Let and be groups. If is an isomorphism, show that for all . Deduce that any two isomorphic groups have the same number of elements of order for each . Is the result true if is only assumed to be a homomorphism?
Suppose first that and that . Then we have , so that , a contradiction. Now suppose that and . Then we have ; since is injective, so that , a contradiction. So either and are both infinite (hence equal) or both finite.
Suppose now that and are both finite; say and . We have , so that . Likewise, we have , and since is injective, . Thus . Hence .
Now any group isomorphism gives, for all , a bijection on the set of elements in of order and those of order in , so these two sets have the same cardinality.
Note that the above proof depends on being an injection. The result does not hold if is merely a homomorphism; in particular, it does not hold for the trivial homomorphism from any nontrivial group to the trivial group.