Math4302 Modern Algebra (Lecture 17)

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 $f(a\ker \phi)=\phi(a)$

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

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

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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 $a\ker(\phi)\mapsto \phi(a)$

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 $[a]\mapsto ([a\mod 3],[0])$. $\ker(\phi)=\{0,3,6,9,12,15\}$

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 $\mathbb{Z}_4$.

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