4
$\begingroup$

Suppose $G$ is a finite length (or finite) group, with composition series $1 = G_0 \subseteq \ldots \subseteq G_k = G$. Denote the composition factors of this series (up to isomorphism) by the ordered list $(A_1, A_2, \ldots A_k)$. Suppose that $1 = G'_0 \subseteq \ldots \subseteq G'_k = G$ is a different composition series, which composition factors $(A'_1, \ldots A'_k)$.

The Jordan-Holder theorem tells us that these factors are unique up to reordering. So, I can think of $(A'_1, A'_2, \ldots A'_k)$ as a permutation of $(A_1, \ldots, A_k)$. In particular, this is some element in $\sigma \in S_k$. I will write $(A'_1, A'_2, \ldots A'_k) = \sigma (A_1, \ldots, A_k)$ The jist of my question is this: is $\sigma^n (A_1, \ldots, A_k)$ also the composition factors of some series for $G$. More generally, if we let $N$ be the subgroup of $S_k$ generated by all the $\tau \in S_k$ so that $\tau (A_1, \ldots, A_k)$ corresponds to the composition factors of some composition series, is it again the case that, for any $\nu \in N$, $\nu (A_1, \ldots, A_k)$ corresponds to some composition series?

This holds trivially in uninteresting cases:

  1. I can permute the composition factors of a finite abelian group in any way that I like. Likewise for any finite direct sum of simple groups. In this case, $S_k$ acts naturally and transitively on the possible orders of the composition factors of the group, although there may be some kernel due to repeated factors.

  2. Other times, I am totally limited in the extent to which I can permute the composition factors - $S_n$ comes to mind ($n \geq 5$) comes to mind, as does $S_3$. (And also $S_4$, since the composition series is $1 \subset Z_2 \subset Z_2 \times Z_2 \subset A_4 \subset S_4$, which has composition factors $\{Z_2, Z_2, Z_3, Z_2\}$.) So in this case, I would say that the trivial group ($S_1$) acts naturally.

  3. The dihedral groups are also uninteresting in this regard: In $Z_n \rtimes Z_2$ we need to factor $Z_n$ out first, since all the proper normal subgroups are contained in this cyclic subgroup, and we need the quotient to be of prime order. Thus we have the top term fixed and the rest can be permuted as in the finite abelian case.

Is there in general a (natural), transitive group action on the possible orders of the composition factors of a group?

$\endgroup$
  • $\begingroup$ @JackSchmidt No, because there is only one possible order of composition factors, so $\{1\}$ acts transitively on the orders. Sorry if this wasn't clear from my question. $\endgroup$ – Lorenzo May 23 '14 at 3:40
  • 1
    $\begingroup$ I cannot figure out exactly what you are asking. (I am not even sure which examples you are talking about when you say "In the first case" and "In the latter case".) Could you try and formulate your question more clearly? $\endgroup$ – Derek Holt May 23 '14 at 8:23
  • $\begingroup$ @DerekHolt I updated my question. Hopefully it is more clear now. $\endgroup$ – Lorenzo May 23 '14 at 17:32
  • $\begingroup$ Looks interesting now. I think the repeated factors make the question a little complicated, so if you can work on how to handle them, that will make your question easier to answer. For instance dihedral of order 12 has minor issues: the 2 on top cannot be moved, but the 2 in the middle/bottom can be moved. I'm fairly certain you can't just do it by "position" (so the permutation group cannot simply act on {1,2,...,k} in your notation). $\endgroup$ – Jack Schmidt May 23 '14 at 17:45
  • 1
    $\begingroup$ Yes, $C_5 \times S_3$ should be bad. Just using the position, you cannot tell if the factor is "top/middle" (C2), "middle/bottom" (C3), or "top/middle/bottom" (C5). $\endgroup$ – Jack Schmidt May 23 '14 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.