Clarification needed for the proof of the theorem: If $T$ is any subgroup of $G/N$, then $T=H/N,$ where $H$ is a subgroup of $G$ that contains $N$

Question

If $T$ is any subgroup of $G/N$, then $T=H/N,$ where $H$ is a subgroup of $G$ that contains $N$.

Proof: Let $H=\{a\in G; Na\in T\}.$ Since $N$ is a normal subgroup of $G$, and $T$ is a subgroup of $G/N$, then $H$ is a subgroup of $G$. If $a \in N$, then $ae^{-1}=ae=a\in N,$ so $a=e \pmod N$, then $Na=Ne\in T.$ Hence $a\in H$. Therefore, $N\subset H$. Finally, the quotient group $H/N$ consists of all cosets $Na$ with $a\in H$, that is, all $Na\in T$. Thus, $H/N=T.\\$

My question about this proof is that I don't see any demonstration where it is shown $T\subset H/N$. It seems that the entirety of the proof is devoted to show $H/N \subset T$. There is something that either I am not understanding or I have missed. Can someone explain to me how to show $T\subset H/N$. Thank you in advance.

Answer 1

This is written using right cosets so I'll do that too.

A typical element of $T$ looks like $Na$ for some $a \in G$. By definition of $H$ we have $a \in H$. Thus $Na \in H/N$.