Proof of the Correspondence theorem (group theory) I'm trying to prove the correspondence theorem differently than my professor did.  Is this valid?
Theorem: Let $G$ be a group and $N \le G$ be a normal subgroup.  Then there is a $1-1$  correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N$.
Proof: Let *Sub*$ (G/N)$ and *Sub*$(G)$ denote the set of subgroups of $G/N$ and subgroups of $G$ containing $N$ respectively. Define the map $f$: *Sub*$(G)$ $\to$ *Sub*$(G/N)$ by $f(H) = H/N$.  This map is injective since if $f(H) = f(H')$, by the fact that the cosets partition a group,
$$H = \bigcup_{a \in H} aN = \bigcup_{a \in H'} aN = H'$$
To show that $f$ is surjetive, let $K \le  G/N$ be a subgroup.  We show that $V = \cup_{\alpha \in K} \alpha$ is a subgroup.  Let $x,y \in V$.  It follows that $x = an $ and and $y = bn'$ where $a,b \in G$ and $n, n' \in N$.  So $xy = (an)(bn') = abn''n$ for some $n'' \in N$ , and hence $xy \in V$.  Furthermore, for any $an \in V$ $(an)^{-1}  =n^{-1}a^{-1} \in V$, so $V$ is a subgroup which clearly contains $N$, and $f(V) = K$. 
 A: You get into trouble here:

To show that $f$ is surjetive, let $K \le  G/N$ be a subgroup.  We show that $V = \cup_{\alpha \in K} \alpha$ is a subgroup.  Let $x,y \in V$.  It follows that $x = an $ and and $y = bn'$ where $a,b \in G$ and $n, n' \in N$.  So $xy = (an)(bn') = abn''n$ for some $n'' \in N$ , and hence $xy \in V$.  Furthermore, for any $an \in V$ $(an)^{-1}  =n^{-1}a^{-1} \in V$, so $V$ is a subgroup which clearly contains $N$, and $f(V) = K$. 

The fact that you never used the hypothesis that $K$ is a group should make you very suspicious. A minor point first: normality of $N$ in $G$ gives you $(an)(bn')=abn''n'$ for some $n''\in N$, not $abn''n$. (You’re using $nb=bn''$.) Let’s assume that that’s been fixed, so that you know that $xy=abm$ for some $m\in N$ and $a,b\in G$; how does that tell you that $xy\in V$?
In order to show that $xy\in V$, you must find an $\gamma\in K$ such that $xy\in\gamma$. This can be done, but you have to work a bit more. Start with the assumption that $x,y\in V$: this means that there are $\alpha,\beta\in K$ such that $x\in\alpha$ and $y\in\beta$, and there are $a,b\in G$ such that $\alpha=aN$ and $\beta=bN$. Now pick your $n_x,n_y\in N$ such that $x=an_x$ and $y=bn_y$, so that $xy=an_xbn_y=abnn_y$ for some $n\in N$ by normality of $N$ in $G$. Clearly, then, $xy\in abN$, and we now want to show that $abN\in K$. The obvious candidate is $\alpha\beta$, so the natural things to do next is to try to prove that $abN=\alpha\beta$. But that’s just proving that $abN=(aN)(bN)$, which you presumably already proved when constructing quotient groups in the first place, so we’ve shown that $V$ is closed under the group operation.
Your argument for closure under taking inverses has the same problem. You can’t say that $n^{-1}a^{-1}\in V$ unless you can show that there is some $\gamma\in K$ such that $n^{-1}a^{-1}\in\gamma$. If $a\in\alpha\in K$, the obvious candidate is $\alpha^{-1}$, where the inverse is taken in $G/N$, and you can finish off the argument much as I did the one for closure under the operation.
