Let $H$ be a group generated by 2 elements $a,b$ that commute ($ab=ba$) and where each has order $3$ or $1$. Show $H$ has at most order 9.

Question

Edit re the close as duplicate: I think it should be the other way around. the not order 6 seems to be answered with the answer for at most order 9. i propose this be reopened and the other closed.

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Let $H$ be a group with identity $1_H$ that is generated by 2 elements $a,b$ that commute ($ab=ba$) and where each has order $3$ or $1$. In symbols (I hope I translated correctly):

$$H=\langle a,b\rangle \ \text{, where} \ a^3=b^3=1_H=a^{-1}b^{-1}ab$$

I'm supposed to show $H$ has at most order $9$. This is easy to do when we don't have all of $\{a,b,1_H\}$ to be distinct. As for all distinct, I get the $3 \times 3$ multiplication table to give other elements $a^2,b,ba,ba^2,b^2,b^2a,b^2a^2$. Then I can construct a $9 \times 9$ multiplication table to show that we don't get any new elements. Is this a shorter way to do this?

• (Similar problems I've had recently: Given 2 elements $x,y \in S_4$, is there any particular trick to determining the subgroup generated by them? and How do I show $A_4$ is the only subgroup of order 12 of $S_4$ ? (Do not use Lagrange, index, cosets, normal subgroups, quotient groups))

Answer 1

Every element of $H$ is a finite product of $a,b$ and $a^{-1}$ and $b^{-1}$. Fortunately $a^{-1}$ is $a^2$ and $b^{-1}$ is $b^2$. It follows every element of $H$ is a finite product of $a$'s and $b$'s. Since $a$ and $b$ commute we can show by induction over the number of factors that every expression is of the form $a^nb^m$. To do this take an expression $(x_1x_2\dots x_s)x_{s+1}$ and see it is equal to $(a^nb^m)x_{s+1}$ and do the case when $x_{s+1}$ is $a$ or $b$.

Finally since every $a^n$ is equal to one of $1,a,a^2$ and every $b^m$ is equal to one of $1,b,b^2$ you will be done.

Answer 2

Since, by definition of a presentation, the presentation

$$P=\langle a,b\mid a^3,b^3, ab=ba\rangle$$

defines a group that maps onto $H$, and that group defined by $P$ is $\Bbb Z_3\times \Bbb Z_3$, we must have that $\lvert H\rvert\le 9$.