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.


1 Answer 1


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.

  • 1
    $\begingroup$ 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 $i^p \equiv i \bmod p$. $\endgroup$
    – Derek Holt
    May 14, 2014 at 22:18
  • $\begingroup$ Right, but there are also the cases where both exponents are greater than 1. $\endgroup$
    – Nishant
    May 15, 2014 at 1:36
  • 1
    $\begingroup$ I had to think for a moment about the sentence beginning "This means..." But the argument is clear and exactly at the level I was looking for. $\endgroup$ May 16, 2014 at 0:17
  • $\begingroup$ @DerekHolt I don't see why $yxy^{-1} = x^{2}\implies y^{3}xy^{-3} = x^{8}.$ Shouldn't it be $y^{3}xy^{-3} = x^{6}?$ $\endgroup$
    – Student
    Nov 6, 2019 at 17:31
  • 1
    $\begingroup$ Each conjugation by $y$ squares the group element it applies to, so conjugating thrice gives $x\mapsto x^2\mapsto x^4\mapsto x^8$. $\endgroup$
    – Nishant
    Nov 7, 2019 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.