Question

Let G, H be groups with identity \(e_G, e_H\)
Let f: G--> be an isomorphism
Show a) \(f(e_G)=e_H\) I got this part
b) \(\forall g\in G\) \(f(g^{-1} = (f(g))^{-1}\)
Please help

Answer 1

@ganeshie8

Answer 2

If \(a\in G\), to show that \(b=a^{-1}\), you just have to show that: $$ab=ba=e_G.$$ What is: $$f(g)\cdot f(g^{-1})=?$$ $$f(g^{-1})\cdot f(g)=?$$

Answer 3

What I confuse is f(g^-) = (f(g))^- or not? Do I have to prove it?

Answer 4

You need to prove it (and this is the process to do it).

Answer 5

oh!! so, after proving they are equal, nothing is hard because all your expression = e_H, right?

Answer 6

yes. Since $$f(g)\cdot f(g^{-1})=f(g^{-1})\cdot f(g)=e_H,$$ and inverses are unique, we have that $$f(g)^{-1}=f(g^{-1}).$$

Answer 7

As a remark, we actually don't need for \(f\) to be an isomorphism for this to be true. This is valid for any homomorphism as well.

Answer 8

Even the \(f(e_G)=e_H\) part. That's true for any homomorphism.

Answer 9

My T.A said that, too. And he also said :"Later on, you can use it but not now. Now, you have to prove it." :)

Answer 10

Is there any way we can prove the problem without using it? I have test at 10. I need it now, please, give me the whole thing. I have to understand the process, not just memorize the note ( I have the proof on my notebook, but I don't understand)

Answer 11

The way mentioned above is really the only way.

Answer 12

Whaaaat??? ONLY one?? oh oh!! But it 's ok, I got it. Thanks for the help :)

Answer 13

\(f(g^-) *f(g) = f(g^-*g) = f(e_G) = e_H\)
\((f(g))^-*f(g) = e_H\)
So that \(f(g^- ) = (f(g))^-\)
Yeeeees!! It makes sense to me.

Answer 14

Thank you so much, lalala... just stuck at this, now, I got it, lalala...

Answer 15

finally somebody who knows what you're asking !