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So a group G is soluble if and only if it has a subnormal series $$ \{ 1\} =G_0 \ \triangleleft \ G_1 \ \triangleleft \ ... \ \triangleleft \ G_n=G $$ where all quotient groups $G_{i+1}/G_i $ are abelian.

My confusion is that surely elements of $G_{i+1}/G_i $ are of the form $ {g+ G_i} $ where $g\in G_i $. Now for this to be abelian we must have $(g_1+ G_i )+(g_2+G_i)=(g_2+G_i) +(g_1+G_i) $ which by definition of quotient group addition means $(g_1+g_2)+G_i =(g_2+g_1)+G_i $ which surely just boils down to the group $G_{i+1} $ being abelian. Am I missing something obvious here?

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  • $\begingroup$ Of course, addition is commutative, but the group law is not addition in general, just $x\circ y$. So the elements of the quotient, the cosets, are of the form $gG_i$. $\endgroup$ – Dietrich Burde Apr 15 at 13:18
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You are. Take for instance $\mathfrak{S}_3$ which is famously nonabelian ($(1 2)(2 3) = (123)$, $(23)(12) = (132)$), then $\mathfrak{S}_3/\mathfrak{A}_3 \simeq \mathbb{Z/2Z}$ is abelian.

The thing you're missing is that when $G/H$ is abelian, $g_1g_2H = g_2g_1H$ does not imply $g_1g_2=g_2g_1$; it implies $g_1g_2g_1^{-1}g_2^{-1} \in H$, but that's pretty much it; if $H$ is large enough, there are plenty of things that $g_1g_2g_1^{-1}g_2^{-1}$ could be that are different from $e$.

Actually, there is an elementary statement which is very important :

Let $H$ be a normal subgroup of $G$. Then $G/H$ is abelian if and only if $[G,G]\subset H$.

where for subgroups $K,L$, $[K,L]$ is the subgroup generated by $\{[k,l], k\in K, l\in L\}=\{klk^{-1}l^{-1}, k\in K, l\in L\}$

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