Math4302 Modern Algebra (Lecture 16)

Group

Normal subgroup

Suppose $H\leq G$ , then the following are equivalent:

$aH=Ha$ for all $a\in G$
$aHa^{-1}= H$ for all $a\in G$
$aha^{-1}\subseteq H$ for all $a\in G$
the operation $(aH)(bH)=abH$ is well defined for all $a,b\in G$ , on the set of left coset of $H\leq G$

Then $H\trianglelefteq G$

If $H\trianglelefteq G$ , then the set of left coset of $H\leq G$ is a group under the operation $(aH)(bH)=abH$

$G/H$ is factor (or quotient) group of $G$ by $H$

identity $H$ = $eH$

Example

If $|H|=\frac{1}{2}|G|$ , then $H$ is a normal subgroup of $G$ , then $G/H$ is isomorphic to $\mathbb{Z}_2$

Let $\phi:G\to G'$ be a homomorphism, then $\ker(\phi)\trianglelefteq G$

$\mathbb{Z}/5\mathbb{Z}\trianglelefteq \mathbb{Z}$

And $\mathbb{Z}/5\mathbb{Z}$ is isomorphic to $\mathbb{Z}_5$

$G/G$ is isomorphic to trivial group

$G/\{e\}$ is isomorphic to $G$

$\mathbb{R}/\mathbb{Z}$ is isomorphic to $S^1$

$\mathbb{Z}_3\times\mathbb{Z}_6/\langle (1,1)\rangle$ is isomorphic to $\mathbb{Z}_3$

$\langle (1,1)\rangle=\{(1,1),(2,2),(0,3),(1,4),(2,5),(0,0)\}$

Recall

The lagrange theorem, if $G$ is finite and $H\leq G$ , then $|H| | |G|$ .
If $G$ is finite, abelian, $d||G|$ , then $G$ has a subgroup of order $d$ .

We will show that 2. is not true if $G$ is not abelian. (consider $A_4$ with order 12, have no subgroup of order 6)

Proof

Suppose $H\leq A_4$ , and $|H|=6$ . Then $A_4/H$ is normal in $A_4$ , (since $|A_4/H|=2$ ), and $A_4/H$ is isomorphic to $\mathbb{Z}_2$ .

In other words, every element in $A_4/H$ has either order 1 or 2.

So for any $\sigma\in A_4$ , $(\sigma H)(\sigma H)=\sigma^2 H$ . Therefore $\sigma^2=H$ .

But $\sigma=(1,3)(1,2)\in A_4$ and $\sigma^2=(1,3,2)\in H$ .

Similarly, $(1,3,2)(1,2,4)\dots(1,4,3)$ are all even permutations, making $|H|\geq 8$ , that is a contradiction.

Fundamental homomorphism theorem (first isomorphism theorem)

If $\phi:G\to G'$ is a homomorphism, then the function $f:G/\ker(\phi)\to \phi(G)$ , ( $\phi(G)\subseteq G'$ ) given by $f(a\ker(\phi))=\phi(a)$ , $\forall a\in G$ , is an well-defined isomorphism.

Proof

First we will prove the well definedness and injectivity of $f$ .

We need to check the map will not map the same coset represented in different ways to different elements.

Suppose $a\ker(\phi)=a'\ker(\phi)$ , then $a^{-1}b\in \ker(\phi)$ , this implies $\phi(a^{-1}b)=e'$ so $\phi(a)=\phi(b)$ .

Reverse the direction to prove the converse.

Other properties are trivial.