A homomorphism is injective if and only if its kernel is trivial.

[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

 

[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:

f(gg’)=f(g)f(g’)=f(g)²=f(gg).
Thus, gg’=gg.

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.

 

[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'.

 

[micromass]

That is correct!