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?