**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.2**

Let $G$ and $H$ be groups. If $\varphi : G \rightarrow H$ is an isomorphism, show that $|\varphi(x)| = |x|$ for all $x \in G$. Deduce that any two isomorphic groups have the same number of elements of order $n$ for each $n \in \mathbb{Z}^+$. Is the result true if $\varphi$ is only assumed to be a homomorphism?

Solution: Suppose first that $|\varphi(x)| = \infty$ and that $|x| = n < \infty$. Then we have $$\varphi(x)^n = \varphi(x^n) = \varphi(1) = 1,$$ so that $|\varphi(x)| < |x|$, a contradiction. Now suppose that $|x| = \infty$ and $|\varphi(x)| = n < \infty$. Then we have $$\varphi(x^n) = \varphi(x)^n = 1 = \varphi(1);$$ since $\varphi$ is injective, $x^n = 1$ so that $|x| < |\varphi(x)|$, a contradiction. So either $|x|$ and $|\varphi(x)|$ are both infinite (hence equal) or both finite.

Suppose now that $|x|$ and $|\varphi(x)|$ are both finite; say $|x| = n$ and $|\varphi(x)| = m$. We have $$\varphi(x)^n = \varphi(x^n) = \varphi(1) = 1,$$ so that $m \leq n$. Likewise, we have $$\varphi(1) = 1 = \varphi(x)^m = \varphi(x^m),$$ and since $\varphi$ is injective, $x^m = 1$. Thus $n \leq m$. Hence $|x| = |\varphi(x)|$.

Now any group isomorphism $\varphi : G \rightarrow H$ gives, for all $n$, a bijection on the set of elements in $G$ of order $n$ and those of order $n$ in $H$, so these two sets have the same cardinality.

Note that the above proof depends on $\varphi$ being an injection. The result does not hold if $\varphi$ is merely a homomorphism; in particular, it does not hold for the trivial homomorphism from any nontrivial group to the trivial group.