NoteNextra-origin/content/Math4302/Math4302_L17.md

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'

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.

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)

Examples for application of theorem

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)

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|?

Solution

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\}

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

Solution

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.

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

Solution

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

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.