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

Question

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?

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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!

Answer 1

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.)

Answer 2

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 $|x|\leq |\phi(x)|$.