Group isomorphisms preserve the order of an element

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?

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. \blacksquare

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