Prove that additive order is preserved by isomorphisms

Question

I want to show that $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$ are not isomorphic by using the fact that $\mathbb{Z}_4$ has one element of additive order 4 (the largest additive order), while $\mathbb{Z}_2 \times \mathbb{Z}_2$ has no elements of additive order 4. Can someone help me get started in showing that additive order is a property preserved by isomorphisms?

Answer 1

In general, if $f: A \to B$ is a homomorphism of groups and $a\in A$, then $ord(f(a)) \le ord(a)$ because $a^n=1$ implies $f(a)^n=f(a^n)=f(1)=1$.

When $f$ is an isomorphism with inverse $g$, we get $$ ord(a) = ord(gf(a)) \le ord(f(a)) \le ord(a) $$ and so $ord(a)=ord(f(a))$.

Answer 2

Let $\phi: G \to H$ be an isomorphism. Let $x \in G$, and write $e_R$ for the identity of a group $R$.

Then $$x^n = e_G$$ so $$\phi(x^n) = \phi(e_G)$$

Homomorphisms (not just isos) preserve the identity, so the right-hand side is $e_H$. Likewise homomorphisms (not just isos) preserve products, so the left-hand side is $\phi(x)^n$.

Therefore the order of $\phi(x)$ divides $n$, the order of $x$.

But now replace $\phi$ by $\phi^{-1}$, $G$ by $H$, and $H$ by $G$ throughout. This tells us that the order of $\phi^{-1}(\phi(x)) = x$ divides the order of $\phi(x)$.

Since the two orders both divide each other (and they're positive), they must be equal.