If $G$ is an abelian group and $H$ be any subgroup, then why is $G/H$ also a group? 
If $G$ is an abelian group and $H$ be any subgroup, then why is $G/H$ also a group?

I get that every subgroup of an abelian group is normal, but how can I use that to prove that $G/H$ is a group?
 A: Because it is a quotient group.
A less glib answer is that the group operation $\cdot$ is $$\begin{align}gH\cdot g'H&=gHg'H \\ &\stackrel{(1)}{=}(gg')H\end{align}$$ for each $g,g'\in G$. Checking that $(G/H, \cdot)$ is a group is routine and $(1)$ holds because $G$ is abelian (and, in particular, $H$ is thus normal in $G$). Here is a proof that it is indeed a group.
A: The quotient group G/H is the set of all left cosets gH, where g is an element of G, and with the binary operation gH * g'H = (gg')H.
 It is closed under * ; for a given g and g' in G, gg' is in G since it is a group. So, for two given cosets gH and g'H in G/H, gH * g'H = (gg')H is in G/H. It has an identity element; eH *gH = (eg)H = gH for all g in G. So, eH is our identity in G/H. Every element has an inverse; let g' be the inverse of g in G. Then, gH * g'H = (gg')H = eH. So, for a given element gH, g'H is its inverse in G/H. * is Associative; let g, g', and g'' be in G. Then, (gH * g'H) * g''H = (gg'')H * g''H = ((gg')g'')H = (g(g'g''))H =
= gH * (g'g'')H = gH * (g'H * g''H).
So, it satisfies the group axioms and is therefore a group.
