[proofplan] We first verify the convention-sensitive point: although $H$ is only assumed to be a subgroup, it is automatically normal because conjugation in an [abelian group](/page/Abelian%20Group) fixes every element. Hence the [quotient group](/theorems/790) $G/H$ is well-defined. We then take two arbitrary cosets in $G/H$ and compute their products in both orders using the quotient operation and commutativity in $G$. [/proofplan]

[step:Show that the subgroup $H$ is normal in $G$]Let $g \in G$ be arbitrary. We prove $gHg^{-1}=H$, where \begin{align*} gHg^{-1}=\{ghg^{-1}:h\in H\}. \end{align*} For every $h \in H$, commutativity in $G$ gives \begin{align*} ghg^{-1}=gg^{-1}h=h. \end{align*} Thus every element of $gHg^{-1}$ lies in $H$, and every $h \in H$ is equal to $ghg^{-1}$ for the same $h \in H$. Therefore $gHg^{-1}=H$. Since $g \in G$ was arbitrary, $H \trianglelefteq G$, so $G/H$ is a well-defined quotient group.[/step]

[guided]The only subtle point is that quotient groups are defined for normal subgroups. The theorem assumes merely that $H \le G$, so we must prove normality from the abelian hypothesis. Fix an arbitrary element $g \in G$. To show normality, we prove that conjugating $H$ by $g$ leaves $H$ unchanged. Define \begin{align*} gHg^{-1}=\{ghg^{-1}:h\in H\}. \end{align*} Now take an arbitrary $h \in H$. Since $G$ is abelian, $g$ and $h$ commute, so $gh=hg$. Therefore \begin{align*} ghg^{-1}=hgg^{-1}=h. \end{align*} This shows that conjugation by $g$ fixes every element of $H$ pointwise. Hence the set of all conjugates $ghg^{-1}$ with $h \in H$ is exactly the original set $H$: \begin{align*} gHg^{-1}=H. \end{align*} Because this holds for every $g \in G$, the subgroup $H$ is normal in $G$. Therefore the quotient group $G/H$ is defined.[/guided]

[step:Compute the product of two arbitrary cosets in both orders] Let $aH$ and $bH$ be arbitrary elements of $G/H$, where $a,b \in G$. The quotient group operation is defined by \begin{align*} (aH)(bH)=(ab)H. \end{align*} Since $G$ is abelian, $ab=ba$. Hence \begin{align*} (aH)(bH)=(ab)H=(ba)H=(bH)(aH). \end{align*} Thus the arbitrary elements $aH$ and $bH$ commute in $G/H$. [/step]

[step:Conclude that the quotient group is abelian] We have shown that for every pair of elements $aH,bH \in G/H$, \begin{align*} (aH)(bH)=(bH)(aH). \end{align*} Therefore every pair of elements of $G/H$ commutes. Hence $G/H$ is an abelian group. [/step]