Commutator subgroup does not consist only of commutators?

Let $G$ be a group, $G'=[G, G]$.

"Note that it is not necessarily true that the commutator subgroup $G'$ of $G$ consists entirely of commutators $[x, y], x, y \in G$ (see [107] for some finite group examples)."

Quoted from http://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf page 8.

Anybody can provide the examples? I can't find the book [107].

• math.stackexchange.com/questions/7811/… Commented Oct 30, 2010 at 19:45
• By definition, G' is the subgroup generated by elements of the form [x,y]. Commented Oct 30, 2010 at 20:00
• I think the title needs to be modified: "contains all" should be replaced with "consist of" Commented Oct 30, 2010 at 20:19
• This is basically a textbook question, illustrating the problem of over-reliance on readily available online sources that don't cover all the ground or include the best references. Actually, a more interesting research-level question goes in the other direction: how to show efficiently (without the classification) that every element of a finite nonabelian simple group actually is a commutator. Commented Oct 30, 2010 at 21:25
• In fact, I can prove by ''elementary'' methods the following: that if G is finite and if k(G) < 2|G:G'|, then G' is ''made'' of only commutators. For those ''in characters'': whenever G has less nonlinear irreducible (complex) characters than linear ones, G' consists of only commutators. Just think of Q_8, or of the abelian ones. I plan to publish this, but, who on earth will accept it? Of course, it's just a sufficient condition for G' to consist of only commutators - see the above comments on the ''simples''. Marian
– user47717
Commented Mar 3, 2014 at 9:04

The problem is whether the commutator subgroup may contain elements that are not commutators. One example are the free groups. For instance, in the free group of rank $4$, freely generated by $x$, $y$, $z$, and $w$, the element $[x,y][z,w]$ of the commutator subgroup cannot be written in the form $[a,b]$ for some $a,b$ in the group.

The smallest finite examples are groups of order 96; there's two of them, nonisomorphic to each other. (This was a result in Robert Guralnick's thesis). See this Math Stack exchange question for a description of these groups, and some references.

• How can one prove that [x,y][z,w] is not [a,b] in a free group? Commented Oct 31, 2010 at 23:33
• @mathreader: I could cheat and say that we know it's not a simple commutator because there are groups in which the product of two commutators is not a commutator, so you can simply map the free group appropriately, but I suspect that is not the answer that you want (-: One can just work head-long into it: $[a,b]=a^{-1}b^{-1}ab$. If there is no cancellation, you cannot have this equal to $x^{-1}y^{-1}xyz^{-1}w^{-1}zw$; so either the last letter of $a$ matches with the first of $b$, the first of $a$ with the first of $b$ (cancellation in $b^{-1}a$), etc. Checking leads to impossibility. Commented Nov 1, 2010 at 1:31
• "One can just work head-long into it: $[a,b]=a^{-1}b^{-1}ab$. If there is no cancellation, you cannot have this equal to $x^{-1}y^{-1}xyz^{-1}w^{-1}zw$; so either the last letter of $a$ matches with the first of $b$, the first of $a$ with the first of $b$ (cancellation in $b^{-1}a$), etc. Checking leads to impossibility. – Arturo Magidin" Can you give some hint on how to get contradiction? It seems to me too many cases to consider
– user10631
Commented Nov 7, 2010 at 3:52
• @Ding: First, assume $a$ and $b$ are cyclically reduced (not of the form $d^{-1}a'd$ in reduced form). Write $a=d'a'd$ and $b=d^{-1}b'd'^{-1}$, where $ab=d'a'b'd'^{-1}$, $a^{-1}b^{-1}=d^{-1}a'^{-1}b'^{-1}d$, $d'a'd$, and $d^{-1}b'd'^{-1}$ are all reduced. So $[a,b]= d^{-1}a'^{-1}b'^{-1}dd'a'b'd'^{-1}$; since $a$ is cyclically reduced, there can be no cancellation in $dd'$, so this is reduced. If none is the empty word, then this would require $d=x$ and $d=y$, which is impossible. So at least one is empty; keep going. If not cyclically reduced, consider that case. This is not necessarily (cont) Commented Nov 7, 2010 at 4:39
• Answering my own comment above, I would like to mention the article of Wicks: Wicks, N. J. Commutators in free products. J. London Math. Soc. 37 1962 433–444. where the author proves that an element W of a free group is a single commutator, iff it is conjugate to a cyclically reduced word of the form XYZX^-1Y^-1Z^-1. This gives a practical algorithm to determine that [a,b][c,d] is not a commutator. Commented Sep 12, 2015 at 6:14

In $SL_2(\mathbb R)$ the element $-I$ is not a commutator. Proof: If $ABA^{-1}B^{-1}=-I$ then $ABA^{-1}=-B$, whence $B$ has trace $0$. Wlog, then, $B$ is a standard $90$ degree rotation matrix. Now $B^{-1}AB=-A$ forces $A$ to be such that its determinant has the form $-x^2-y^2$. Contradiction.

• Nice, I did not know. In the other side: Gotô proved in [Morikuni Gotô, A theorem on compact semi-simple groups, J. Math. Soc. Japan, 1949 vol. 1 pp. 270-272] that every element in a semi-simple compact Lie group is a commutator. Commented Jan 12, 2012 at 8:18
• Why should it be rotation matrix? If we change the basis should we change other matrix ? Commented Oct 13, 2017 at 16:24
• Yes. If $B$ has trace $0$ and determinant $1$ then for some $C$ the matrix $B'=CBC^{-1}$ is a standard rotation matrix. Let $A'$ be $CAC^{-1}$. Then $AB=-BA$ implies $A'B'=-B'A'$, ... Commented Oct 13, 2017 at 23:45

No-one has mentioned that which elements of the commutator subgroup are actually commutators can be determined from the character table, so I will. The commutator subgroup consists of those conjugacy classes in the kernel of every linear character of course, but it can also be shown that $g \in G$ is a commutator if and only if

$${\sum _\chi} \frac{\chi(g)}{ \chi(1)} \neq 0$$

where the sum is over all irreducible characters $\chi$. This is exercise 3.10 in Isaacs' book on character theory. For the groups of order 96 it's reasonably practical to work out the character table (though not much fun).

• For any element of the commutator subgroup, you can even work the fewest number of commutators required to write out the element. I believe this is also in Isaacs's, and one can read an algorithmic version of it in GAP's CommutatorLength function. Commented Dec 6, 2010 at 20:24

Let me add a quick topological proof that a product of two commutators is not a commutator in a free group.

Solutions to the equation $[a,b]=[c,d][e,f]$ in a free group $F$ correspond naturally to homomorphisms $\Gamma\to F$ where $\Gamma=\langle a,b,c,d,e,f\mid [a,b]=[c,d][e,f]\rangle$. Notice that $\Gamma$ is the fundamental group of $\Sigma$, the orientable surface of genus three!

We will realise $F$ as the fundamental group of a graph $X$. So we can interpret the question as being about homotopy classes of continuous maps $\Sigma\to X$. Next we use a Folklore Theorem, which seems to date back to a circle of ideas explored by Stallings and Zieschang.

Folklore Theorem. Let $U$ be a handlebody with $\partial U=\Sigma$ and let $\iota:\Sigma\to U$ be the inclusion map. Up to free homotopy, any continuous map $f:\Sigma\to X$ factors as $\Sigma\stackrel{\alpha}{\to}\Sigma\stackrel{\iota}{\to}U\to X$ where $\alpha$ is a self-homeomorphism of $\Sigma$.

The theorem is not very hard to prove; the idea is to pull back midpoints of edges of $X$ and observe that this gives you an embedded multicurve in $\Sigma$ that dies.

In our case, $\pi_1U$ is a free group of rank three, and so any elements $a,b,c,d,e,f$ of $F$ which satisfy $[a,b]=[c,d][e,f]$ must generate a subgroup of rank at most three. For a counterexample, therefore, take $F$ to be a free group of rank four generated by $c,d,e,f$.

My first instinct was to vote to close this question, as it is a familiar one that has been addressed in other places, including the wikipedia article commutator subgroup. When I looked at this article, though, I saw that the claims that it made about noncommutators were not supported by any references. (Then I remembered that I myself substantially rewrote this article a couple of years ago and lamented the lack of references for these facts on the discussion page! But, as is often the case, nothing happened with this.)

One good reference to an infinite family of finite groups in which the commutator subgroup contains noncommutators is

Isaacs, I. M. Commutators and the commutator subgroup. Amer. Math. Monthly 84 (1977), no. 9, 720–722.

http://alpha.math.uga.edu/~pete/Isaacs77.pdf

Also Mariano's answer on math.SE where he uses GAP to find these two groups of order $$96$$ that people often speak of seems especially valuable: it shows how in these modern days it may be easier simply to recompute something for yourself than to try to track down a reference.

$${}{}$$

• Upon closer inspection, I see that the article I mention above is also referred to in Mariano's math.SE answer. Still, my answer does have one small virtue over his, and for that reason I'll leave it up for now. Commented Nov 1, 2010 at 8:41
• My own pedagogical view is that it's pointless to define commutator subgroups abstractly without at least raising the question posted here and making sure students know that examples (even though not totally obvious) exist to show that not every product of commutators need be a commutator. Sources like the Math Monthly are helpful, including the paper by Marty Isaacs cited already and a further one by Phyllis Cassidy giving concrete examples. Anyway, the elusive book [107] is not written by a leading expert and isn't a primary source. Commented Nov 2, 2010 at 22:19
• In fact, the construction in my Monthly article shows that Commented Mar 10, 2014 at 22:36
• Sorry---I accidentally hit return and prematurely terminated my comment. I wanted to say that even if G is perfect, so G' = G, it may happen that not every element is a commutator. Also if H is any group other than an abelian two-generator group, then for all sufficiently large abelian groups A, the wreath product A wr H provides an example where not every element in the commutator subgroup is a commutator. This appears in Problem 4A.12 of my group theory book. Commented Mar 10, 2014 at 22:49

Rotman, The Theory Of Groups, 2nd ed., page 38 attributes the following example to R Carmichael, An Introduction to the Theory of Groups of Finite Order:

Let $G$ be the subgroup of $S_{16}$ generated by the following eight elements: \eqalign{(ac)(bd);&(eg)(fh);\cr(ik)(jl);&(mo)(np);\cr(ac)(eg)(ik);&(ab)(cd)(mo);\cr(ef)(gh)(mn)(op);&(ij)(kl).\cr} Then the commutator subgroup is generated by the first four elements above, and is of order 16. Moreover, $$\alpha=(ik)(jl)(mo)(np)$$ is in the commutator subgroup, but is not a commutator.

A simple yet useful way of getting examples is to look at step $2$ nilpotent groups. Hence we consider central extensions $1\to A\to H\to B\to1$, where $A$ and $B$ are abelian. Taking commutators induces a map $\Lambda^2B\rightarrow A$ (and any such map occurs for some extension). The image of this map is the commutator subgroup and the image of the pure tensors $b\wedge b'$ is the set of actual commutators. Hence, as soon as the image of $\Lambda^2B$ is larger than the image of the pure tensors we have an example.

As a first example we let $B$ be an elementary abelian $p$-group ($p$ a prime) and $A=\Lambda^2B$ with the commutator map being the identity. The set of pure tensors contains $0$ and is stable under multiplication by $(\mathbb Z/p)^\ast$ so we may as well consider the image in the projective space $\mathbb P(\Lambda^2B)$. The set of pure tensors are then just the image of the Plücker map from the Grassmannian of two-dimensional subspaces of $B$. Hence we are almost OK as soon as that image is a proper algebraic subset as is the case exactly when the dimension of $A$ (as $\mathbb Z/p$-vector space) is at least $4$. "Almost" as in principle the image could still contain all $\mathbb Z/p$-points. This is not the case however. For that we may look at the case of subspaces contained in a fixed $4$-dimensional subspace so we are reduced to the $4$-dimensional case. In that case the image of the Plücker embedding is a smooth quadric and there are always points outside such a quadric (concretely $u\wedge u'+v\wedge v'$ where $u,u',v,v'$ is a basis).

In particular looking at the case when $B$ is $4$-dimensional we get that $\Lambda^2B$ is $6$-dimensional and hence we get a group of order $p^{10}$. We can get it down a little by trying to divide out $\Lambda^2B$ by some subspace $V$. Letting $V$ be $1$-dimensional this corresponds to a projection from $V$ as a point of $\mathbb P(\Lambda^2B)$ and we want to choose $V$ so that there is a point of $\mathbb P(\Lambda^2B/V)$ outside the image of the Grassmannian. If we pick $V$ on the Grassmannian, then there are no such points. If we pick $V$ off the Grassmannian, then the map from the Grassmannian to $\mathbb P(\Lambda^2B/V)$ is a double cover so roughly half of the points of $\mathbb P(\Lambda^2B/V)$ lie in the image. Sauf erreur, the precise number of points off the image is $(p^4-p^2)/2$ which is always strictly positive and hence we get an example of order $p^9$ for all $p$. As any linear $\mathbb Z/p$-subspace of $\mathbb P(\Lambda^2B)$ of dimension at least $2$ meets the Grassmannian in rational points which do not all lie in some linear subspace this is the smallest examples that can be obtained in this way.

E. Kowalski just put up a nice blogpost about this subject. In particular, he gives the interesting counter-example $A_5 \ltimes (\mathbb{Z}/2)^4$ and asks for a conceptual proof that it works.

The following is a variant of Thorsten's answer. It doesn't add anything mathematically, but I think it is a clearer presentation than any of the answers above. I'm motivated to post this by someone reasking the question earlier today, and my thought that none of the answers seem like something I'd want to present in a class.

Let $k$ be a field of characteristic not $2$. Let $V$ be a vector space over $k$ with dimension at least $4$. Let $\bigwedge^{\bullet} V$ be the $k$-algebra generated by $V$, modulo the relations $u v=-vu$ for $u$ and $v \in V$. If students have already had some differential geometry, or an unusually good linear algebra course, they may already know this construction.

I claim that the unit group of $\bigwedge^{\bullet} V$ is a counterexample. Specifically, let $u$, $v$, $w$ and $x$ be linearly independent in $V$. Then $(1+u)(1-u) = 1-u^2 = 1$ so $1+u$ is a unit of $\bigwedge^{\bullet} V$, and the same for $1+v$, $1+w$ and $1+x$. We have $$(1+u)(1+v)(1+u)^{-1}(1+v)^{-1} = (1+u)(1+v)(1-u)(1-v) = 1 + 2 u v$$ so $$(1+2uv)(1+2wx) = 1+2(uv+wx)+4uvwx$$ is in the commutator group of $(\bigwedge^{\bullet} V)^{\times}$.

On the other hand, computations like the above show that any commutator in $(\bigwedge^{\bullet} V)^{\times}$ is of the form $1+ st + (\mbox{higher order terms})$ for some $s$ and $t$ in $V$. And $2(uv+wx)$ is not of rank one in $\bigwedge^2 V$. So $1+2(uv+wx)+4uvwx$ is in the commutator subgroup but is not a commutator.

• "none of the answers seem like something I'd want to present in a class" - de gustibus non est disputandum, I suppose.
– HJRW
Commented Jan 12, 2012 at 7:50
• What does it mean to be "of rank one in $\wedge^2 V$" ? Commented Mar 3, 2014 at 12:54
• Oh yes, you mean in the sense of tensor rank. Commented Mar 3, 2014 at 12:56

Sorry for adding one more answer, but here's a simple argument based on using that if the center is large enough then the commutator map has too small image to cover the derived subgroup, and even showing that, in suitable (varying) finite groups, the commutator length can be unbounded.

Let $p$ be an odd prime. Consider the free 2-step-nilpotent Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ on $n$ generators. It has dimension $n+n(n-1)/2$, and its center has dimension $n(n-1)/2$ and coincides with the derived subalgebra.

The law $x\ast y=x+y+\frac12[x,y]$ defines a group law (this is the Baker-Campbell-Hausdorff formula), defining a group $G$ of order $p^{n+n(n-1)/2}$ (and exponent $p$).

The center $Z$ of $G$ has order $p^{n(n-1)/2}$ and coincides with the derived subgroup. The commutator map $G\times G\to [G,G]$ factors through $G/Z\times G/Z$, and hence its image contains at most $p^{2n}=|G/Z|^2$ elements. As soon as $2n<n+n(n-1)/2$ (i.e., $n\ge 4$), this implies that there are non-commutators.

Better, this shows that the set of products of $k$ commutators has cardinal $\le p^{2nk}$, and therefore does not cover the commutator group if $2nk<n+n(n-1)/2$, that is, $n\ge 4k$.

Notes:

1) This example (at least the existence of product of commutators that are not commutators) is certainly covered by Torsten's answer; however I provide a different argument.

2) More complicated elaborations, also based on the center being large, yield finite perfect groups with arbitrary large commutator width.

3)One can directly define $G$ as the relatively free group on $n$ generators in the variety of 2-step-nilpotent groups of exponent $p$, but then one needs to check by hand the statement on the derived subgroup and center.