# Homework Help: A homomorphism is injective if and only if its kernel is trivial.

1. Jul 4, 2012

### jmjlt88

I was a little curious on if I did the converse of this biconditonal statement correctly. Thanks in advance! =)

Proposition: Suppose f:G->H is a homomorphism. Then, f is injective if and only if K={e}.
Proof:
Conversely, suppose K={e}, and suppose f(g)=f(g’). Now, if f(g)=f(g’)=e, then it follows that g=g’=e since the kernel is trivial. Otherwise, assume f(g)=f(g’)≠e. Then, since f is a homomorphism, we have
(1) f(gg’)=f(g)f(g’).
By our assumption f(g)=f(g’). Hence, f(g)f(g’)=f(g)2. Then,
(2) f(gg’)=f(g)f(g’)=f(g)2=f(gg).
Thus, gg’=gg. By using our left cancellation law, we obtain g=g’ and hence f is injective as required.
QED

2. Jul 4, 2012

### micromass

No, it's incorrect. This is easily seen since you never really use that the kernel is trivial in your second part.

The fundamental mistake is here:

From $f(gg^\prime)=f(gg)$, we can of course not deduce that $gg^\prime=gg$ unless we know that the function is injective! But this is exactly what we want to prove.

3. Jul 4, 2012

### jmjlt88

Thank you! Thought about it. Now, here's what I got.

If f(g)=f(g'), then f(g)f(g')-1=e=f(g)f(g'-1). Since f is a homomorphism, this implies f(gg'-1)=e. Since the kernel is trivial, its follows that gg'-1=e. Multiplying on both sides on the right by g', we obtain g=g'.

4. Jul 4, 2012

### micromass

That is correct!

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