It goes without saying that the name in the title tentatively refers to a series whose name one does not know yet and probably in the future I may come with a post titled "The x-sequence" or "The x-function" ,etc. I do not know whether the quotient series I am going to construct is known as a "descending quotient series". Okay, it's enough.

Let $G$ be a non-abelian non-simple finite group. Let $H_0$ be one of the smallest proper normal subgroups of $G$. Now if $G/H_0$ is non-simple, call it $G_1$ and let $H_1$ be one of the smallest normal subgroups of $G_1$. Next if $G_1/H_1$ is non simple, call it $G_2$ and let $\ldots$ Continue the process until a simple group is found (which is always possible). I was wondering whether

  1. The length of the series is unique for a group.

  2. Two non-isomorphic groups cannot have isomorphic series.

Thanks. (My apologies if this turns out a trivial question.)

  • 1
    $\begingroup$ When you say "smallest", do you mean "has minimal order", or "is minimal with respect to inclusion"? $\endgroup$ Feb 9, 2011 at 18:45
  • $\begingroup$ Could you clarify what is meant by $H_0$ is "one of the smallest proper normal subgroups", i.e. does this mean $H_0$ is as small as possible without being the trivial subgroup? Or does it mean that no subgroup of $H_0$ is normal in $G$? $\endgroup$
    – ARupinski
    Feb 9, 2011 at 18:46
  • $\begingroup$ @Zev, I thought 'smallest' has the universal meaning of 'smallest order' and it is in this sense that I am using it. Had it been w.r.t inclusion, then we will have the Noetherian/Artinian groups which are not my targets here. @ARupinski, since $H_0$ is proper, it is, by definition, different from the improper/trivial subgroup. $\endgroup$
    – Unknown
    Feb 9, 2011 at 19:02
  • $\begingroup$ @Elohemahab: I understood $H_0$ is not the trivial subgroup, but thanks for clarifying which notion of minimal you are using (which is what I was really asking, albeit in a roundabout way). $\endgroup$
    – ARupinski
    Feb 9, 2011 at 21:16
  • $\begingroup$ Since group extensions are not unique, it is (relatively) easy to build up two groups from the same smaller groups which become your 'smallest proper normal subgroup', which are not isomorphic. $\endgroup$
    – David Roberts
    Feb 10, 2011 at 0:07

2 Answers 2


This is called a chief series or principal series. The quotients of the terms in the series are called chief factors, and the length of the series is called the chief length. Non-isomorphic groups can have isomorphic chief factors. It depends on your definition of isomorphism of chief series whether non-isomorphic groups can have isomorphic chief series (for instance, if G itself is part of the data of the chief series, then G is obviously determined by its series).

The chief length is a group isomorphism invariant. There is a version of Schreier's refinement theorem that applies to chief series. Groups can have many chief series, though some groups have only one, like S4 and SL(2,3) (which have the same chief factors but in a different sort of order).

  • $\begingroup$ @Jack, the terms of my series are quotients subgroups but not necessarily subgroups of the original group $G$. I suppose chief series has a different definition. $\endgroup$
    – Unknown
    Feb 10, 2011 at 0:09
  • $\begingroup$ Your Hi are called the chief factors. They are minimal normal subgroups of the quotients of a chief series. $\endgroup$ Feb 10, 2011 at 0:16
  • $\begingroup$ @Jack, from the post, $H_0 \triangleleft G$, $H_1 \triangleleft G/H_0$, $H_2 \triangleleft (G/H_0)/H_1$, $H_3 \triangleleft ((G/H_0)/H_1)/H_2 ,\ldots$ each $H_i$ being minimal w.r.t order. However, this does not mean $H_n\subset H_{n-1} \subset \ldots \subset H_1 \subset H_0$. But for a chief series(notation as in the Wikipedia article you linked), $1=N_0\subset N_1 \subset \ldots \subset N_n = G$. Am I missing anything? I still hold that the two series are different. $\endgroup$
    – Unknown
    Feb 10, 2011 at 0:44
  • $\begingroup$ Your Hi are called the chief factors of the chief series. $\endgroup$ Feb 10, 2011 at 2:27
  • 3
    $\begingroup$ If this sounds strange, you might not have embraced the lattice isomorphism theorem. H1 = N1/N0, H2 = (N2/N0)/(N1/N0) ≅ N2/N1, etc. $\endgroup$ Feb 10, 2011 at 2:33

Isn't it the case that the group of the square and the quaternion group both have a normal subgroup of order 2 with quotient group Klein4, thus non-isomorphic groups with identical series?

  • $\begingroup$ In general all p-groups of fixed size have series in which every factor is of order p. $\endgroup$ Feb 9, 2011 at 23:10
  • $\begingroup$ Thank you very much. That answers Q2. I suspected a counterexample with groups of low order. $\endgroup$
    – Unknown
    Feb 10, 2011 at 0:19

Your Answer

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

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