Proof verification: For isomorphism $\phi : G\to H$, prove that $G$ has subgroup of order $n$ iff $H$ has subgroup of order $n$. Could someone please verify my following proof?

For isomorphism $\phi : G\to H$ of the groups $G$ and $H$, prove that $G$ has a subgroup of order $n$ iff $H$ has a subgroup of order $n$.

Proof: Let $K\leq G$ with $|K|=n$. Then $\phi (K)$ is the image of $K$ under $\phi$. By the definition of $\phi$, $\phi(K) \subset H$. Since $\phi$ is an isomorphism that maps a group to a group, $\phi(K)$ is a group. Then $\phi(K) \leq H$. Since $\phi$ is bijective and $\phi(K)=K$ (?), $|\phi(K)|=|K|=n$. Then $H$ has a subgroup of order $n$.

I don't know a good way to express $\phi(K)=K$ because I think $\phi(K)=K$ is incorrect.

Let $I\leq H$ with $|I|=n$. Then $\phi (I)$ is the image of $I$ under $\phi$. Since $\phi$ is an isomorphism, the inverse $\phi ^{-1}: H \to G$ exists. Then by the definition of $\phi ^{-1}$,  $\phi ^{-1}(I) \subset G$. Since $\phi ^{-1}$ is an isomorphism that maps a group to a group, $\phi ^{-1}(I)$ is a group. Then $\phi^{-1}(I) \leq G$. Since $\phi^{-1}$ is bijective and $\phi^{-1}(I)=I$ (?), $|\phi^{-1}(I)|=|I|=n$. Then $G$ has a subgroup of order $n$.
 A: Let me copy-past the first half of your proof. 
Proof: Let $K\leq G$ with $|K|=n$. Then $\phi (K)$ is the image of $K$ under $\phi$. By the definition of $\phi$, $\phi(K) \subset H$. Since $\phi$ is an isomorphism that maps a group to a group, $\phi(K)$ is a group. 

I wouldn't say "since $\phi$ is an isomorphism that maps a group to a group", the good reason for $\phi(K)$ to be a group is simply "since $\phi$ is a group morphism" and then you may continue with "it sends groups to groups", and end up with "thus $\phi(K)$ is a group". In your formulation, it is not clear whether you use the "iso" part or the "morphism" part in isomorphism. Furthermore, it sounds like "mapping a group to a group" is something that may or may not be verified an isomorphism but happens to be verified for $\phi$ whereas it is always true.   

Then $\phi(K) \leq H$. Since $\phi$ is bijective and $\phi(K)=K$ (?), $|\phi(K)|=|K|=n$. 

As you probably guessed it given the question mark $\phi(K)=K$ has absolutely no reason at all to be true or to even make sense given that they are not subset of the same sets a priori. I would write instead "Since $\phi$ is bijective, $|\phi(K)|=|K|=n$." Remark that the very definition of cardinality imposes that the cardinality of $K$ and $\phi(K)$ have to be the same. 

Then $H$ has a subgroup of order $n$.

I would rather write, "Thus $\phi(K)$ is a subgroup of $H$ of order $n$". 

Let $I\leq H$[...]

You seem to have copy-pasted the first part of the answer and switch $G$ and $H$, $\phi$ to $\phi^{-1}$ and $K$ to $I$. This a good idea. Why don't you just write that one can apply what you have just proved to $\phi^{-1}:H\to G$? This is a completely legitimate thing to write at this point (i.e. when you have done the first part correctly) and much more time efficient. 

