# Group theory, isomorphism

1. Oct 16, 2009

Let G be a finite group. For all elements of G (the following holds: g^2=e(the idendity.) So , all except the idendity have order two.

Proof that G is isomorphic to a finite number of copies of Z_2 ( the group of adittion mod 2, Z_2 has only two elements (zero and one).)

I can try to tell you what I have already tried, but please, can someone give a hint in the right direction..? I basically need a bijection from G to (Z_2)^m, but no idea how I can do it( In particular: they should have the same size as they are finite.)

2. Oct 16, 2009

Define $$\phi : G \rightarrow Z_2$$, so that it is forced to be surjective; i.e. tell the identity where to go. Show $$\phi$$ is a hom. Then look at the kernel. There is only one possibility.

3. Oct 16, 2009

I am sorry, I am not able to see it. I keep on thinking about the fact that m products of Z_2 has order 2^m, which is supposed to be equal to n( the order of the group G..)

I dont even get your advice: How can we tell the identity where to go!? The identity(of G) should always go to idendity(of Z_2). ..

Sorry, I am confused...

4. Oct 18, 2009

No one who can provide some insights..?

5. Oct 21, 2009

Show the group is isomorphic to $$Z_2 \times Z_2 \times \ldots$$, so

G must be commutative since it is the case that forall g, $$g = g^{-1}$$ (since gg=1)
In particular, $$ab = (ab)^{-1} = b^{-1}a^{-1} = ba (by assumption)$$ So that G is commutative.

Next, you could use the fundamental theorem of finite abelian groups, says that any abelian finite group is isomorphic to $$Z_{m_1} \times Z_{m_2} \times \ldots$$

Next, look to intuitionize what is happening, look at subgroups generated by a single non identity element. There will be the |G|- 1 of them and they will all have order 2. There is only one group of order 2 up to isomorphism. What is the next step?

6. Oct 21, 2009