On page 104, Dummit and Foote discusses the extension problem and makes the following remark:

Part (2) of the Hölder Program, sometimes called the extension problem, was rather vaguely formulated. A more precise description of "putting two groups together" is: given groups A and B, describe how to obtain all groups G containing a normal subgroup N such that $N\cong B$ and $G/N\cong A$. For instance, if $A = B = Z_{2}$, there are precisely two possibilities for G, namely, $\mathbb{Z}_4$ and $V_{4}$ (see Exercise 1 0 of Section 2.5) and the Hölder program seeks to describe how the two groups of order 4 could have been built from two $\mathbb{Z}_{2}$ without a priori knowledge of the existence of the groups of order 4. This part of the Hölder Program is extremely difficult, even when the subgroups involved are of small order. For example, all composition factors of a group G have order 2 if and only if $|G|=2^{n}$ , for some n (one implication is easy and we shall prove both implications in Chapter 6). It is known, however, that the number of nonisomorphic groups of order $2^{n}$ grows (exponentially) as a function of $2^{n}$, so the number of ways of putting together groups of 2-power order together is not bounded.

I do not understand how do they conclude that 'number of ways of putting together groups' is not bounded.

  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$
    – Shaun
    Jun 1 at 11:13
  • $\begingroup$ Thanks @Shaun, I've added a bit more of the quote to give context. However, I am not sure what work needs to be included. Like, I simply do not see the logical link between the premises and the conclusion. $\endgroup$
    – Jaspreet
    Jun 1 at 17:50

1 Answer 1


I find this statement a bit vague. I guess it means something like if $N$ and $M$ are finite $p$-groups, then there is no absolute bound on the number of isomorphism classes of groups $G$ with a normal subgroup $K \cong N$ and $G/K \cong M$. Any such bound is a function of $|N|$ and $|M|$.

That is really not surprising. What is more impressive is the rate of increase. Graham Higman proved that if $N$ and $M$ are elementary abelian of orders $p^{n/3}$ and $p^{2n/3}$, then there are at least $p^{2n^3/27 + O(n^2)}$ possible $G$.

  • $\begingroup$ At the risk of sounding childish, may I ask why is the rate impressive? I am too novice in group theory to be able to appreciate this right away. $\endgroup$
    – Jaspreet
    Jun 2 at 4:05
  • 2
    $\begingroup$ @Jaspreet It tells us that the number of groups of order $p^n$ is "about" $p^{2n^3/27}$, which is generally regarded as being an impressively large number. (The upper bound was proved later by Charles Sims and has since been improved, but the point is that $2n^3/27$ iss the principal term in the exponent.) $\endgroup$
    – Derek Holt
    Jun 2 at 7:41

Not the answer you're looking for? Browse other questions tagged .