I was surprised to see that one talks about the subgroup generated by the commutators, because I thought the commutators would form a subgroup. Some research told me that it's because commutators are not necessarily closed under product (books by Rotman and Mac Lane popped up in a google search telling me). However, I couldn't find an actual example of this. What is one? The books on google books made it seem like an actual example is hard to explain.

Wikipedia did mention that the product $[a,b][c,d]$ on the free group on $a,b,c,d$ is an example. But why? I know this product is $aba^{-1}b^{-1}cdc^{-1}d^{-1}$, but why is that not a commutator in this group?

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    $\begingroup$ For the case of finite groups see math.stackexchange.com/questions/7811/… . $\endgroup$ – Qiaochu Yuan Aug 26 '11 at 4:22
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    $\begingroup$ The fact that the example you gave is not a commutator follows from the existence of any other example, because if $w,x,y,z$ are elements of any group, there is a homomorphism from the free group on $\{a,b,c,d\}$ sending $a$ to $w$, $b$ to $x$, etc., and such a homomorphism sends commutators to commutators. In other words, although there might be a nice direct proof, that example is more or less a restatement of the fact that examples exist. $\endgroup$ – Jonas Meyer Aug 26 '11 at 4:31
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    $\begingroup$ @Jonas: you should add an explanation to the wikipedia page! $\endgroup$ – Mariano Suárez-Álvarez Aug 26 '11 at 4:34
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    $\begingroup$ Pet peeve: It's "Mac Lane", not MacLane. $\endgroup$ – Arturo Magidin Aug 26 '11 at 4:38
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    $\begingroup$ Some more references: A paper of Martin Isaacs, courtesy of Pete Clark; this MO question; and this question on this very site. $\endgroup$ – Arturo Magidin Aug 26 '11 at 4:48

I. D. MacDonald gives reasonable examples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]

If you have access to JSTOR, it is at http://www.jstor.org/stable/2323464

In particular, he proves by a simple counting argument the nice theorem that

if $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.

Here $Z(G)$ is the center of $G$ and $G'$ its derived subgroup.

  • $\begingroup$ Thanks, I'll try to go to a library or something to get jstor access. $\endgroup$ – Jaymes Aug 26 '11 at 4:30
  • $\begingroup$ I seem to answer this differently each time it pops up :) $\endgroup$ – Mariano Suárez-Álvarez Aug 26 '11 at 4:31
  • $\begingroup$ @Jaymes: send me an email (google for my web page using my name: you'll find my address there) and I can send you the paper. $\endgroup$ – Mariano Suárez-Álvarez Aug 26 '11 at 4:32

A direct answer to the free groups question:

It is easy to see that $$[a, b][c, d]$$ is not a commutator in $F(a, b, c, d)$, as a commutator is of the form $wPw^{-1}Q$ for some words $P$ and $Q$ of the same length, $|P|=|Q|$ (it just so happens $Q=P^{-1}$). Quite clearly $$a^{-1}b^{-1}abc^{-1}d^{-1}cd$$ is not of this form. This is because $ba$ is not a subword of your commutator, and $a$ and $a^{-1}$ are split by a word of length $1$, and $1<5$.

An alternative proof would be the fact that $1\neq [U, V]$ is never a proper power of any word $W$, with $U, V, W\in F_n$ for all $n>1$. So, for example, $[a, b]^2$ is not a commutator. This is a result of Schutzenberger (1959). However, the proofs of this result which exist seem quite involved (or, at least, the proofs that I've found). But the result is pretty...

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    $\begingroup$ Although I much prefer to "it holds because it holds in some homomorphic image" proof... $\endgroup$ – user1729 Aug 26 '11 at 11:35
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    $\begingroup$ Could you give more detail about your first argument? You seem to be saying the element can't be expressed as $wPw^{-1}Q$ with $|P|=|Q|$ but it can; take $w=a^{-1}cb$, $P=b^{-1}c^{-1}b^{-1}cb$, $Q=bc^{-1}d^{-1}cd$. $\endgroup$ – stewbasic Mar 27 '18 at 5:44

I had minor problems convincing myself of the fact that the group described by Geoff exists (See also Derek Holt's answer to the previous version of this question), most notably that it has the prescribed order. So I spent some time on it, and want to share this more concrete version. Hopefully I didn't fumble this.

Inside the group of upper triangular 3x3 matrices (entries from $F_p$) we have the (often used) matrices $$ A=\left(\begin{array}{ccc}1&1&0\\0&1&0\\0&0&1\end{array}\right),\quad B=\left(\begin{array}{ccc}1&0&0\\0&1&1\\0&0&1\end{array}\right),\quad C=\left(\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right), $$ satisfying the relations $A^p=B^p=C^p=1, [A,B]=C, [A,C]=[B,C]=1$.

Using these we can realize that group as $m\times m$ upper triangular matrices, where $m=3n(n-1)/2$, using $n(n-1)/2$ blocks (sized 3x3) along the diagonal. Label the blocks with pairs of indices $(i,j), 1\le i<j\le n$. The generator $x_i$ has matrix $A$ in any block with label $(i,y), y>i$, matrix $B$ in any block with label $(x,i),x<i$ and the identity matrix in the other blocks. Consequently the commutator $[x_i,x_j]$ has the matrix $C$ in the block labelled $(i,j)$ and the identity matrix elsewhere.

The entire group $G$ then consists of matrices with blocks $$ g_{i,j}=\left(\begin{array}{ccc}1&u_i&v_{i,j}\\0&1&u_j\\0&0&1\end{array}\right)=A^{u_i}C^{v_{i,j}-u_ju_i}B^{u_j}, $$ where the $n(n+1)/2$ coefficients $u_i,v_{i,j}$ are arbitrary elements of $F_p$.

If there is a simpler concrete description of this group, I'm all ears :-).

Edit: Anyway, we have $x_i^p=1$ for all $i$, $[x_i,x_j]^p=1$ for $i<j$ and all the commutators are central. The commutator subgroup consists of all those matrices with $u_i=0$ for all $i$. A commutator of two elements $[(g_{i,j}),(g'_{i,j})]$ has $v_{i,j}=u_iu'_j-u_ju'_i$, and there are too few of those.


In fact, one can go further than the (implicit) statement in the question. For any positive integer $m,$ there is a finite group $G$ and an element $x \in [G,G]= G^{\prime}$ such that $x$ can't be expressed as a product of fewer than $m$ commutators.

One way to see this is to use a group $G$ of a form which often comes up in this sort of question. Let $p$ be an odd prime, and let $G = \langle x_1,x_2,\ldots,x_n: x_i^{p} = [x_i,x_j]^{p} = [x_i,x_j,x_k] = 1 \rangle.$ This is a finite $p$-group of order $p^{\frac{n(n+1)}{2}}$ with $|Z(G)| = p^{\frac{n(n-1)}{2}}$ and $G^{\prime} = Z(G).$ The number of commutators in $G$ is at most $[G:Z(G)]^{2} = p^{2n},$ using the fact noted in the paper of MacDonald mentioned in the other answers. Since $G^{\prime}$ is an elementary Abelian $p$-group of order $p^{\frac{n(n-1)}{2}}$, there are elements in $G^{\prime}$ which can't be expressed as a product of fewer than $\frac{n-1}{4}$ commutators.

This is discussed in a paper of R. Guralnick, as is the example of order $96$ mentioned in another answer. Recently, D. Segal has obtained good upper bounds for the number $h$ of commutators needed to express an element of $G^{\prime}$ as a product of $h$ commutators when $G$ is a finite solvable group.

There are many other directions to pursue here: there is an interesting character-theoretic formula due to Burnside, which states that if $G$ is a finite group, then an element $x \in G$ can be expressed as a product of $t$ commutators if and only if $\sum_{i=1}^{k} \frac{\chi_i(x)}{\chi_i(1)^{2t-1} }\neq 0$, where $\chi_i : 1 \leq i \leq k$ are the complex irreducible characters of $G$.

  • $\begingroup$ Some of this answer duplicates parts of Derek Holt's answer in the link given by Qiaochu in his comment to the question. $\endgroup$ – Geoff Robinson Aug 27 '11 at 7:02

See the exercise 2.43 in the book "An introduction to the theory of group"- Joseph Rotman (4th ed.).

He also had made a nice remark:

The first finite group in which product of two commutators is not a commutator has order 96.


Here is an explicit example, paraphrased from the paper of Martin Isaacs linked in the comments. Let $S_3$ be the group of permutations of a 3-element set, and consider the group ring $\mathbb Z_2[S_3]$, which is an abelian group under addition and admits an action of $S_3$ by right multiplication. Thus we can construct the semidirect product $G=S_3\ltimes\mathbb Z_2[S_3]$. Explicitly the operation is $$ (w_1,x_1)(w_2,x_2)=(w_1w_2,x_1w_2+x_2). $$ We have $$\begin{eqnarray*} [(w_1,x_1),(w_2,x_2)] &=&(w_1,x_1)(w_2,x_2)(w_1^{-1},x_1w_1^{-1})(w_2^{-1},x_2w_2^{-1})\\ &=&([w_1,w_2],x_1(w_2+1)w_1^{-1}w_2^{-1}+x_2(w_1^{-1}+1)w_2^{-1}) \end{eqnarray*}$$ For any $w_1,w_2\in S_3$, the element $$ (1,w_1+w_2)=[(1,w_1),(w_2^{-1}w_1,0)] $$ is a commutator. For $S\subseteq S_3$, let $\bar S=\sum_{s\in S}s\in\mathbb Z_2[S_3]$. Let $T=\{1,(123)\}$. Then $$ (1,\bar G+\bar T)=(1,(12)+(13))(1,(23)+(321)) $$ is a product of commutators. Suppose it equals $[(w_1,x_1),(w_2,x_2)]$. Then $[w_1,w_2]=1$ and $$ \bar G+\bar T=x_1(w_2+1)w_1^{-1}w_2^{-1}+x_2(w_1^{-1}+1)w_2^{-1}. $$ Now $w_1,w_2$ generate an abelian subgroup $H<S_3$, and $w_1\bar H=w_2\bar H=\bar H$. Thus $$ (\bar G+\bar T)\bar H=0. $$ By considering each possibility for $H$, we see that this is impossible.


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