# Group is soluble if and only if quotient is abelian

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?

• 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 $gG_i$. – Dietrich Burde Apr 15 at 13:18

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 $$e$$.
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 $$\{[k,l], k\in K, l\in L\}=\{klk^{-1}l^{-1}, k\in K, l\in L\}$$