# 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?

• In general, if G is any group and H is a normal subgroup of G, then G/H is unambiguous and has a natural group operation. You should try to define this operation and prove it really makes G/H into a group! Sep 28 '19 at 4:56
• $G/H$ is then an abelian group, see this post. Sep 28 '19 at 14:41

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.
• Because $G$ is commutative, $gHg'H = gg'HH = gg'H$. Sep 28 '19 at 6:18