1
$\begingroup$

As part of a larger proof, I need to use the fact that for a homomorphism $\phi : A \rightarrow B$, where $A$ is cyclic and $a \in A$ its generator, $\phi(a^{-n}) = \phi(a^n)^{-1}$. All I came up with is that $\phi(a^{-n}) = \phi((a^n)^{-1})$ and then I hit a wall. Is it true in general that $\phi(a^{nm}) = \phi(a^n)^m$, even for $m < 0$?

I'm also not sure if it's important that $A$ is cyclic and $a$ is its generator.

Thanks in advance for any help.

$\endgroup$
0

2 Answers 2

2
$\begingroup$

Since $\phi(a^{n})\phi(a^{-n})=\phi(a^na^{-n})=\phi(1)=1$, then $\phi(a^{-n})=\phi(a^n)^{-1}$ since inverse of $\phi(a^{n})$ is unique.

$\endgroup$
1
  • 1
    $\begingroup$ And indeed, $A$ being cyclic with generator $a$ has nothing to do with this, it holds for all group elements. $\endgroup$ Jan 17, 2017 at 18:17
1
$\begingroup$

because $\phi$ is homomorphism group then $\phi ( (a^n)^{-1}))=\phi((a.a...a)^{-1})=\phi(a^{-1}...a^{-1})=\phi(a^{-1})...\phi(a^{-1})= \phi(a)^{-1}...\phi(a)^{-1}=(\phi(a)...\phi(a))^{-1}=(\phi(a^n))^{-1} $

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .