Prove that for isomorphism $\phi: G\to H$, if $K\leq G$, then $\phi (K)=\left \{ \phi(k):k\in K \right \}\leq H$.

Question

Could someone please verify whether my following proof is correct?

Prove that for isomorphism $\phi: G\to H$, if $K\leq G$, then $\phi (K)=\left \{ \phi(k):k\in K \right \}\leq H$.

Proof:

Subset: By definition of $\phi$, $\phi(K) \subset H$.

Identity: Since $K\leq G$, $e_{G}\in K$. Then $\phi(e_{G})=e_{H}\in H$. Then $e_{H}=\phi(e_{G})\in \phi(K)$.

Inverse: Since $\phi$ is an isomorphism, its inverse $\phi ^{-1}$ exists. Then for all $\phi(k)\in \phi(K)$, there exists an inverse $\phi^{-1}(k)$.

Closure: For $k,k'\in K$, then $\phi(kk')\in \phi(K)$ since $K$ is a group. Then $\phi(kk')=\phi(k)\phi(k')\in \phi(K)$.

I am unsure whether closure is correct! Is this proof sufficient?

Answer 1

Your proof is correct apart from the bit about inverses. To show that inverses exist, take any $\phi(k)\in\phi(K)$. Since $K$ is a group, each $k\in K$ has an inverse, $k^{-1}$. You can check that $(\phi(k))^{-1}=\phi(k^{-1})$.