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?

  • $\begingroup$ How can I make the normal subgroup triangle? $\endgroup$ – Yorch Mar 21 '14 at 14:00
  • $\begingroup$ Only the quotients between subgroups with indices differing by $1$ (otherwise it would imply that the group itself was abelian). $\endgroup$ – Tobias Kildetoft Mar 21 '14 at 14:01
  • $\begingroup$ Oh right... that should have been obvious, thanks. $\endgroup$ – Yorch Mar 21 '14 at 14:02
  • 3
    $\begingroup$ normal subgroup triangle is lhd ($\lhd$) or unlhd ($\unlhd$), left hand diamond is probably what it stands for. $\endgroup$ – Jack Schmidt Mar 21 '14 at 14:15
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    $\begingroup$ @TobiasKildetoft You should post that as an answer (so this isn't listed as an "unanswered question") $\endgroup$ – 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.

Here is a good way to think of solvability:

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.
So where a general group is "build from" simple groups, a solvable group is "build from" finite groups 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).
If one takes a normal series which is as "fine" as possible (so we cannot put further subgroups in between the chosen ones and still get a normal series), then one can check (somewhat harder exercise) that the quotients are elementary abelian, so they are in fact vector spaces over a prime field, and this has a lot of important consequences when one studies representations.


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