Let $G$ be a group and let $G'$ be the derived subgroup, defined as the subgroup generated by the commutators of $G$.

Is there an example of a finite group $G$ where not every element of $G'$ is a commutator? $G'$ is only generated by commutators, but with all of the properties of commutators (ie: what happens under conjugation, exponentation, etc) I can't think of an example.


5 Answers 5


For any prime $p$ and $n>1$, there are nilpotent groups $G$ of class 2 and order $p^{n(n+1)/2}$ with generators $a_i$ $(1 \le i \le n)$, $b_{ij}$ $(1 \le i < j \le n)$, such that $[a_i,a_j] = b_{ij}$, the $b_{ij}$ are all central in the group, and all generators have order $p$.

Then $G'$ is the group of order $p^{n(n-1)/2}$ generated by the $b_{ij}$.

In any group, we have $[ax,by] = [a,b]$ when $x,y$ are central in the group, so $G$ has at most $p^{2n}$ distinct elements that are commutators.

Hence, for any fixed $k>0$, by choosing $n$ sufficiently large we can find $G$ such that not all element of $G'$ are products of at most $k$ commutators.

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    $\begingroup$ Nice! (I was waiting for you to show up :) ) $\endgroup$ Oct 26, 2010 at 21:55

For a nice source of examples, see

I. M. Isaacs: Commutators and the Commutator Subgroup. The American Mathematical Monthly. Vol. 84, No. 9 (Nov., 1977), pp. 720-722

which you can get from JSTOR.

Using GAP I find that the two examples of order 96, which is the minimal possible order, are the groups generated by the following two lists of permutations on $32$ elements.

[ (3,8,6)(4,7,5)(9,27,17)(10,28,18)(11,30,22)(12,29,21)(13,26,23)(14,25,24)(15,31,20)(16,32,19), (1,17,7,23)(2,18,8,24)(3,19,5,21)(4,20,6,22)(9,26,15,32)(10,25,16,31)(11,28,13,30)(12,27,14,29), (1,9,5,13)(2,10,6,14)(3,11,7,15)(4,12,8,16)(17,25,21,29)(18,26,22,30)(19,27,23,31)(20,28,24,32), (1,5)(2,6)(3,7)(4,8)(9,13)(10,14)(11,15)(12,16)(17,21)(18,22)(19,23)(20,24)(25,29)(26,30)(27,31)(28,32), (1,3)(2,4)(5,7)(6,8)(9,11)(10,12)(13,15)(14,16)(17,19)(18,20)(21,23)(22,24)(25,27)(26,28)(29,31)(30,32), (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32) ]


[ (3,7,5)(4,8,6)(9,25,17)(10,26,18)(11,31,21)(12,32,22)(13,27,23)(14,28,24)(15,29,19)(16,30,20), (1,17)(2,18)(3,19)(4,20)(5,21)(6,22)(7,23)(8,24)(9,25)(10,26)(11,27)(12,28)(13,29)(14,30)(15,31)(16,32), (1,9)(2,10)(3,11)(4,12)(5,13)(6,14)(7,15)(8,16)(17,25)(18,26)(19,27)(20,28)(21,29)(22,30)(23,31)(24,32), (1,5,2,6)(3,8,4,7)(9,13,10,14)(11,16,12,15)(17,21,18,22)(19,24,20,23)(25,29,26,30)(27,32,28,31), (1,3,2,4)(5,7,6,8)(9,11,10,12)(13,15,14,16)(17,19,18,20)(21,23,22,24)(25,27,26,28)(29,31,30,32), (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32) ]

(They have GAP ids $(96, 3)$ and $(96, 203)$ respectively.)

Later: I used the following very simple-minded code to find the groups

chk := function (g)
  local comms, sub;
  comms := Set(List(Cartesian(g, g), p -> p[1]*p[2]*p[1]^-1*p[2]^-1));
  sub := Set(DerivedSubgroup(g));
  return Size(comms) = Size(sub);    
examples := AllSmallGroups(96, chk, false);

This works because I remembered there were examples of order 96. To get the permutation representations I did, for example,

  • 1
    $\begingroup$ Thanks! Would you mind outlining your method for finding these in GAP? I am somewhat familiar with the program. $\endgroup$
    – Abelsh
    Oct 25, 2010 at 16:28

The "simplest" example is the free group on two generators, but is perhaps not the easiest to play with unless you are comfortable with free groups.

A nice survey of the problem of commutators vs. commutators subgroups in finite groups is:

  • Kappe, Luise-Charlotte, and Morse, Robert Fitzgerald. On commutators in groups. Groups St. Andrews 2005, Vol. 2, pp. 531-558. London Math. Soc. Lecture Notes Ser., 340, Cambridge University Press, 2007. MR 2331612 (2008g:20068)

The smallest groups in which the commutator subgroup does not equal the set of commutators have order 96; in fact, there are two non-isomorphic groups of order 96 in which the set of commutators does not equal the commutator subgroup. (This result also appeared in Robert Guralnick's 1977 thesis, "Expressing group elements as products of commutators", UCLA).

Kappe also has an upcoming joint paper in which she discusses the same question, restricted to $p$-groups.

Another interesting item is the Ore Conjecture, which stated that if $G$ is a finite nonabelian simple group, then the set of commutators equals the commutator subgroup. Its proof was recently completed, and it appears in:

  • Liebeck, Martin W.; O'Brien, E. A.; Shalev, Aner; Tiep, Pham Huu. The Ore conjecture. J. Eur. Math. Soc. (JEMS) 12 (2010), no. 4, 939–1008. MR MR2654085. A PDF preprint can be found at O'Brien's website.
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    $\begingroup$ Is the proof of the Ore conjecture a verification based on classification? (One can state the conjecture without the "nonabelian" :) ) $\endgroup$ Oct 25, 2010 at 17:12
  • $\begingroup$ @Mariano: I'll add a link to the preprint from O'Brien's site. From the introduction, it seems to me that it does rely on the classification at least to some extent. $\endgroup$ Oct 25, 2010 at 17:22

Keith Dennis has also done some work on this problem. In the paper:

Dennis, R. K.(1-CRNL); Vaserstein, L. N.(1-PAS) Commutators in linear groups. K-Theory 2 (1989), no. 6, 761–767.

they consider conditions under which all elements of the commutator subgroup are products of at most 2 commutators. I seem to recall that Dennis did computer calculations to find the smallest group for which the commutator subgroup contains elements that are not commutators (and my recollection is that the size of the group was around 100 elements) but I can't see anything on-line about that.

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    $\begingroup$ 96 is the minimal size :) $\endgroup$ Oct 25, 2010 at 15:30
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    $\begingroup$ Cool! I'm surprised I remembered this to within 4%. My entire knowledge of this is based on a two-minute conversation in the tea room back in 2000... $\endgroup$
    – Dan Ramras
    Oct 25, 2010 at 18:26

An explicit example is $G = (V \times Q) \rtimes C$ where $V=\langle \varepsilon \rangle \times \langle \varepsilon \rangle \cong C_2 \times C_2$, $Q$ is the quaternion group $Q_8=\{\pm 1, \pm i, \pm j, \pm k\}$ and $C_3 \cong C =\langle t \rangle$ acts "diagonally" on both $V$ and $Q$. It acts on $V$ so that $VC \cong A_4$ and it acts on $Q$ by cyclically permuting $i,j,k$: $i \to j \to k \to i$, so that $QC \cong SL(\mathbb{F}_3^2)$.

We have $|G|=96$. It is a nice exercise to compute the character table of $G$, which is of size $12 \times 12$. If you do so, you will realize that the element $x=((1,\varepsilon),-1) \in V \times Q = G'$ satsifies $\sum_{i=1}^{12} \chi_i(x)/\chi_i(1)=0$ so it is not a commutator.

(This group has GAP id $(96, 203)$.)


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