Proving the correspondence theorem for groups without the inverse map

Question

I have been trying to prove the correspondence theorem by showing that the map from subgroups of G to subgroups of G/N containing N is a bijection. I am aware that constructing an inverse map does the job but instead of that, I want to prove directly that this map is a bijectjon (by showing it is one to one and onto). I got stuck at showing that it is 1-1. That is, if A/N=B/N, then A=B. Any help?

Answer 1

What is wrong with 'let $b\in B$. If $Nb$ lies in $B/N=A/N$ then $Nb=Na$ for some $a\in A$. Thus $1.b=n.a$ for some $n\in N$. But $N\leq A$ so $na\in A$. Thus $b\in A$ so $B\leq A$. By symmetry, $A\leq B$ so $A=B$'?

Answer 2

You could show that $A\subseteq B$ and $B\subseteq A$ (a standard method to show equality of sets).

If $A/N=B/N$, then for all $a\in A$ there is a $b\in B$ such that $aN=bN$, which in turn means that there exist $n_1,n_2\in N$ such that $an_1=bn_2$. Use the fact that $N\subseteq B$ to show that $a\in B$ and then $A\subseteq B$. Now switch $A$ and $B$ in your argument to get $B\subseteq A$. And then you have $A=B$.