Math4302 Modern Algebra (Lecture 20)

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 $G'\leq N$.

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.

Let $G$ be a group,

$G$ acts on $G$ by $g\cdot x\coloneqq g h$, $g,x\in G$

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$G$ 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.

$e\cdot x=exe^{-1}=x$

$$\begin{aligned} g_2\cdot (g_1\cdot x)&= g_2\cdot (g_1xg_1^{-1})\\ &= g_2g_1xg_1^{-1} g_2^{-1}\\ &= g_2g_1xg_1^{-1} g_2^{-1}\\ &= (g_2 g_1)(x)(g_1^{-1}g_2^{-1})\\ &= (g_2 g_1)(x) \end{aligned}$$

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Take $S_n$ acts on $\{1,2,\ldots,n\}$ via $\sigma\cdot x\coloneqq \sigma(x)$.

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$GL(n,\mathbb{R})$ (general linear group) acts on $\mathbb{R}^n$ by $A\cdot x\coloneqq A x$, $A\in GL(n,\mathbb{R}), x\in \mathbb{R}^n$

$\sigma_g$ 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 $\sigma_{g_1}\cdot \sigma_{g_2}=g_1\circ(g_2\circ x)=(g_1g_2)\circ x=\sigma_{g_1g_2}\circ x$.