Show finite $\langle x,y,z,t\rangle$ with relations $y^x=y^2$, $z^y=z^2$, $t^z=t^2$, $x^t=x^2$ is trivial. This is Exercise 3.2.8 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search on Approach0, it is new to MSE, although this similar question that is in fact from a previous exercise of the book is the first hit.
The Details:
Presentations are defined in a previous chapter of Robinson's book; however, many of the tools available in combinatorial group theory are not discussed; I cannot, for instance, use Tietze transformations in this exercise.
The tools available are:

*

*Basic facts about free groups.


*Basic facts about reduced words.


*The normal form of an element.


*The projective property of free groups.


*von Dyck's Theorem.


*Basic facts about finitely presentated groups; viz., B.H. Neumann's theorem and P. Hall's theorem.
This list is not exhaustive. There's a whole section of the book on varieties but I doubt it's relevant.
The chapter the exercise in question is in is about constructions of groups. There's a lot about simple groups, which might be relevant as Robinson states before the exercise . . .

The next two exercises constitute G. Higman's construction of a finitely generated infinite simple group.

The Question:

Suppose that $H=\langle x,y,z,t\rangle$ is a finite group in which the following relations hold: $y^x=y^2$, $z^y=z^2$, $t^z=t^2$, and $x^t=x^2$. Prove that $H=1$.

There is the following . . .

$\color{red}{{\rm Hint}}$: Let $i,j,k,l$ be the orders of $x, y,z,t$: show that $j\mid 2^i-1$, $k\mid 2^j-1$, etc. Using the fact that if $n>1$, the smallest prime divisor of $2^n-1$ exceeds the smallest prime divisor of $n$, show that $i=j=k=l=1$.

Thoughts:
I'm not sure what to do. Even if I had Tietze transformations at my disposal, I would still be at a loss.

In the interest of providing context, here is a quick Q & A.

What are you studying?

A postgraduate research degree in combinatorial group theory.

What kind of approaches (to similar problems) are you familiar with?

I'm aware that the isomorphism problem is undecidable. There is no one-size-fits-all approach.

What kind of answer are you looking for? Basic approach, hint, explanation, something else?

I'm looking for a strong hint or a full solution, please.

Is this question something you think should be able to answer? Why or why not?

Perhaps, with enough time, yes. I think that my experience with group presentations is sufficient for me to understand what's going on.
However, I have been stuck on this problem for a few days to no avail. The hint is very number-theoretic and, therefore, not something I am especially confident with.
I want to move on to the next exercise in the book. Spending too much time on this one might have diminishing returns.

Please help :)
 A: Here is a sketch proof - more detail on request.
$y^x = y^2 \Rightarrow y^{x^i} = y^{2^i}$ for all $i \ge 0$. So if $x$ has order $i$ then $y^{2^i} = y$, so the order $j$ of $y$ divides $2^i-1$.
Similarly, $k|2^j-1$, $l|2^k-1$ and $i|2^l-1$. Note that this implies that $i,j,k,l$ are all odd.
Now suppose that $i>1$ and let $p$ be the smallest prime dividing $2^i-1$. So $2^i \equiv 1 \mod p$. We also have $2^{p-1} \equiv 1 \mod p$ (since $p \ne 2$), so the multiplicative order of $2$ mod $p$ divides both $i$ and $p-1$, and hence $i$ is divisible by a smaller prime than $p$.
So, since $j|2^i-1$, the smallest prime dividing $j$ is bigger than the smallest prime dividing $i$, similiarly for $k$ and $j$, $l$ and $k$, and $i$ and $l$, contradiction.
So $i=j=k=l=1$.
Incidentally, the group
$$ G = \langle x,y,z,t \mid y^x=y^2,z^y=z^2,t^z=t^2, x^t=x^2\rangle$$
defined by this presentation is a well known example of Graham Higman of an infinite group with no nontrivial finite quotients.
This exercise shows that $G$ has nontrivial finite quotient groups. In fact $G$ itself is infinite. To prove that, you really need to be familiar with the basic theory of group extensions.
The (isomorphic) groups
$\langle x,y,z \mid y^x=y^2,z^y=z^2\rangle$ and  $ \langle z,t,x \mid t^z=t^2, x^t=x^2\rangle$ are both double HNN extensions of a cyclic groups, and (in both cases) their subgroup $\langle x,z \rangle$ is free.
The group $G$ is their amalgamated free product.
