I have to prove following theorem.
Let $\phi : G \rightarrow H$ be an isomorphism of two groups. Then following statements are true.
$1. \ \ \ \ \phi^{-1} : H \rightarrow G \ \ \text{is an isomorphism}.$
$2. \ \ \ \ \vert G \vert = \vert H \vert.$
$3. \ \ \ \ \text{If G is abelian, then H is abelian.}$
$4. \ \ \ \ \text{If G is cyclic, then H is cyclic.}$
$5. \ \ \ \ \text{If G has a subgroup of order n, then H has a subgroup of order n.}$
So far I have proved $1-3$ but I need help for $4$ and $5$
$G$ is cyclic if and only if there exists a surjective group homomorphism $e:\mathbb Z\to G$. Since $\phi$ is bijective, $\phi\circ e:\mathbb Z\to H$ is a surjective group homomorphism if and only if $e:\mathbb Z\to G$ is a surjective group homomorphism. Consequently, $G$ is cyclic if and only if $H$ is cyclic.
Let $K\subseteq G$ be a subgroup. Then $\phi(K)\subseteq H$ is a subgroup and $[K:1]=[\phi(K):1]$ because $\phi$ is bijective. Consequently, $G$ has subgroup of order $n$ if and only if $H$ has subgroup of order $n$.