# Elementary approach to proving that a group of order 9 is Abelian

The trick here is to provide an elementary solution; I'll explain what I mean.

Prove that a group of order 9 must be Abelian. The standard approach is to use the class equation to show that any $p$-group has a non-trivial center. From that, it's easy to show that any group of order $p^2$ is Abelian. If not, pick an element $a$ not in the center and look at its centralizer. This includes the center and $a$, so it has at least $p+1$ elements; hence it's the whole group by Lagrange; hence $a$ is in the center, a contradiction.

Okay, but the problem is given as an exercise very early in Herstein, before any of this is discussed. Other than Lagrange and some easy consequences, all we really have to work with is the fundamental homomorphism theorem; in fact, the exercise is included in the set at the end of the section introducing the theorem. So what I'm looking for is an approach that uses only very basic group properties and, especially, one that applies the homomorphism theorem.

Let $G$ be a group of order $9$. If there's an element of order $9$, the group is cyclic, so assume all nonidentity elements have order $3$.
Let $x$ be a nonidentity element and let $y\in G-\langle x\rangle$. Then the elements of $G$ are $x^iy^j$, where $i, j\in \left\{0, 1, 2\right\}$. We are done if we show $xy=yx$. Consider $yx$. It must equal one of the $x^iy^j$. By cancellation, both $i$ and $j$ must be nonzero.
If $yx=x^2y$, then $yxy^{-1}=x^2$. This means that $y^3xy^{-3}=x^8=x^2\neq x$, a contradiction since $y^3=1$. Similarly, $yx\neq xy^2$. Now, if $yx=x^2y^2$, then $yx=x^{-1}y^{-1}=(yx)^{-1}$, which means $yx$ has order $1$ or $2$, clearly impossible.
• I think essentially the same argument works with $p^2$ in place of $9$ for any prime $p$. If $yxy^{-1}=x^i$ then $x = y^pxy^{-p} = x^{i^p} = x^i$, since $i^p \equiv i \bmod p$. May 14, 2014 at 22:18
• @DerekHolt I don't see why $yxy^{-1} = x^{2}\implies y^{3}xy^{-3} = x^{8}.$ Shouldn't it be $y^{3}xy^{-3} = x^{6}?$ Nov 6, 2019 at 17:31
• Each conjugation by $y$ squares the group element it applies to, so conjugating thrice gives $x\mapsto x^2\mapsto x^4\mapsto x^8$. Nov 7, 2019 at 18:09