# Math4302 Modern Algebra (Lecture 20)
## Groups
### Commutator of a group
Let $G$ be a group and $a,b\in G$, $[a,b]=aba^{-1}b^{-1}$.
Let $G'\leq G$ be the subgroup of $G$ generated by all commutators of $G$, $G'=\{[a_1,b_1][a_2,b_2]\ldots[a_n,b_n]\mid a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n\in G\}$.
Last time we shed that $G'$ is a normal subgroup of $G$ and $G/G'$ is abelian.
#### Proposition for commutator subgroup
If $N\trianglelefteq G$ is a normal subgroup of $G$ and $G/N$ is abelian, then $G'\leq N$.
Proof
We have $aNbN=bNaN$ for all $a,b\in G$.
so $abN=baN$, $a^{-1}b^{-1}abN=N$, so $a^{-1}b^{-1}\in N$, so for every $a,b\i G$, since $a^{-1},b^{-1}\in N$, $(a^{-1})^{-1}(b^{-1})^{-1}a^{-1}b^{-1}\in N$, so $[a,b]\in N$.
So $G'\leq N$.
Example
Consider $G=S_3$. find $G'$.
_the trivial way is to enumerate all the commutators, but we can do better than that applying the proposition above to check if any normal group has an abelian factor group to narrow down the search_
Let $N=\{e,\rho,\rho^2\}$, $\rho=(1,2,3)$, then $|N|=|G|/2$, so $N$ is normal and $|G/N|=2 so $G/N\simeq \mathbb{Z}_2$ si abelian, so $G'\subseteq N$.
Now let $\rho=(1,2,3),\sigma=(1,2)$, $[\rho,\sigma]=(1,2,3)(1,2)(1,3,2)(1,2)=(1,3,2)$
So $\rho^2=(1,3,2)$ is in $G'$ and $\rho=(1,3,2)^{-1}\in G$, therefore $N\subseteq G'$.
So $G'=N$.
> Few additional exercises to for $n\geq 5$, we have $G'=A_n$. (relates to simple subgroup properties.) You may check it out.
### Group acting on a set
#### Definition for group acting on a set
Let $G$ be a group, $X$ be a set, $X$ is a $G$-set or $G$ acts on $X$ if there is a map
$$
G\times X\to X
$$
$$
(g,x)\mapsto g\cdot x\, (\text{ or simply }g(x))
$$
such that
1. $e\cdot x=x,\forall x\in X$
2. $g_2\cdot(g_1\cdot x)=(g_2 g_1)\cdot x$
> There is always a trivial action defined on $X$ by $g\cdot x=x$ satisfying the two properties.
Example
Let $G$ be a group,
$G$ acts on $G$ by $g\cdot x\coloneqq g h$, $g,x\in G$
---
$G$ acts on $G$ via conjugation, $g\cdot x\coloneqq g x g^{-1}$, $g,x\in G$
Let's check the two properties are satisfied.
$e\cdot x=exe^{-1}=x$
$$
\begin{aligned}
g_2\cdot (g_1\cdot x)&= g_2\cdot (g_1xg_1^{-1})\\
&= g_2g_1xg_1^{-1} g_2^{-1}\\
&= g_2g_1xg_1^{-1} g_2^{-1}\\
&= (g_2 g_1)(x)(g_1^{-1}g_2^{-1})\\
&= (g_2 g_1)(x)
\end{aligned}
$$
---
Take $S_n$ acts on $\{1,2,\ldots,n\}$ via $\sigma\cdot x\coloneqq \sigma(x)$.
---
$GL(n,\mathbb{R})$ (general linear group) acts on $\mathbb{R}^n$ by $A\cdot x\coloneqq A x$, $A\in GL(n,\mathbb{R}), x\in \mathbb{R}^n$
#### Group action is a homomorphism
Let $X$ be a $G$-set, $g\in G$, then the function
$$
\sigma_g:X\to X,x\mapsto g\cdot x
$$
is a bijection, and the function $\phi:G\to S_X, g\mapsto \sigma_g$ is a group homomorphism.
Proof
$\sigma_g$ is onto: If $y\in X$, let $x=g^{-1}y$, then $\sigma_g(g^{-1}\cdot y)=g\cdot (g^{-1}\cdot y)=(gg^{-1})\cdot y=e\cdot y=y$.
$\sigma_g$ is one-to-one: If $\sigma_g(x_1)=\sigma_g(x_2)$, then $g\cdot x_1=g\cdot x_2$.
So $g^{-1}\cdot (g\cdot x_1)=g^{-1}\cdot (g\cdot x_2)=x_1=x_2$.
Then we need to show that $\phi$ is a homomorphism.
$$
\phi(g_1g_2)=\phi(g_1)\cdot \phi(g_2)
$$
Note that $\phi(g_1g_2)=\sigma_{g_1g_2}$, $\phi(g_1)=\sigma_{g_1}$, $\phi(g_2)=\sigma_{g_2}$.
For every $x\in X$, $\sigma_{g_1g_2}(x)=(g_1g_2)\cdot x$, $\sigma_{g_1}(x)=g_1\cdot x$, $\sigma_{g_2}(x)=g_2\cdot x$. By the second property of $G$-sets, we have $\sigma_{g_1}\cdot \sigma_{g_2}=g_1\circ(g_2\circ x)=(g_1g_2)\circ x=\sigma_{g_1g_2}\circ x$.
> [!NOTE]
>
> $\phi$ as above is general not injective and not surjective.
>
> If $G$ acts trivially on $x$ ($g\cdot x=x,\forall g\in G$), then $\phi(g)$ is the identity function for all $g\in G$.
Define a relation on $X$ by $x\sim y\iff y=g\cdot x$ for some $g\in G$.
This equivalence relation is well-defined.
- Reflexive: $x\sim x$, take $e\cdot x$
- Symmetric: $x\sim y\implies y\sim x$ ($g^{-1}\in G$, $g^{-1}\cdot (g\cdot x)=g^{-1}\cdot y$)
- Transitive: $x\sim y, y\sim z\implies x\sim z$ take $x=g_1\cdot y=g_1\cdot (g_2\cdot z)=g_1g_2\cdot z$ and $g_1g_2\in G$.
This gives a orbit for $x\in X$!