Is true that : group of exponent 4 implies that $[[[x,y],y],y]= \text{identity}$?

Question

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.

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

Update:

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 $$[x^{2^{n-1}},y^{2^{n-1}}]_{n}= \text{identity}$$

where

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

Answer 1

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...

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$^{\ast}$ 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).

Answer 2

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.