representation of a finite group

syj

1. The problem statement, all variables and given/known data

Prove that a representation of a finite group G is faithful if and only if its image is isomorphic to G.

2. Relevant equations



3. The attempt at a solution

micromass

What did you try already?? If you show us where you're stuck, then we'll know where to help...

syj

I am not very eloquent when it comes to proofs.
So I am just going to lay out what I know.

Let the representation be noted as F, and the image of G'
if F is a faithful representation then ker{F}={1_G}

Can I conclude then by the first isomorphism theorem that G is isomorphic to G'?





I know that for an "if and only if" proof there are two directions. If I can get the first direction of the proof, I can easily get the other direction.

micromass

Indeed, the first isomorphism theorem does the trick!!

syj

Ok, so is this enough:

If f is faithful then ker{f}={1_G}
therefore by the first isomorphism theorem, G[itex]\cong[/itex]G'

If G[itex]\cong[/itex]G' then by the first isomorphism theorem ker{f}={1_G}
therefore by the definition of a faithful representataion, f is faithful.

it seems so plain.
lol.
too plain to be complete.
but if it is, i am one happy girl ;)

micromass

This is true (but only for finite groups), but you might want to explain in some more detail.

The rest is ok!

syj

can you please explain how i should expand further?
I am told that G is finite in the question.
thanks

micromass

Well, you know that

[tex]G\cong G/\ker(\phi)[/tex]

Why does that imply that [itex]\ker(\phi)=\{1\}[/itex] ??

Think of the order...

syj

ok,
am i making sense here:

a corollary to the first isomorphism theorem says:

|G:ker([itex]\varphi[/itex]|=|[itex]\varphi[/itex](G)|

from this can I conclude:
|[itex]\frac{G}{ker(\varphi)}[/itex]|=|G'|

and then conclude:
ker([itex]\varphi[/itex])={1_G}

micromass

Indeed, that works!!

syj

wooo hoooo !!!!
i am the happiest girl in the world!!
until the next proof comes my way ... at which time I shall bug u some more!
thanks so much.