9

Arpit Kansal showed here that a group $G$ in which $x\mapsto x^3$ is an isomorphism is Abelian. He first showed that we have $a^3b^3=b^3a^3$ for all $a,b\in G$ (only using that $x\mapsto x^3$ is homomorphism) and then used injectivity of $x\mapsto x^3$ to get $ab=ba$ for all $a,b\in G$.

Is there a non-Abelian group $G$ in which $x\mapsto x^3$ is a homomorphism?

Community
  • 1
principal-ideal-domain
  • 6,324

3 Answers3

6

Smallest example: there exists a non-abelian group of order $27$, and of exponent 3 ($x^3=1$ for all $x \in G$), which can be constructed as all $3 \times 3$ upper diagonal matrices with 1's on the diagonal and entries in the field of 3 elements. Note that the proof of Arpit depends only on the injectivity of the power map.

Nicky Hekster
  • 52,147
  • Can you prove that there is no group with less elements and my desired property? – principal-ideal-domain Sep 17 '15 at 19:12
  • By "it depends only on the injectivity of the power map" you mean that each group in which $x\mapsto x^3$ is injective is Abelian? – principal-ideal-domain Sep 17 '15 at 19:15
  • @principle-ideal-domain: the smallest example with exponent $3$, must be sought among 3-groups, for if you allow other prime divisors of the order of $G$, then certainly the exponent is larger than $3$. However, a group of order $3$, or $9$ is abelian. So $27$ would be the first order of an eligible non-abelian group. To your second remark, yes that is almost exactly right: if $x \rightarrow x^3$ is an injective homomorphism of $G$ to itself then $G$ is abelian. – Nicky Hekster Sep 17 '15 at 20:27
3

Let $G$ be a non-Abelian group such that order of its element be $3$, for example see here or this, then obvious that $x\rightarrow x^3$ is a homomorphism. For complete of this type group see this. About extend of this problem: there is non-Abelian group such that order of its element be $n$; this completely related to the "Burnside's problem".

Community
  • 1
3

As suggested by @Nicky Hekster Consider the Heisenberg group of all matrices of the form $$\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right),$$ where $x,y,z\in\mathbb{Z}/3\mathbb{Z}$.This group has exponent $3$ i.e. for every $g \in G$ we have $g^3=e$. Just to make sure that $G$ is non abelian consider the following matrices and show that these matrices don't commute. $$\left(\begin{array}{ccc} 1 & 1 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1 \end{array}\right),$$ $$\left(\begin{array}{ccc} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{array}\right)$$

Arpit Kansal
  • 10,489