2
$\begingroup$

Let $G$ and $H$ be groups, $a,b\in G$ and $f: G\to H$ be a homomorphism. We were tasked to show that

a) $|ab|=|ba|$

b) if $f(a)$ has finite order in $H$, then $|a|$ is either infinite or $|f(a)|$ divides $|a|.$

I assume b could be answered when I have showed that a holds, but what I only have in my scratch solution is that since $f$ is a homomorphism then $f(ab)=f(a)f(b)$. I'm stuck. Just a hint on how to continue would be much appreciated.

Thank you very much.

$\endgroup$
  • 2
    $\begingroup$ If $a^n=e_G$, then $(f(a))^n=e_H$ $\endgroup$ – J. W. Tanner Sep 24 at 18:28
  • 3
    $\begingroup$ Actually, those two questions are independent of each other. $\endgroup$ – José Carlos Santos Sep 24 at 18:28
  • $\begingroup$ Thanks for pointing that out, Sir Jose. Can I humbly ask for a hint on how to be able to have a kickstart for a proof in $b$? Thank you. $\endgroup$ – Eigenvector Sep 24 at 18:32
  • 1
    $\begingroup$ Sir J.W. Tanner, thank you. It will be a big help. $\endgroup$ – Eigenvector Sep 24 at 18:33
  • 1
    $\begingroup$ Eigenvector, have a look at some duplicates at this site for your question. They have very good answers. $\endgroup$ – Dietrich Burde Sep 24 at 18:43
3
$\begingroup$

Hints:

(a) If $(ab)^n=e_G$, then $(ba)^{n+1}=ba$

(b) If $a^n=e_G$, then $f(a)^n=e_H$, together with $b^n=e\implies |b|\mid n$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Here's what I did for letter b. $\endgroup$ – Eigenvector Sep 24 at 21:03
2
$\begingroup$

If $a^n=e$, then $$f(a)^n\stackrel{(1)}{=}f(a^n)=f(e)\stackrel{(2)}{=}e$$ where $(1)$ follows from the fact that $f(ab)=f(a)f(b)$ ($f(x)^{-1}=f(x^-1), e=f(x)f(x)^{-1}=f(x)f(x^{-1})=f(xx^{-1})=f(e)$)

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Here's what I did for letter b.

Let $f(a)\in H$ be an element of order $k$. Hence, $f(a)^k=e_H$, where $e_H$ is the identity element in $H$. Since $f:G\to H$ is a homomorphism, then \begin{align*} f(a^k)&=f(a)^k\\ &=e_H \end{align*}

and hence $a^k=e_G$. Therefore, it must be noted that $a\in G$ must at least have an order $k$. Hence, $a^m=e_G$ requires that $m=kq$, which is equivalent to saying that $|f(a)|=k$ must divide $|a|=m$. On the other hand, notice that \begin{align*} f(a^{-k})&=f(a)^{-k}\\ f(a^{-k})f(a)^k&=f(a)^{k}f(a)^k\\ f(a^{-k})f(a^k)&=e_H\\ f(a^{-k}a^k)&=e_H\\ f(a^0)&=e_H \end{align*}

hence $a^0=e_G$, which means $|a|$ is infinite.

Any thoughts?

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.