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A group $G$ is generated by two elements $a$ and $b$ such that for any $g\in G:$ $g^3=e$. Show that $G$ is finite.

I don't understand how this is so. I don't know if $G$ is abelian, so I can construct more and more elements of the group of the form $abababababababab...$ which can't be simplified. So how do I know $G$ is finite?

(If it was abelian, I could simply list all the elements $e,a,a^2,b,b^2,ab,a^2b,ab^2,a^2b^2$.)

Leo 254
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    You can't use what you say : $ab$ is of order $3$, so $ababab = e$. – Kevin Quirin Oct 01 '15 at 14:33
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    Since $a^2 = a^{-1}$ and $b^2 = b^{-1}$ you can just consider words of the form $x_1 y_1 x_2 y_2 \ldots$ where each $x_i$ is either $a$ or $a^{-1}$ and each $y_i$ is either $b$ or $b^{-1}$ (or vice versa). And note each such word has order 3. – user2566092 Oct 01 '15 at 14:57
  • FWIW: You mention parenthetically that this would be easy if the group were abelian; although this is not the case for the question at hand, I thought I'd mention that in a ring for which every element $r$ has the property that $r^3 = r$, it is the case that the ring must be commutative. This is a classic puzzle/exercise; you can find a write-up in MSE 67148. – Benjamin Dickman Oct 02 '15 at 01:28

1 Answers1

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Every element of $G$ can be written as product of zero or more factors $\in\{a,b,a^{-1},b^{-1}\}$, i.e., for $g\in G$ there exist $n\in\mathbb N_0$ and $x_i\in\{a,b,a^{-1},b^{-1}\}$, $1\le i\le n$, such that $g= x_1x_2\ldots x_n$. For $g\in G$ let $l(g)$ be the smallest $n$ that can be used.

Fix $g\in G$ and let $g=x_1x_2\ldots x_n$ with $n=l(g)$, $x_i\in\{a,b,a^{-1},b^{-1}\}$.

Lemma 1. If $1\le k<n-2$ then $x_{k+2}=x_k^{\pm1}$.

Proof. As $aa$ can be shortend to $a^{-1}$ and so on, we know that $x_{k+1}$ "uses the other letter" than $x_k$, hence $x_{k+2}$ "uses the same letter" again as $x_k$ and hence can be either the same as or the inverse of $x_k$. $_\square$

Lemma 2. If $1\le k\le n-3$ then $x_k=x_{k+2}^{-1}$. If $2\le k\le n-2$ then $x_k=x_{k+2}^{-1}$.

Proof. Otherwise, with $u=x_k$, $v=x_{k+1}$ we have one of the following for $x_kx_{k+1}x_{k+2}x_{k+3}$: $$\underbrace{uvuv}_4=(uv)^2=(uv)^{-1}=\underbrace{v^{-1}u^{-1}}_2$$ or $$\underbrace{uvuv^{-1}}_4=uvuv\cdot v=\underbrace{v^{-1}u^{-1}v}_3$$ contradicting minimality of $n$. The second claim follows by symmetry / considering the reverse word. $_\square$

This allows us to estimate the number of elements in $G$ with given length:

  • There is one element, $e$, of length $0$
  • There are up to four elements of length $1$
  • There are up to $8$ elements of length $2$: Choose one of four first symbols, then choose one of two symbols with "the other letter" as second.
  • There are up to $16$ elements of length $3$: Continuing from above, there are two choices for the third symbol (by lemma 1)
  • If $n>3$ then there are at most $8$ elements of length $n$: Four choices for the first symbol, two for the second, then $x_3,\ldots, x_{n-1}$ are determined by lemma 2 (first sentence) and $x_n$ by lemma 2 (second sentence).

Especially, elements of length $7$ are of the form $$\begin{align}\underbrace{xyx^{-1}y^{-1}xyx}_7 &=(xyx^{-1}y^{-1})^2yx^2\\ &=(xyx^{-1}y^{-1})^{-1}yx^{-1}\\ &=\underbrace{yxy^{-1}x^{-1}yx^{-1}}_6\end{align}$$ and hence cannot occur; nor can greater lengths. We conclude $$ |G|\le 1+4+8+16+8+8+8=53.$$ As $|G|$ must be a power of $3$ (Sylow), we conclude $$ |G|\le 27.$$

Hagen von Eitzen
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