# Group homomorphism

• Nov 11th 2011, 04:55 PM
dwsmith
Group homomorphism
Let $\phi :G\to H$ be a group homomorphism.
(1) phi is injective iff ker phi is 1
(2) $|G:ker\phi |=|\phi (G)|$

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

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

Is the above correct?

How do I prove part 2?
• Nov 11th 2011, 05:06 PM
Drexel28
Re: Group homomorphism
This isn't quite right. What if $H$ has more than one element of order two? But, you have just proven it in your second line. You know that $\phi(1)=1$ and if $\phi(a)=1$ then by the DEFINITION of injectivity $a=1$. Your converse is correct though, good job!

Part two is just a restatement (in a sense) of the first isomorphism theorem. Know it?
• Nov 11th 2011, 05:15 PM
dwsmith
Re: Group homomorphism
Quote:

Originally Posted by Drexel28
This isn't quite right. What if $H$ has more than one element of order two? But, you have just proven it in your second line. You know that $\phi(1)=1$ and if $\phi(a)=1$ then by the DEFINITION of injectivity $a=1$. 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?
• Nov 11th 2011, 05:45 PM
Deveno
Re: Group homomorphism
Quote:

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

proof of (1)
$\Rightarrow$
Suppose phi is injective. Let $a\in ker\phi$.
Then $\phi(a)=1$

as alex pointed out, you're essentially done here, by the injectivity of φ (because φ(1_G) = 1_H, right?)

Quote:

$\phi(a^{-1})=\phi^{-1}(a)=1^{-1}=1=\phi(a)$
Since $\phi$ is injective, $a=a^{-1}$
Therefore, $1_G\mapsto 1_H$
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.

Quote:

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

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)?