NoteNextra-origin/content/Math4302/Math4302_L16.md

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

Example

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)\}

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)

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.