TITLE: Is $G/H$ always a subgroup of $G$?
QUESTION [6 upvotes]: Given a normal subgroup $H$ of a finite group $G$, is there always an injective homomorphism
  $$\varphi:G/H\to G?$$
  In other words, is $G/H$ a subgroup of $G$?

If we pick an arbitrary representative element of each coset, we can get an injective map $\varphi:G/H\to G$, but I am not sure if we can always choose the $g_i$'s to get a homomorphism.
The requirement would be that if $\{g_i\}$ is the chosen set of representatives of the cosets of $G$ by $H$, then for each $i,j$ we have that $g_ig_j$ is the chosen representative for the coset $g_ig_jH$, i.e. $g_ig_jH=g_kH$ for a unique $k$. Can we always choose some $g_i$ to satisfy that?

REPLY [6 votes]: No. Consider the quaternion group $Q_8 = \{1, -1, i, -i, j, -j, k, -k\}$. The centre of $Q_8$ is $\{1, -1\}$ and $Q_8/\{1, -1\} \cong \mathbb{Z}_2\times\mathbb{Z}_2$, but every order four subgroup of $Q_8$ is cyclic and hence isomorphic to $\mathbb{Z}_4$. As $\mathbb{Z}_4 \not\cong \mathbb{Z}_2\times\mathbb{Z}_2$, there is no subgroup of $Q_8$ isomorphic to $Q_8/\{1, -1\}$.