Hw5 Problem 8

### Problem 8

Let $\phi : G \rightarrow G'$ be an isomorphism of a group $\langle G, * \rangle$ with the group $\langle G',*' \rangle$. Prove: If $H$ is a subgroup of $G$, then $\phi [H] = \{ \phi (h) \ h\in H \}$ is a subgroup of $G'$. That is, an isomorphism carries subgroups into subgroups.

Solution

(closed)

Let $a,b \in H$ such that $\phi (a) , \phi (b) \in \phi [H]$.
Then $(a*b) \in H$ since $H \leq G$, and is therefore closed.
Since $\phi$ is an isomorphism $\phi (a) *' \phi (b) = \phi (a*b) \in \phi [H]$, so $\phi [H]$ is closed under $*'$.

(identity)

By Theorem 3.14 since $\phi$ is an isomorphism $e' = \phi (e) \in \phi [H]$.

(inverse)

Let $a \in H$ such that $\phi (a) \in \phi [H]$. Then $a^{-1} \in H$ by definition of a subgroup. It follows that

(1)
\begin{align} e' = \phi (e) = \phi (a^{-1}*a ) = \phi (a^{-1})* \phi (a) \end{align}

then,

(2)
\begin{align} \phi (a)^{-1} = \phi (a ^{-1}) \in \phi [H] \end{align}

Therefore, since all conditions hold, $\phi [H] = \{ \phi (h) \ h\in H \}$ is a subgroup of $G'$.