If $f$ is an isomorphism from a group $(G,*)$ to a group $(G',\#)$ and $a\in G$, then $|a|=|f(a)|$

Question

If $f$ is an isomorphism from a group $(G,*)$ to a group $(G',\#)$ and $a\in G$, then prove $|a|=|f(a)|$.

How would I prove this? The class I'm taking is abstract algebra.

Answer 1

Suppose that $|a|=n$, then noting that $f(e_G)=e_{G'}$ we compute $$f(a^n)=f(a)^n=e_{G'}$$ Furthermore, if $f(a)^m=e_{G'}$ for some $m<n$, then $f(a)^m=f(a^m)=e_{G'}$ which contradicts that $f$ is a bijection.

Answer 2

You prove this by showing that $a^n$ is the identity in $G$ if and only if $(f(a))^n$ is the identity in $G'$.

Answer 3

Let a in G, suppose $a^n=1$, $f(a^n)=f(a)^n=1$ thus $ord(f(a))\leq ord(a)$ the same argument shows $ord(f^{-1}(f(a))=ord(a)\leq ord(f(a))$ thus $ord(a)=ord(f(a))$