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Is there a perfect group in which not every element is a commutator?

By a well-known fact, it must have order at least $96.$

By Ore's conjecture (now a theorem), it must be infinite or non-simple.

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    $\begingroup$ The same question has been asked for Lie algebras, see for example here. $\endgroup$ Feb 19 at 9:25

3 Answers 3

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A good source to look for examples is I.M.Isaacs' paper "Commutators and the commutator subgroup", The American Mathematical Monthly, Vol. 84, No. 9 (Nov., 1977), pp. 720-722 (3 pages), available from JSTOR.

The main theorem is:

Theorem. Let $U$ and $H$ be finite groups, $U$ abelian and $H$ nonabelian. Let $G=U\wr H$ be their wreath product. Then $G'$ contains a noncommutator if $$\sum_{A\in\mathscr{A}} \left(\frac{1}{|U|}\right)^{[H:A]}\leq \frac{1}{|U|},$$ where $\mathscr{A}$ is the set of maximal abelian subgroups of $H$. In particular this condition holds if $|U|\geq|\mathscr{A}|$.

Towards the end he notes that if $H$ is simple and $U$ an abelian group, then the wreath product $G=U\wr H$ satisfies $G'=G''$, so if $U$ is large enough then $G'$ is a perfect group in which not every element is a commutator.

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    $\begingroup$ I just did a computer calculation, and the smallest example, with $H=A_5$ and $|U|=2$, works. The group $G'$ is an extension $2^4:A_5$ of order $960$, and only $840$ of its elements are commutators. $\endgroup$
    – Derek Holt
    Feb 19 at 10:55
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As already mentioned, the answer is yes, including for finite groups, and including for finitely presented simple groups.

For instance, finitely presented simple groups constructed by Hyde and Lodha (arXiv link) have non-commutators (furthermore they have elements $g$ such that, denoting $c$ the commutator length, $\liminf c(g^n)/n>0$).

For finite groups, one way is to find perfect groups $G$ with center $Z$ such that $|G/Z|^2<|G|$, i.e., $|Z|>|G|^{1/2}$. Indeed the commutator map induces a map $(G/Z)^2\to G$ whose image is the set of commutators. (I don't remember where this is done and I guess I wrote it down somewhere on MO, these are (nilpotent)$\rtimes$(simple) groups.)

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  • $\begingroup$ Philip Hall, in his paper on classifying $p$-groups up to isoclinism, discusses the commutator map $(G/Z)^2\to [G,G]$. $\endgroup$ Feb 20 at 19:05
  • $\begingroup$ @ArturoMagidin Indeed these are semidirect products $P\rtimes S$ with $P$ $p$-group and $S$ simple, with the constraint that $|Z(P)^S|\ge (|P|.|S|)^{1/2}$, where $Z(P)^S$ is the set of fixed points of $S$ on the center $Z(P)$, and also satisfying that the coinvariants of $S$ on $P/[P,P]$ is the trivial group (to ensure that $P\rtimes S$ is trivial. Are such things discussed in Hall's paper, or does he stick to $p$-groups (without outer action)? $\endgroup$
    – YCor
    Feb 20 at 19:10
  • $\begingroup$ He introduces isoclinism in general, but sticks to finite $p$-groups; I think I misunderstood what your "this" refered to... $\endgroup$ Feb 20 at 19:11
  • $\begingroup$ The Hyde--Lodha examples are overkill! Probably the easiest infinite example is the free product $A_5*A_5$: all hyperbolic groups admit non-trivial quasimorphisms. Or you can use topologial work of Culler to get an explicit uniform lower bound on the stable commutator length. $\endgroup$
    – HJRW
    Feb 21 at 7:39
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    $\begingroup$ @HJRW Yes it's a quotient (since $A_5$ is a maximal subgroup therein and is non-normal, so it generates along with any distinct conjugate). This can at least provide one proof (possibly appealing for finite group theorists with no background in geometric group theory) of the fact that there is a non-commutator in $A_5*A_5$. $\endgroup$
    – YCor
    Feb 21 at 7:53
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In addition to Isaacs' Theorem that Arturo mentioned, below is a concrete example of the application of this theorem. It appeared in the paper On Commutators in Groups, by L-C. Kappe, R. F. Morse, Groups St Andrews $2005$, Vol. 2, p.541. However, their calculation in Example 4.5 is wrong and we present the improved one. The example remains correct.

By the way, it seems (GAP calculations needed) that the smallest perfect group that contains an element not being a commutator, has order $960$:

Let $G=C_2 \wr A_5$, the regular wreath product. Then $|G|=2^{60} \cdot 60$. Observe that the maximal abelian subgroups of $A_5$ are exactly its Sylow subgroups. As usual, let $n_p(G)$ denote the number of Sylow $p$-subgroups of $G$. Then $n_2(A_5)=5, n_3(A_5)=10, n_5(A_5)=6$. This yields the sum: $$5(\frac{1}{2})^{15} + 10(\frac{1}{2})^{20} + 6(\frac{1}{2})^{12} \lt 21 \cdot (\frac{1}{2})^{12} \lt \frac{1}{2}.$$ Hence by Isaacs' theorem (and the remark at the end of his paper: if $H$ is simple, $U$ is abelian and $G=U \wr H$, then $G'=G''$), $G'$ is perfect and $G'$ contains a non-commutator. It has order $960.$

Note (added February 20th 2024) (thanks to welcomed comments/verifications of Derek Holt and Arturo Magidin) In 2010, Robert Guralnick improved the result of I. M. Isaacs by showing that if $U$ is any non-trivial Abelian group of finite order and $H$ is a finite group with derived subgroup of order at least $3$, then some element of the derived subgroup of $U \wr H$ is not a commutator. In particular, if $H$ is perfect, then the derived subgroup of $U \wr H$ is perfect and contains elements which are not commutators.

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    $\begingroup$ It was not immediately clear to me what you mean by $G = C_2 \wr A_5$. I assumed at first that you meant the permutation wreath product, which has order $2^5 \times 60$, and its commutator subgroup is precisely the smallest example of order $960$ that you mentioned. But you probably meant the wreath product with the regular representation of $A_5$, in which case $|G| = 2^{60} \times 60$, but $G$ is not perfect, and its perfect commutator subgroup has order $2^{59} \times 60$. $\endgroup$
    – Derek Holt
    Feb 19 at 12:01
  • $\begingroup$ No relation to Marston Morse eh? $\endgroup$ Feb 19 at 12:48
  • $\begingroup$ @metoo No, nor to Inspector Morse. $\endgroup$ Feb 19 at 15:20
  • $\begingroup$ @DerekHolt Write $G=U \wr H$. It is not the regular wreath product, but (citing Isaacs) $G$ has a normal (the base) subgroup $B$ of all functions $B \rightarrow H$. Multiplication is pointwise. Also $H \subseteq G$, $G=BH$ and $B \cap H=1$. If $f \in B, h \in H$, then $h^{-1}fh=f^h \in B$, with $f^h(x)=f(xh^{-1})$. Still, you remark worries me since that implies that the example given in the Kappe-Morse paper is entirely wrong ... $\endgroup$ Feb 19 at 15:27
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    $\begingroup$ @NickyHekster: Isn't that the regular wreath product? E.g., Rotman talks about the wreath product $D\wr_{\Omega} Q$ where $\Omega$ is a (finite) $Q$-set, and later says "A special case of the wreath product construction has $\Omega=Q$ regarded as a $Q$-set acting on itself by left multiplication. In this case, we write $W=D\wr_r Q$ and we call $W$ the regular wreath product." That's exactly the construction Isaacs considers. $\endgroup$ Feb 19 at 16:40

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