# If $\phi: G \to H$ is an isomorphism, prove that $|\phi(x)| = |x|$ for all $x \in G$

The question from number 2 on page two of this document:

If $$\phi: G \to H$$ is an isomorphism, prove that $$|\phi(x)| = |x|$$ for all $$x \in G$$. Deduce that any two isomorphic groups have the same number of elements of order $$n$$ for each $$n \in \Bbb Z^+$$. Is the result true if $$\phi$$ is only assume to be a homomorphism?

The beginning of the solution states:

Since $$\phi: G \to H$$ is a homomorphism, $$\phi (x^n) = \phi(x)^n$$ for all $$n \in \Bbb Z^+$$. (I assume this is true because we can alternatively write it as $$\phi (x*x*x...x) = \phi(x) * \phi(x) * \phi(x) * ... * \phi(x)$$ where the right hand side is computed in $$H$$ and the left hand side is computed in $$G$$)

When $$n = 0$$, we have $$\phi(1_G) = 1_H)$$.

If $$|x| = m$$ is finite, then $$x^m = 1_G$$ and so $$\phi(x)^m = 1_H$$, which shows $$|\phi (x)| \leq |x|$$ (since isomorphism implies bijection).

I see how if $$x$$ has finite order, then $$x^m = 1_G$$, but why can we deduce that $$\phi(x)^m = 1_H$$ if $$x^m = 1_G$$?

Thanks for the help!

An isomorphism sends only the identity to the identity, so $$\phi(x)^m = \phi(x^m) = \phi(1_G) = 1_H.$$
By the way, not only do you have $$|\phi(x)| \leq |x|$$ but you actually have $$|\phi(x)| \mid |x|$$, i.e. the order of $$\phi(x)$$ divides the order of $$x$$. (This requires just a bit of work.)
Suppose $$\phi(x)^n=1_H$$, where $$n$$ is the order of $$\phi(x)$$. Then $$\phi(x^n)=1_H$$. Since $$\phi$$ is injective, we have $$x^n=1_G$$. Hence $$|x|\leq |\phi(x)|$$.