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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?

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    $\begingroup$ 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! $\endgroup$ Sep 28 '19 at 4:56
  • $\begingroup$ $G/H$ is then an abelian group, see this post. $\endgroup$ Sep 28 '19 at 14:41
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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.

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

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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Sep 28 '19 at 6:25
  • $\begingroup$ Since we are defining the operation between cosets by means of representatives, I think we need firstly to prove that the result is independent of the choice of the these latter. $\endgroup$
    – user615081
    Sep 28 '19 at 10:29

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