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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.

calc ll
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3 Answers3

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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.

Arturo Magidin
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    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. – Derek Holt Feb 19 '24 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.)

YCor
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  • Philip Hall, in his paper on classifying $p$-groups up to isoclinism, discusses the commutator map $(G/Z)^2\to [G,G]$. – Arturo Magidin Feb 20 '24 at 19:05
  • @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)? – YCor Feb 20 '24 at 19:10
  • He introduces isoclinism in general, but sticks to finite $p$-groups; I think I misunderstood what your "this" refered to... – Arturo Magidin Feb 20 '24 at 19:11
  • 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. – HJRW Feb 21 '24 at 07:39
  • @HJRW I was explicitly wanting to provide finitely presented simple examples. (OP mentioned they have to be infinite or non-simple, so I wanted to provide both finite non-simple and infinite simple examples.) I'm aware of these easier examples when one consider non-simple groups. By the way, although recent, the Hyde-Lodha examples are not complicated (of the same league as the classical Thompson groups). – YCor Feb 21 '24 at 07:41
  • @YCor: fair enough! I think it's good to record that there are easier constructions, though. (And $A_5*A_5$ could have something to do with the smallest finite example, which looks like it might be a quotient.) – HJRW Feb 21 '24 at 07:45
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    @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 $A_5*A_5$. – YCor Feb 21 '24 at 07:53
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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.

Arturo Magidin
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Nicky Hekster
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    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 $2^{59} \times 60$. – Derek Holt Feb 19 '24 at 12:01
  • No relation to Marston Morse eh? – calc ll Feb 19 '24 at 12:48
  • @metoo No, nor to Inspector Morse. – Nicky Hekster Feb 19 '24 at 15:20
  • @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 ... – Nicky Hekster Feb 19 '24 at 15:27
  • You can't mean functions $B \to H$. Perhaps $H \to U$? That would give you the standard regular wreath product, which is not perfect for $U$ abelian, but its commutator subgroup is perfect. – Derek Holt Feb 19 '24 at 15:50
  • @DerekHolt Thanks for pointing out my typo. Of course the $B \rightarrow H$ should be $H \rightarrow U$. I will amend my answer tomorrow according to your insights! – Nicky Hekster Feb 19 '24 at 16:19
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    @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. – Arturo Magidin Feb 19 '24 at 16:40
  • Yes it is the regular one! – Nicky Hekster Feb 19 '24 at 17:08
  • I think the only error you have is the claim that $G=G'$; Isaacs says you will get $G'=G''$. Peter Neumann has a paper where he describes the commutator of the regular wreath product (restricted and unrestricted, though that doesn't matter for finite groups). I'll see if I can access it tomorrow from the office. – Arturo Magidin Feb 20 '24 at 05:28
  • In Peter Neumann's On the structure of the standard wreath product of groups, Math. Zeitschrift 84 (1964) 343-373, as a corollary of what he does (which includes restricted and unrestricted wreath products of possibly infinite groups) it follows that for nontrivial finite groups $A$ and $B$, the commutator of $A\wr B$ is generated by the subgroup of $A^B$ of elements $f$ with $\prod_{b\in B}f(b)\in A'$ (which he shows is well defined) and the subgroup $B'$ (Cor. 4.5). For $C_2\wr A_5$, it would be generated by the tuples with an even number of nontrivial coordinates and all of $A_5$. – Arturo Magidin Feb 20 '24 at 19:02