Let be a group homomorphism.
(1) phi is injective iff ker phi is 1
(2)
proof of (1)
Suppose phi is injective. Let .
Then
Since is injective,
Therefore,
Suppose ker phi is 1. If , then
So
Is the above correct?
How do I prove part 2?
This isn't quite right. What if has more than one element of order two? But, you have just proven it in your second line. You know that and if then by the DEFINITION of injectivity . Your converse is correct though, good job!
Part two is just a restatement (in a sense) of the first isomorphism theorem. Know it?
as alex pointed out, you're essentially done here, by the injectivity of φ (because φ(1_G) = 1_H, right?)
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.
Since is injective,
Therefore,
i suspect you meant (φ(a))^-1, however, all that does is show that a is either 1, or of order 2.
is there an isomorphism between G/ker(φ) and φ(G) (perhaps mapping the coset g(ker(φ)) --> φ(g) would work)?
Suppose ker phi is 1. If , then
So
Is the above correct?
How do I prove part 2?