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This question is inspired by a recent question by user Jaymes about the set of commutators not being a group.

Following a link to MO, Gerry Myerson posted an example from Carmichael. I quote

Let $G$ be a 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.

However, no proof is given. Can somebody offer a proof of this?


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up vote 4 down vote accepted

This is probably not the cleverest way, but it is a very explicit and "follow-your-nose" way of doing it.

  1. Clearly $\alpha$ lies in the subgroup generated by the first four elements: it's just the product of the third and fourth.

  2. The first four elements are all commutators: $$\begin{align} (ac)(bd) &= {}[(ac)(eg)(ik),(ab)(cd)(mo)]\\ (eg)(fh) &= [(ac)(eg)(ik),(ef)(gh)(mn)(op)]\\ (ik)(jl) &= [(ac)(eg)(ik),(ij)(kl)]\\ (mo)(np) &= [(ab)(cd)(mo),(ef)(gh)(mn)(op)]. \end{align}$$

  3. The subgroup generated by the first four elements is normal: note that they commute with all elements (they are pairwise disjoint permtuations, so they commute among themselves, and conjugating by any of the other four generators leaves them invariant). So in fact, it is a central subgroup.

  4. The first four elements are the only nontrivial commutators of the form $[x,y]$ with $x$ and $y$ elements of the generating set. Simply verify that the remaining two commutators are trivial: $$\begin{align*} {}[(ab)(cd)(mo),(ij)(kl)] &= 1;\\ {}[(ef)(gh)(mn)(op),(ij)(kl)] &= 1. \end{align*}$$

  5. The commutator subgroup is generated the commutators of the form $[x,y]$ with $x$ and $y$ in the given generating set. Use the identity $$[xy,zt] = [x,t]^y[y,t][x,z]^{yt}[y,z]^t$$ and the fact that all commutators of two generating elements are central to decompose any commutator into a product of commutators of the given form.

  6. In fact, in this group, the commutator bracket induces a bilinear map $G/N\times G/N \to N$, where $N$ is the subgroup generated by these four elements (since the commutator subgroup is central). The simple commutators are precisely the image of this map. The equation above shows that the commutator bracket is bilinear (since commutators are central), and $[x,c]=1$ when $c$ is central, so the map $G\times G\to [G,G]$ factors through $G/Z(G)\times G/Z(G)$).

  7. $\alpha$ is not a simple commutator. Call the generators $x_1,\ldots,x_8$, in the order given in the problem. The bilinear map given by the commutator bracket is generated by: $$\begin{align*} {}[\;\overline{x_5},\overline{x_6}\;] &= x_1\\ {}[\;\overline{x_5},\overline{x_7}\;] &= x_2\\ {}[\;\overline{x_5},\overline{x_8}\;] &= x_3\\ {}[\;\overline{x_6},\overline{x_7}\;] &= x_4 \end{align*}$$ and all other brackets are trivial. To show that $\alpha = x_3x_4$ is not a simple commutator, note that we can restrict ourselves to elements that are of the form $x_5^ax_6^bx_7^cx_8^d$, as every element is of this form modulo the center; and using bilinearity of the commutator bracket, we have: $$ [x_5^{\epsilon_5}x_6^{\epsilon_6}x_7^{\epsilon_7}x_8^{\epsilon_8},x_5^{\eta_5}x_6^{\eta_6}x_7^{\eta_7}x_8^{\eta_8}] = x_1^{\epsilon_5\eta_6+\epsilon_6\eta_5} x_2^{\epsilon_5\eta_7 + \epsilon_7\eta_5} x_3^{\epsilon5\eta_8 + \epsilon_8\eta_5} x_4^{\epsilon_6\eta_7+\epsilon_7\eta_6}$$ with $\epsilon_i,\eta_j=0,1$, and the exponents taken modulo $2$

    In order to get $x_3x_4$, we cannot have both $\epsilon_5$ and $\eta_5$ equal to $0$ (the exponent of $x_3$ does not work out); assume without loss of generality that $\epsilon_5=1$. Then $\eta_6=\eta_5\epsilon_6$ and $\eta_7=\eta_5\epsilon_7$ (so the exponents of $x_1$ and $x_2$ are correct), and the exponent of $x_4$ tells us we cannot have both equal to $0$. So $\eta_5=1$ as well. But this means that $\eta_6=\epsilon_6$ and $\eta_6=\epsilon_6$, so the exponent of $x_4$ is either $0$ if any of them is zero, or $1+1\equiv 0$; hence we cannot get $x_3x_4$ as a simple commutator.

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An alternative is to first establish a normal form: it is not hard to verify that every element can be written uniquely in the form $x_1^{\alpha_1}\cdots x_8^{\alpha_8}$, with $\alpha_i=0$ or $1$, and that the product of two elements will simply add the exponents for $x_5,\ldots,x_8$, and add them "with a twist" on $x_1,\ldots,x_4$ to reflect any commutation being done among $x_5,\ldots,x_8$. This is simply a collection process using the fact that the group is nilpotent of class two. – Arturo Magidin Aug 27 '11 at 19:42
Thanks Professor Magidin. – Waldott Aug 28 '11 at 22:23

I am not sure if this is a good enough of a proof for you, but I can work it out in Mathematica and confirm that with explicit computations. If this not helpful, I would be happy to delete the answer.

Mathematica session snapshot

The copy and paste ready code follows:

Commutator[a_, b_] := 
 PermutationProduct[a, b, PermutationPower[a, -1], 
  PermutationPower[b, -1]]

ru = {"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5, "f" -> 6, 
   "g" -> 7, "h" -> 8, "i" -> 9, "j" -> 10, "k" -> 11, "l" -> 12, 
   "m" -> 13, "n" -> 14, "o" -> 15, "p" -> 16};

generators = {
   Cycles[{{1, 3}, {2, 4}}], Cycles[{{5, 7}, {6, 8}}],
   Cycles[{{9, 11}, {10, 12}}], Cycles[{{13, 15}, {14, 16}}],
   Cycles[{{1, 3}, {5, 7}, {9, 11}}], 
   Cycles[{{1, 2}, {3, 4}, {13, 15}}],
   Cycles[{{5, 6}, {7, 8}, {13, 14}, {15, 16}}], 
   Cycles[{{9, 10}, {11, 12}}]

g = PermutationGroup[generators];


g1 = PermutationGroup[Take[generators, 4]];


el = Cycles[{{9, 11}, {10, 12}, {13, 15}, {14, 16}}];

MemberQ[GroupElements[g1], el]

AllCommutators = 
 DeleteDuplicates[Commutator @@@ Tuples[GroupElements[g], {2}]]

Complement[GroupElements[g1], AllCommutators]
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Thanks for your answer Sasha. – Waldott Aug 28 '11 at 22:24

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