1. ## Group homomorphism

Let $\displaystyle \phi :G\to H$ be a group homomorphism.
(1) phi is injective iff ker phi is 1
(2) $\displaystyle |G:ker\phi |=|\phi (G)|$

proof of (1)
$\displaystyle \Rightarrow$
Suppose phi is injective. Let $\displaystyle a\in ker\phi$.
Then $\displaystyle \phi(a)=1$
$\displaystyle \phi(a^{-1})=\phi^{-1}(a)=1^{-1}=1=\phi(a)$
Since $\displaystyle \phi$ is injective, $\displaystyle a=a^{-1}$
Therefore, $\displaystyle 1_G\mapsto 1_H$

$\displaystyle \Leftarrow$
Suppose ker phi is 1. If $\displaystyle \phi(x)=\phi(y)$, then $\displaystyle \phi^{-1}(y)\phi(x)=1$
$\displaystyle \Rightarrow\phi(y^{-1}x)=1$ So $\displaystyle y^{-1}x\in ker\phi$
$\displaystyle y^{-1}x=1\Rightarrow y=x$

Is the above correct?

How do I prove part 2?

2. ## Re: Group homomorphism

This isn't quite right. What if $\displaystyle H$ has more than one element of order two? But, you have just proven it in your second line. You know that $\displaystyle \phi(1)=1$ and if $\displaystyle \phi(a)=1$ then by the DEFINITION of injectivity $\displaystyle a=1$. Your converse is correct though, good job!

Part two is just a restatement (in a sense) of the first isomorphism theorem. Know it?

3. ## Re: Group homomorphism

Originally Posted by Drexel28
This isn't quite right. What if $\displaystyle H$ has more than one element of order two? But, you have just proven it in your second line. You know that $\displaystyle \phi(1)=1$ and if $\displaystyle \phi(a)=1$ then by the DEFINITION of injectivity $\displaystyle a=1$. Your converse is correct though, good job!

Part two is just a restatement (in a sense) of the first isomorphism theorem. Know it?
How do I do the first part of part one then?

4. ## Re: Group homomorphism

Originally Posted by dwsmith
Let $\displaystyle \phi :G\to H$ be a group homomorphism.
(1) phi is injective iff ker phi is 1
(2) $\displaystyle |G:ker\phi |=|\phi (G)|$

proof of (1)
$\displaystyle \Rightarrow$
Suppose phi is injective. Let $\displaystyle a\in ker\phi$.
Then $\displaystyle \phi(a)=1$
as alex pointed out, you're essentially done here, by the injectivity of φ (because φ(1_G) = 1_H, right?)

$\displaystyle \phi(a^{-1})=\phi^{-1}(a)=1^{-1}=1=\phi(a)$
Since $\displaystyle \phi$ is injective, $\displaystyle a=a^{-1}$
Therefore, $\displaystyle 1_G\mapsto 1_H$
this, on the other hand, is completely incomprehensible to me. in the first place φ^-1(a) doesn't exist, because a is in G, and has no pre-image under φ.

i suspect you meant (φ(a))^-1, however, all that does is show that a is either 1, or of order 2.

$\displaystyle \Leftarrow$
Suppose ker phi is 1. If $\displaystyle \phi(x)=\phi(y)$, then $\displaystyle \phi^{-1}(y)\phi(x)=1$
$\displaystyle \Rightarrow\phi(y^{-1}x)=1$ So $\displaystyle y^{-1}x\in ker\phi$
$\displaystyle y^{-1}x=1\Rightarrow y=x$

Is the above correct?

How do I prove part 2?
is there an isomorphism between G/ker(φ) and φ(G) (perhaps mapping the coset g(ker(φ)) --> φ(g) would work)?