If $G$ and $H$ are isomorphic groups and $K\leq G,$ then H has a subgroup isomorphic to K I know that $G$ and $H$ being isomorphic means there is an isomorphism and therefore they are of the same order. It seems perfectly reasonable that $H$ has a subgroup isomorphic to $K$ but don't know what I need to use to show this.
 A: Well if $\phi:G\to H$ is an isomorphism, then $\phi(K)$ is a subgroup of $H$, and since $\phi$ maps $K$ bijectively onto its image; so $K\cong\phi(K)$.
A: Let $\phi:G\to H$ be an isomorphism between $G$ and $H$. Then $\phi$ is one-to-one and onto (and is a homomorphism of course). The obvious candidate for a subgroup of $H$ would be $\phi(K)$.
$\phi(K)$ is contained in $H$. It remains only to see that $\phi(K)$ is isomorphic to $K$. The homomorphism between $K$ and $\phi(K)$ is clearly $\phi$. Since $\phi$ is one-to-one, it takes $K$ to $\phi(K)$ in a one-to-one fashion and is onto by definition. Hence $\phi(K)$ is your isomorphic subgroup.
A: If $H \cong G$ then there is a map $\phi: H \rightarrow G$ s.t. $\phi$ is a homomorphism and is a bijection. Now show that $\phi$ restricted to $K$ is still a homomorphism and bijective. 
It is clearly still a homomorphism so we just need check one-to-one and onto. But by definition $\phi$ is one-to-one and the restriction does not effect this. We also know that $\phi$ as a function is onto its image. So $\phi(K)$ is isomorphic to $K$.
Does this help?
