NoteNextra-origin/content/Math4302/Math4302_L20.md

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

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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

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!