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

  1. Jul 4, 2012 #1
    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. jcsd
  3. Jul 4, 2012 #2

    micromass

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    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 [itex]f(gg^\prime)=f(gg)[/itex], we can of course not deduce that [itex]gg^\prime=gg[/itex] unless we know that the function is injective! But this is exactly what we want to prove.
     
  4. Jul 4, 2012 #3
    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'.
     
  5. Jul 4, 2012 #4

    micromass

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    That is correct!
     
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