lambda_{g^{-1}}$. Do we show that as follows?
$$\left (\lambda_g(x)\right )^{-1}=\left (gx\right )^{-1}=x^{-1}g^{-1}\ \overset{G\text{ group }}{=} \ g^{-1}x^{-1}$$ But how can we continue? Or do we show that in an other way? :unsure:

 

[I like Serena]

Ahh yes!

As for 2:
I have shown that $\lambda_g$ is a permutation. It is left to show that $\lambda_g^{-1}=\lambda_{g^{-1}}$. Do we show that as follows?
$$\left (\lambda_g(x)\right )^{-1}=\left (gx\right )^{-1}=x^{-1}g^{-1}\ \overset{G\text{ group }}{=} \ g^{-1}x^{-1}$$ But how can we continue? Or do we show that in an other way?

I think we are mixing the inverse of a function with the multiplicative inverse of the group.

Let $y=\lambda_g(x)$.

Then $\lambda_g^{-1}(y) = x$.
And $\left (\lambda_g(y)\right )^{-1}=(gy)^{-1}$.
And $\lambda_{g^{-1}}(y)=g^{-1}y$.
These are different concepts aren't they? (Worried)

We have to verify that $\lambda_g^{-1}(y)=\lambda_{g^{-1}}(y)$, don't we?

$$x^{-1}g^{-1}\ \overset{G\text{ group }}{=} \ g^{-1}x^{-1}$$

I don't think that is generally true. (Worried)
Doesn't the group have to be abelian to be able to swap arguments?

 

[mathmari]

Ok! So let $y=\lambda_g(x)\Rightarrow y=gx$.
Then $\lambda_g^{-1}(y) = x$.
We also have that $\lambda_{g^{-1}}(y)=g^{-1}y=g^{-1}gx=x$.

Since these are equal it follows that $\lambda_g^{-1}(y)=\lambda_{g^{-1}}(y)$ and since $y$ is arbitrary we have that .$\lambda_g^{-1}=\lambda_{g^{-1}}$.

Is that correct? :unsure:

 

[I like Serena]

Since these are equal it follows that $\lambda_g^{-1}(y)=\lambda_{g^{-1}}(y)$ and since $y$ is arbitrary we have that .$\lambda_g^{-1}=\lambda_{g^{-1}}$.

Is that correct?

Yep. (Nod)

 

[mathmari]

As for 5:

To show that $
\gamma_g\in \text{Sym}(G)
$ we have to show that $\gamma_g$ is bijective, or not?Injectivity:

Let $x_1,x_2 \in G$ and let $g \in G$ fixed. Let $\gamma_g(x_1) = \gamma_g(x_2)$ then we have that $gx_1g^{-1}=gx_2g^{-1}$. Since $g \in G$, where $G$ is a group, it follows that $g^{-1} \in G$. So we get $g^{-1}gx_1g^{-1}g=g^{-1}gx_2g^{-1}g\Rightarrow x_1=x_2$.

So $\gamma_g$ is injective.Surjectivity:

Let $y \in G$. Let $x=g^{-1}yg \in G$ then $gxg^{-1}=y \Rightarrow \gamma_g(x) = y$.

So $\gamma_g$ is surjective.Does this mean that $\gamma_g\in \text{Sym}(G)$ ? Is this the set of permutations?For the inverse:

Let $y=\gamma_g(x)\Rightarrow y=gxg^{-1}$. Then $\gamma_g^{-1}(y) = x$. We also have that $\gamma_{g^{-1}}(y)=g^{-1}yg=g^{-1}gxg^{-1}g=x$.

So it follows that $\gamma_g^{-1}(y)=\gamma_{g^{-1}}(y)$ and since $y$ is arbitrary it follows that $\gamma_g^{-1}=\gamma_{g^{-1}}$.Is everything correct?

 

[I like Serena]

Is everything correct?

Yep. All correct. (Nod)