The order of the group $\langle a, b| w^3, w\in\langle a, b\rangle\rangle$ for $w$ being any word.

Question

I came across this group mentioned in passing as finite. Does anyone know the order of the group, and where I can find a proof of this quantity? Replacing $3$ with $n$, does this problem have a name or solution out there? All I was able to do was write identities for $a$ and $b$ for hours, so any help or information will be appreciated.

I suspect that this quantity is nine, but after hours of calculation, I probably made a mistake somewhere.

Answer 1

Prerequisite:

$$\begin{array}{rcl} (a^mb^n)^3 &=& e \\ a^mb^na^mb^na^mb^n &=& e \\ b^na^mb^n &=& a^{3-m}b^{3-n}a^{3-m} \end{array}$$

Then we fill the following table row by row from top to bottom, each row from left to right, being mindful to collisions described above, and creating new rows as we need.

$$\begin{array}{c|c|c} w & w & wa &wa^2& wb &wb^2\\\hline e & 1 & 2 & 3 & 4 & 5 \\\hline a & 2 & 3 & 1 & 6 & 7 \\\hline a^2 & 3 & 1 & 2 & 8 & 9 \\\hline b & 4 & 10 & 11 & 5 & 1 \\\hline b^2 & 5 & 12 & 13 & 1 & 4 \\\hline ab & 6 & 14 & 15 & 7 & 2 \\\hline ab^2 & 7 & 16 & 17 & 2 & 6 \\\hline a^2b & 8 & 18 & 19 & 9 & 2 \\\hline a^2b^2 & 9 & 20 & 21 & 2 & 8 \\\hline ba & 10 & 11 & 4 & 21 & 22 \\\hline ba^2 & 11 & 4 & 10 & 16 & 23 \\\hline b^2a & 12 & 13 & 5 & 24 & 19 \\\hline b^2a^2 & 13 & 5 & 12 & 25 & 14 \\\hline aba & 14 & 15 & 6 & 13 & 25 \\\hline aba^2 & 15 & 6 & 14 & 20 & 26 \\\hline ab^2a & 16 & 17 & 7 & 23 & 11 \\\hline ab^2a^2 & 17 & 7 & 16 & 27 & 18 \\\hline a^2ba & 18 & 19 & 8 & 17 & 27 \\\hline a^2ba^2 & 19 & 8 & 18 & 12 & 24 \\\hline a^2b^2a & 20 & 21 & 9 & 26 & 15 \\\hline a^2b^2a^2 & 21 & 9 & 20 & 22 & 10 \\\hline bab^2 & 22 & 25 & 27 & 10 & 21 \\\hline ba^2b^2 & 23 & 26 & 24 & 11 & 16 \\\hline b^2ab & 24 & 23 & 26 & 19 & 12 \\\hline b^2a^2b & 25 & 27 & 22 & 14 & 13 \\\hline aba^2b^2 & 26 & 24 & 23 & 15 & 20 \\\hline ab^2a^2b & 27 & 22 & 25 & 18 & 17 \end{array}$$

Some collisions that may require more explanation (I assure you that countless hours are spent by me trying to match them):

• $[22]a^2=[27]$ ($bab^2a^2=ab^2a^2b$): $$\begin{array}{rcl} ab^2ab^2ab^2 &=& e \\ ab^2a^2(a^2b^2a^2)a^2b^2 &=& e \\ ab^2a^2(bab)a^2b^2 &=& e \\ (ab^2a^2b)(aba^2b^2) &=& e \\ ab^2a^2b &=& bab^2a^2 \end{array}$$
• $[23]a=[26]$ ($ba^2b^2a=aba^2b^2$): $$\begin{array}{rcl} ab^2ab^2ab^2 &=& e \\ ab(bab)bab^2 &=& e \\ ab(a^2b^2a^2)bab^2 &=& e \\ (aba^2b^2)(a^2bab^2) &=& e \\ aba^2b^2 &=& ba^2b^2a \end{array}$$

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Links to other more group-theoretic and less brute-force methods: $B(2,3) \cong UT(3,3)$ on math overflow, same question on math.SE.