# Considering isomorphism of groups and their properties

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 $5$

• 4. If $x$ is a generator of $G$, then $\phi(x)$ is a generator of $H$. 5.) If $U$ is a subgroup of order $n$ of $G$, then $\phi(U)$ is a subgroup of order $n$ of $H$. – Dietrich Burde Apr 20 '16 at 20:56
• for $5$ let $G'\leq G$ be a sub of order $n$. Prove $\phi(G')$ is a subgroup of $H$ of order $n$. – Yorch Apr 20 '16 at 20:57
• 4. It's easy to prove that if $G=\langle x\rangle$, then $H=\langle\phi(x)\rangle$ using the homomorphism definition. For every element $a\in G$, there exists $m\in\Bbb Z$ such that $a=x^m$ and the homomorphism definition gives $\phi(a)=[\phi(x)]^m$. Since $H=\phi(G)$, conclude using the previous argument. – learner Apr 20 '16 at 21:06
• 5. Suppose there exists a subgroup $U$ of $G$ of order $n$. Our claim is that $\phi(U)$ is a subgroup of $H$ of order $n$. To prove this, consider $x,y\in\phi(U)$, so there exists $x',y'\in U$ such that $x=\phi(x')$ and $y=\phi(y')$. Show that $xy^{-1}\in\phi(U)$ to show that $\phi(U)$ is a subgroup of $H$. To prove that $|\phi(U)|=n$, consider the map $f\colon U\to\phi(U)$ by $f(x)=\phi(x)$ and show that it's bijective to conclude $|\phi(U)|=|U|=n$. – learner Apr 20 '16 at 21:12

1. $G$ 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.
2. 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 $n$.