It is well known that: If the square of every element of a group is the identity then the group is abelian.

Also is known that: In a group, if (for all $x$) the cube of $x$ is the identity (i.e. a group of exponent 3), then the equation $[[x,y],y]=\text{identity}$ holds, where $[x,y]=xyx^{-1}y^{-1}$ (i.e. the commutator of $x$ and $y$).

My question is if the following assert is true:

In a group, if (for all $x$) the fourth power of $x$ is the identity (i.e. a group of exponent 4), then the equation $[[[x,y],y],y]= \text{identity}$ holds.

The following spin-off discussion was asked and answered in this spin-off question.


Using the automated theorem provers Prover9 and Vampire3 it is possible to prove that

group of exponent 4 implies that $[[[x^2,y^2],y^2],y^2]= \text{identity}$

This theorem is also proved by hand using pen and paper.

As @user1729 is indicating is very easy to prove that

group of exponent 4 implies that $[[x^2,y^2],y^2]= \text{identity}$

and then, as @user1729 is indicating, this implies that

group of exponent 4 implies that $[[[x^2,y^2],y^2],y^2]= \text{identity}$.

Now, extending the idea from @user1729 , is easy to prove that

group of exponent 8 implies that $[[[x^4,y^4],y^4],y^4]= \text{identity}$

Also is possible to prove that

group of exponent 16 implies that $[[[[x^8,y^8],y^8],y^8],y^8]= \text{identity}$

group of exponent 32 implies that $$[[[[[x^{16},y^{16}],y^{16}],y^{16}],y^{16}],y^{16}]= \text{identity}$$

In general:

Group of exponent $2^{n}$ implies that $$[x^{2^{n-1}},y^{2^{n-1}}]_{n}= \text{identity}$$


$$[x,y]_1 = [x,y]$$ $$[x,y]_2 = [[x,y],y]=[[x,y]_1,y]$$ $$[x,y]_3 = [[[x,y],y],y]=[[x,y]_2,y]$$ $$[x,y]_n = [[x,y]_{n-1},y]$$

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    $\begingroup$ Perhaps you have in mind a group of exponent 4, not of order 4? $\endgroup$ Aug 17, 2014 at 20:31
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    $\begingroup$ Since the question involves 2 elements, it is maybe sufficient that the property holds in the burnside group on 2 generators for exponent 4. $\endgroup$
    – Myself
    Aug 17, 2014 at 20:37
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    $\begingroup$ Those groups you mentioned are abelian. So for any abelian group no matter what two elements you pick, their commutator is the identity. $\endgroup$
    – Hesky Cee
    Aug 17, 2014 at 21:50
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    $\begingroup$ Those two groups are up to isomorphism all groups of order 4 (i.e. containing 4 elements), and since they are abelian, the answer is obvious, as observed by @Hesky Cee. The exponent of a group is a different property (the minimal $n$ such that $g^n$ equals to the identity element of a group for any $g \in G$), and there are groups of different orders (arbitrary large) of exponent 4. If you mean groups of exponent 4, please edit the question and the title. This would be an interesting question. $\endgroup$ Aug 17, 2014 at 22:06
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    $\begingroup$ Given the amount of progress that's been made in these comments and the question itself, I'd suggest submitting at least some of it as an answer. Right now the question has gotten overcrowded. $\endgroup$ Aug 18, 2014 at 22:06

2 Answers 2


I do not know the answer, but I know how to find it out! The checking yes/no requires a computer.

Consider the group $\langle x, y\rangle$. This is a two-generated group of exponent $4$, and hence must be a homomorphic image of the free Burnside group$^{\ast}$ $B(2, 4)$. Hence, if we can prove that the equality holds for the free Burnside group $B(2, 4)$ then we are sorted.

In general, this would be a difficult problem. For example, so far as I know it is an open problem whether the free Burnside group on two generators of exponent $5$, $B(2, 5)$, is finite or not. However, we can prove that the equality holds for $B(2, 4)$, because $B(2, 4)$ is a finite group of order $2^{12}\:^{\dagger}$ (note: citation includes a presentation of the group).

I will leave the actual checking to someone with more time than myself, or with a computer...

$^{\ast}$ The free Burnside group $B(m, n)$ is defined to be the quotient of the free group on $m$ generators by the normal subgroup generated by all $n^{\text{th}}$ powers. It was an open problem, called Burnside's problem, for a long time whether any of these could be infinite (in 1968 Novikov and Adian proved that they could indeed be infinite). Efim Zelmanov won a fields medal in the 90s for a related problem. If you are interested, I wrote down a proof that $B(m, 3)$ is always finite here, while it is a classic undergraduate problem to prove that $B(m, 2)$ is abelian (and hence finite).

$^{\dagger}$J. J. Tobin, On groups with exponent $4$, Thesis , University of Manchester (1954).

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    $\begingroup$ It is a little hard to distill an actual presentation for $B(2,4)$ from Tobin's work. But now there are much simpler presentations known, like the 9-generator ones, given in Coxeter-Moser's book. I reproduced them in my answer here: mathoverflow.net/a/445712/2164 $\endgroup$
    – mathreader
    Apr 28 at 16:03

The answer is negative. This is proved in Corollary 65 of https://arxiv.org/pdf/1904.10072.pdf. There are a couple of invariants defined there. To any element $w$ in the derived subgroup of the free group of rank 2 one associates an integer number. If this is not divisible by 4, then $w$ is not a product of fourth powers. And then, it is non-trivial in the Burnside group $B(2,4)$. Of course, this can be checked by different methods, using that $B(2,4)$ is finite. In particular, with GAP.

  • $\begingroup$ That's impressive! Can you also tell by your methods if the 5th Engel word [x,y,y,y,y,y] is a product of 4th powers in F(2)? This is known to be true (Havas found an explicit expression), but the next in row, 6th Engel word is unknown to be equal to a product of 5th powers, a question equivalent to the Burnside problem for exponent 5. $\endgroup$
    – mathreader
    Apr 28 at 14:46

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