Isomoprhic proof

[Hansel13]

GIVEN PROBLEM:
Let ?: G --> H be an isomorphism between groups G and H and let S be a subgroup of G. Prove that ?(S) is a subgroup of H.

notation: ?(S) = {h?H: ?(x)=h for some x?G}

My Solution:

So by the subgroup thrm., we need to show that ?(S) has an identity, inverses, and closure.

1. First we need to show that the identity of H is in ?(S)
Let h = e[sub:39rgzyih]h[/sub:39rgzyih] for all h?H
Then ?(e[sub:39rgzyih]g[/sub:39rgzyih]) = e[sub:39rgzyih]h[/sub:39rgzyih]
So the identity of H is in ?(S) .

2. Next we need to show that if a??(S), then a[sup:39rgzyih]-1[/sup:39rgzyih]??(S)
No clue how to prove this, help?

3. Lastly, we must show that if a,b??(S), then ab??(S)
I could show you my original work, but it is really bad, and my teacher marked all this as wrong. :/


I got a bad grade in this proof, and I have en exam coming up, so I want to be able to understand isomorphism proofs better! So if anyone could help me out with this proof, I'd appreciate it. It's a lot easier proving subgroups, but proving isomorphisms as a subgroup is a pain!

THANKS

 

[daon]

Your identity inclusion is fine.

Suppose Phi(a) and Phi(b) are in Phi(S) which imply a,b are in S. That 1/b is a member of S comes from it being a sub-group. Thus a/b is an element of S.

Hence Phi(a/b) is an element of Phi(S), and Phi(a/b)=Phi(a)/Phi(b). We have Phi(a) and Phi(b) belong to Phi(S) => Phi(a)/Phi(b) belong to Phi(S), we are done.

 

[daon]

What is in my post above is the "2-step" proof. If you need to show it the way you're doing it:

If Phi(a) is an element of Phi(S) then there is some element a in S which gets mapped to it. So a belonging to S implies 1/a does as well. Hence Phi(S) contains Phi(1/a) = 1/Phi(a).)

Closure is simple: Phi(ab) = Phi(a)Phi(b). If Phi(a), Phi(b) belong to Phi(S), there are a,b that get mapped in S to them. Hence ab is in S => Phi(ab) is in Phi(S).

 

[Hansel13]

Thank you, makes perfect sense. Sometimes I just over complicate things and just need to be reminded the basics.