# Group With an Endomorphism That is "Almost" Abelian is Abelian.

Suppose a finite group has the property that for every $x, y$, it follows that

\begin{equation*} (xy)^3 = x^3 y^3. \end{equation*}

How do you prove that it is abelian?

Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3.

• You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field $\mathbb Z/3\mathbb Z$ has exponent three, so your equation holds, but it is not abelian. Jul 28, 2010 at 21:47
• (There are lots of examples: the most famous ones are the Burnside groups B(m,3), which you'll find described at en.wikipedia.org/wiki/…; the group in the first comment is B(2,3)) Jul 28, 2010 at 21:49
• @Mariano, why don't you give that as the answer and then it can be accepted? Otherwise it looks as though nobody has answered the question.
– bryn
Jul 29, 2010 at 6:57
• By the way, your statement becomes true if you change 3 by 2. Jul 29, 2010 at 14:54
• I wrote a short proof here . Steve
– user641
Aug 4, 2010 at 8:17

You don't, as the group is not necessarily abelian! The group of upper triangular 3-by-3 matrices with ones along the diagonal and coefficients in the three-element field $\mathbb {Z}/3\mathbb{Z}$ has exponent three, so your equation holds, but it is not abelian.

There are lots of examples: the most famous ones are the Burnside groups $B(m,3)$: the group I described above is $B(2,3)$.

On the other hand, if the order of your group is not a multiple of 3 then it must be abelian!

You can read a proof here

• Urgh! Why do they write proofs in such a complicated way?! Jul 29, 2010 at 16:25
• Yeah! This is what I had in mind! Thanks! :)
– user218
Jul 29, 2010 at 18:07

Both proofs so far are for finite groups. However, the problem (with the complete assumptions) holds for not-necessarily-finite groups, provided that the group have no element of order $$3$$.

Here is a proof:

From $$(ab)^3 = a^3b^3$$, we immediately conclude that $$(ba)^2 = a^2b^2$$ by cancellation. We also conclude that cubes commute with squares, since $$b^2a^2=(ab)^2 = (ab)^3(ab)^{-1}= a^3b^3b^{-1}a^{-1} = a^3b^2a^{-1}$$, hence $$b^2a^3=a^3b^2$$.

Now, consider the cube of a commutator, \begin{align*} [a,b]^3 &= (a^{-1}b^{-1}ab)^3\\ &= a^{-3}b^{-3}a^3b^3\\ &= a^{-3}b^{-3}b^2a^3b\\ &= a^{-3}b^{-1}a^3b\\ &= [a^3,b]. \end{align*} In particular, for any square we have $$[a,x^2]^3 = [a^3,x^2]=1$$, because cubes commute with squares. But since $$G$$ has no elements of order $$3$$, we conclude that $$[a,x^2]=1$$. Thus, we conclude that every square in $$G$$ is actually central.

But that means that $$(ab)^2 = b^2a^2 = a^2b^2$$ for all $$a,b\in G$$. And this condition is well known to imply that the group is abelian: $$abab=aabb$$ yields $$ba=ab$$.

In particular, this holds for a finite group whose order is not divisible by $$3$$.