# Solvable Group, which Quotients need to be Abelian?

In Wikipedia it says a group $G$is solvable if it has a subnormal series

$\{e\}=G_1\lhd G_2,\dots \lhd G_n=G$ where $G_i$ is a normal subgroup of $G_{i+1}$ and all the factor groups are abelian.

My question is does this only mean the quotient groups $G_i/G_{i-1}$ or the quotients between any pair of Groups in the sequence?

• How can I make the normal subgroup triangle? – Jorge Fernández Hidalgo Mar 21 '14 at 14:00
• Only the quotients between subgroups with indices differing by $1$ (otherwise it would imply that the group itself was abelian). – Tobias Kildetoft Mar 21 '14 at 14:01
• Oh right... that should have been obvious, thanks. – Jorge Fernández Hidalgo Mar 21 '14 at 14:02
• normal subgroup triangle is lhd ($\lhd$) or unlhd ($\unlhd$), left hand diamond is probably what it stands for. – Jack Schmidt Mar 21 '14 at 14:15
• @TobiasKildetoft You should post that as an answer (so this isn't listed as an "unanswered question") – Chris Brooks Mar 21 '14 at 21:08

As I mentioned in a comment, only those quotients where the indices differ by $1$ are required to be abelian, as otherwise, this would require that the group itself was abelian.
If the group has a composition series (so for example if it is finite), this is a subnormal series where all the quotients (again, with indices differing by $1$) are simple. Now, if the group is solvable, one can check (and it is a nice exercise to do this) that the quotients in the composition series are cyclic of prime order.
Another thing to think about is the requirement that each $G_i$ be normal in $G_{i+1}$ (rather than being normal in $G$). Another good exercise is to show that if $G$ is solvable then there is in fact a normal series (i.e. where each $G_i$ is indeed normal in $G$) with abelian quotients (hint: Consider commutator subgroups).