8
$\begingroup$

Which finite groups $G$ have a unique composition series? I don't mean in the sense of the Jordan-Holder theorem, but rather actually unique.

Some examples are the cyclic groups $C_{p^n}$ and the symmetric group $S_3$ (which has the unique normal subgroup $A_3$).

$\endgroup$
7
  • 1
    $\begingroup$ The symmetric group $S_n$ is an example for every $n\neq 4$, isn't it? Other examples: simple groups, $\operatorname{SL}(2,q)$ for $q\geq 4$. $\endgroup$ Nov 21, 2014 at 22:29
  • $\begingroup$ In other words, you ask which finite groups $G$ have $N(G)$ totally ordered, where $N(G)$ is the set of normal subgroups. Some discussion is here: people.bath.ac.uk/masgcs/problem/commentary10.html $\endgroup$
    – YCor
    Nov 21, 2014 at 22:55
  • 3
    $\begingroup$ @YCor: The property here is stronger: for example, $S_4$ has $N(G)$ totally ordered, namely $1< V_4 < A_4 < S_4$, but $V_4$ has three different composition series. In fact, the set of subnormal subgroups must be totally ordered (in which case it happens that every subnormal subgroup is normal). $\endgroup$ Nov 21, 2014 at 23:12
  • 1
    $\begingroup$ When $G$ has only two composition factors, the composition series is unique if and only if $G$ is not the product of two simple groups. So any nonsplit extension of a simple group by another simple group is an example. $\endgroup$
    – spin
    Nov 22, 2014 at 10:04
  • 2
    $\begingroup$ @spin: Note that the (now proved) Schreir conjecture tells us that you can't have an extension of a finite non-Abelian simple group acted on non-trivially by another finite non-Abelian simple group. $\endgroup$ Nov 22, 2014 at 11:02

1 Answer 1

15
$\begingroup$

Such a group $G$ must have a unique minimal normal subgroup $M,$ which must itself be simple by the uniqueness of the composition series. If $G$ is a $p$-group, it follows by induction that $G$ is cyclic ($G/M$ is cyclic by induction, so $G$ is Abelian, and hence cyclic by the uniqueness of $M$). Suppose then that $G$ is not a $p$-group.

If $G$ is solvable, then it follows from the above ( and the uniqueness of the composition series) that $F(G)$ is a cyclic $p$-group. Then $G/F(G)$ is Abelian, hence cyclic of order $q^{m}$ for some prime $q \neq p$ (and in fact $q^{m}$ divides $p-1).$

Suppose then that $G$ is not solvable. Note that $G/G^{\prime}$ is a cyclic $p$-group for some prime $p,$ by the uniqueness of the composition series. More generally, $G/G^{(\infty )}$ has one of the structures above, where $G^{(\infty)}$ is the terminal member of the derived series of $G.$

We next describe the structure of $G^{(\infty)},$ so, for ease of exposition, we assume from now on that $G$ is perfect. Then since $F(G)$ is cyclic, we have $F(G) = Z(G)$, which is a (possibly trivial) cyclic $p$-group for some prime $p.$ Furthermore, $F(G)$ is the largest solvable normal subgroup of $G$.

Then, as above, $G/Z(G)$ has a unique minimal normal subgroup, which is non-Abelian simple, and has trivial centralizer. Using the Schreier conjecture, it follows that $G$ is quasi-simple in the case under consideration.

In summary, a general finite group $G$ with a unique composition series has one of the following structures:

$G$ may be a cyclic $p$-group for some prime $p.$

$F(G)$ may be a cyclic $p$-group, and $G$ is the semidirect product of $F(G)$ with a cyclic $q$-group for some prime $q \neq p.$

$G$ is not solvable, $F^{\ast}(G)$ is quasi-simple, and $Z(G)$ is a cyclic $p$-group for some prime $p.$ Furthermore, $G/F^{\ast}(G)$ is solvable with a unique composition series, so with one of the structures above (note also that $G/F^{\ast}(G)$ is a subgroup of the outer automorphism group of the simple group $F^{\ast}(G)/Z(G)$). (Note that any group with any of these structures does indeed have a unique composition series).

$\endgroup$
2
  • $\begingroup$ What does $F(G)$ stand for? $\endgroup$
    – Nishant
    Nov 22, 2014 at 5:25
  • 2
    $\begingroup$ $F(G)$ is the largest nilpotent normal subgroup of the finite group $G$, the Fitting subgroup of $G$. Also, $F^{\ast}(G)$ is the generalized Fitting subgroup of $G$. $\endgroup$ Nov 22, 2014 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.