- #1
jmjlt88
- 96
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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
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