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?

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- November 11th 2011, 04:55 PMdwsmithGroup homomorphism
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? - November 11th 2011, 05:06 PMDrexel28Re: Group homomorphism
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? - November 11th 2011, 05:15 PMdwsmithRe: Group homomorphism
- November 11th 2011, 05:45 PMDevenoRe: Group homomorphism
as alex pointed out, you're essentially done here, by the injectivity of φ (because φ(1_G) = 1_H, right?)

Quote:

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.

Quote:

Suppose ker phi is 1. If , then

So

Is the above correct?

How do I prove part 2?