# Math4302 Modern Algebra (Lecture 17)

## Subgroup

### Normal subgroup

#### Fundamental theorem of homomorphism

If $\phi: G\to G'$ is a homomorphism, then the map $f:G/\ker(\phi)\to\phi(G)$ given by $f(a\ker(\phi))= \phi(a)$ is well defined and is an isomorphism.

$\ker(\phi)\trianglelefteq G$, and $\phi(G)=\{\phi(a)|a\in G\}\leq G'$

<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.

The injective property is trivial.

Next, we will show that $f$ is a homomorphism.

$$
\begin{aligned}
    f(a\ker \phi b\ker \phi)&=f(ab\ker \phi)\\
    &=\phi(ab)\\
    &=\phi(a)\phi(b)\text{ since $\phi$ is homomorphism}\\
    &=f(a\ker \phi)f(b\ker \phi)\\
\end{aligned}
$$

We also need to show $f$ is surjective:

If $\phi(a)\in \phi(G)$, then $f(a\ker \phi)=\phi(a)$

</details>

<details>
<summary> Examples for application of theorem</summary>

If $\phi$ is injective, then $\ker \phi=\{e\}$, so $G\simeq \phi(G)\leq G'$

---

If $\phi$ is surjective, then $\phi(G)=G'$, so $G/\ker\phi \simeq G'$

---

Let $\phi:G\to G'$ be any homomorphism between two groups.

Then there exists a surjective map that $G\to G/\ker(\phi)$ by $a\mapsto aN$.

And there exists a injective map that $G/\ker(\phi)\to G'$ by $a\ker(\phi)\mapsto \phi(a)$

</details>

> [!TIP]
>
> In general, if $N\trianglelefteq G$, then we have a homomorphism $\phi:G\to G/N$ where $a\mapsto aN$. $\ker(\phi)=N$

If $\phi:G\to G'$ is a non-trivial homomorphism, and $|G|=18$ and $|G'|=15$, then what is $|\ker\phi|$?

<details>
<summary>Solution</summary>
Note that $G/\ker(\phi)\simeq \phi(G)$ since $|G'|=15$, then $|G/\ker(\phi)|=1,3,5$ or $15$. But $|G/\ker(\phi)|=\frac{|G|}{|G/\ker(\phi)|}=\frac{18}{1,3,5,15}$ so $|G/\ker(\phi)|=1$ or $3$.

Since $\phi$ is not trivial, $G\neq \ker(\phi)$, so $|G/\ker(\phi)|=3$

So $|\ker(\phi)|=6$.

Example: $\mathbb{Z}_{18}\to \mathbb{Z}_3\times \mathbb{Z}_5$ by $[a]\mapsto ([a\mod 3],[0])$. $\ker(\phi)=\{0,3,6,9,12,15\}$

</details>

What is $\mathbb{Z}_{12}/\langle 4\rangle$? \langle 4\rangle=\{0,4,8,\}$

<details>
<summary>Solution</summary>
The quotient of every cyclic group is also cyclic.

Let $G=\langle a\rangle $ and $N\trianglelefteq G$, then $G/N=\langle aN\rangle$ if $bN\in G/N$, then $b=a^k$ for some $k$. So $bN=a^k N=(aN)^k$

So $\mathbb{Z}_{12}/\langle 4\rangle$ is a cyclic group of order $4$, so it is isomorphic to $\mathbb{Z}_4$.
</details>

What is $\mathbb{Z}\times \mathbb{Z}/\langle (1,1)\rangle$? Let $N=\langle (1,1)\rangle$.

<details>
<summary>Solution</summary>

This is isomorphic to $\mathbb{Z}$, sending the addition over one axis $\mathbb{Z}$. Show the kernel is $N$

</details>

#### There is only two group of order 4

Every group of order $4$ is isomorphic to either $\mathbb{Z}_4$ or $(\mathbb{Z}_2\times \mathbb{Z}_2,+)$

If $|G|=4$ and there is an element of order $4$ in $G$. then $G$ is cyclic, so $G\simeq \mathbb{Z}_4$.

Otherwise since $|\langle a\rangle|||G|=4$ every $a\neq G$. Let $G=\{e,a,b,c\}$, Then $a^2=b^2=c^2=e$, and $ab=c (ab\neq e,ab\neq a,ab\neq b)$, so $G$ is isomorphic to $(\mathbb{Z}_2\times \mathbb{Z}_2,+)$

So we can use this property to define $G\to \mathbb{Z}_2\times \mathbb{Z}_2$ by $\phi(e)=(0,0), \phi(a)=(0,1), \phi(b)=(1,0), \phi(c)=(1,1)$ order of $a,b,c$ does not matter.
