Exercise 1.6.2 (Two isomorphic groups have the same number of elements of order $n$)

Proof. Let $x$ be an element in $G$, and $d$ an integer. Then by definition of the order of $x$,

Since $\phi$ is a homomorphism $x^k=e\Rightarrow \phi {(x)}^k=\phi (x^k)=\phi (e)=e^{\prime }$ (see Ex. 1).

Conversely, since $\phi$ is an isomorphism, ${\phi }^{-1}$ is also an isomorphism, so $\phi {(x)}^k=e^{\prime }\Rightarrow e={\phi }^{-1}(e^{\prime })={\phi }^{-1}(\phi {(x)}^k)=x^k$. This shows that for all $x\in G$, and for all $k\in \mathbb{Z}$

Then

So for all $x\in G$,

Suppose now that $G\simeq H$, so that there is an isomorphism $\phi :G\to H$. Suppose that there are exactly $m$ elements of order $d$ for some integer $d\ge 1$, say $x_1,x_2,\dots ,x_m$. Then $\phi (x_1),\phi (x_2),\dots \phi (x_m)$ have order $d$ and are distinct elements of $H$. If $y\in H\setminus \{\phi (x_1),\phi (x_2),\dots \phi (x_m)\}$, then $y=\phi (x)$ for some $x\notin \{x_1,x_2,\dots ,x_m\}$, so $|x|\ne d$, which implies $|y|=|\phi (x)\ne d$. This shows that there are exactly $m$ elements in $h$ of order $d$, namely $\phi (x_1),\phi (x_2),\dots ,\phi (x_m)$.

Two isomorphic groups have the same number of elements of order $d$ for each integer $d\ge 1$.

This property is false if $\phi$ is only assumed to be a homomorphism. As a counterexample, consider the homomorphism $\phi :\mathbb{Z}∕4\mathbb{Z}\to \mathbb{Z}∕2\mathbb{Z}$ defined by $\phi ({[0]}_4)=\phi ({[2]}_4)={[0]}_2,\phi ({[1]}_4)=\phi ({[3]}_4)={[1]}_2$. There is an element of order $4$ in $\mathbb{Z}∕4\mathbb{Z}$, but no element of order $4$ in $\mathbb{Z}∕2\mathbb{Z}$. □