# What does a homomorphism have to do with the order of an element in a group? [closed]

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.

• If $a^n=e_G$, then $(f(a))^n=e_H$ – J. W. Tanner Sep 24 at 18:28
• Actually, those two questions are independent of each other. – José Carlos Santos Sep 24 at 18:28
• 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. – Eigenvector Sep 24 at 18:32
• Sir J.W. Tanner, thank you. It will be a big help. – Eigenvector Sep 24 at 18:33
• Eigenvector, have a look at some duplicates at this site for your question. They have very good answers. – Dietrich Burde Sep 24 at 18:43

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 $$b^n=e\implies |b|\mid n$$

• Here's what I did for letter b. – Eigenvector Sep 24 at 21:03

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 $$f(ab)=f(a)f(b)$$ ($$f(x)^{-1}=f(x^-1), e=f(x)f(x)^{-1}=f(x)f(x^{-1})=f(xx^{-1})=f(e)$$)

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?