Quotient a group by a proper subgroup. If you have a group $G$ and a proper subgroup $H$ inside of the group. Then is $H$ a proper subgroup of the quotient group $G/H$? 
 A: In order to be a subgroup of a group, one must first be a subset. However, $H$ is not a subgroup of $G/H$ for any normal subgroup $H$ because $G/H$ contains as its elements cosets in $G$. For instance $H$ is the identity of $G/H$ because $H$ is the class of the identity $e\in G$. The identity in $H$ is $e$ though and so if it was a subgroup then it would share the same identity as $G/H$. This is not the case.
It is possible for $H$ to be isomorphic to a proper subgroup of $G/H$ and this is a more interesting question. The easiest example is if $H$ is the trivial subgroup generated by the identity. In this case $G/\langle e\rangle=\{\{g\}\mid g\in G\}$ and so the subgroup $\{\{e\}\}\in G/\langle e\rangle$ is isomorphic to the subgroup $\{e\}$ in $G$. (note the double brackets in the notation used for the trivial subgroup in $G/\langle e\rangle$. It's important).
A: No, $H$ is an element of $G/H$, your question is like asking whether $2$ is a subgroup of $\mathbf{Z}$.
A: You have the natural mapping:
$$v\colon G \rightarrow G/H$$
If H is a normal subgroup, then $G/H$ (set of cosets) is a group. There you can identify $H$ with the $0$ element.
