# Math4302 Modern Algebra (Lecture 16)

## Group

### Normal subgroup

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

1. $aH=Ha$ for all $a\in G$
2. $aHa^{-1}= H$ for all $a\in G$
3. $aha^{-1}\subseteq H$ for all $a\in G$
4. 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 $H$=$eH$

<details>
<summary>Example</summary>

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

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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 $\mathbb{Z}_5$

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$G/G$ is isomorphic to trivial group

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

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

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

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

</details>

Recall

1. The lagrange theorem, if $G$ is finite and $H\leq G$, then $|H| | |G|$.
2. 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)

<details>
<summary>Proof</summary>

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.

</details>

#### 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.

<details>
<summary>Proof</summary>

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.

</details>
