Determine order of $\langle(15)(37),(23)(45)(67)\rangle\subset S_7$

Question

Say we have $\sigma,\tau\in S_7$, where $\sigma=(15)(37)$ and $\tau=(23)(45)(67)$. Then $\operatorname{ord}(\sigma\tau)=12$. Consider $\langle\sigma,\tau\rangle$. I've shown that each element in $\langle\sigma,\tau\rangle$ can be written as a product of alternating $\tau$ and $\sigma$, using the fact that $\operatorname{ord}(\sigma)=\operatorname{ord}(\tau)=2$. I'm now asked to determine the order of $\langle\sigma,\tau\rangle$.

Since $\sigma\tau$ has order 12, we know there are at least 12 elements; $( \sigma\tau)^k$. We have another 12 elements, since $\sigma(\sigma\tau)^k$ cannot equal any of the $(\sigma\tau)^k$, because the parities are different. Now I'm unsure about $(\sigma\tau)^k\tau$. How can I prove (or disprove) that we can't have $(\sigma\tau)^k\tau=\sigma(\sigma\tau)^l$, for some $k,l\in\mathbb Z$? This would yield: \begin{align} (\sigma\tau)^k\tau&=\sigma(\sigma\tau)^l\\ \tau&=(\sigma\tau)^{12-k}\sigma(\sigma\tau)^{l-k}=(\sigma\tau)^{12-k-(l-k)+1}\sigma=(\sigma\tau)^{13-l}\sigma=\sigma(\tau\sigma)^{13-l}. \end{align} Now I don't really know if it proves or disproves anything. Any help?

Answer 1

I would approach this as follows.

• Indeed, $\beta:=\sigma\tau=(154)(2763)$ is of order $12$.
• But, observe the placing of $1$ and $5$ (resp. $3$ and $7$) in the two disjoint cycles of $\sigma\tau$. This is something that you may recall from dealing with $S_3$ as well as the dihedral group of order eight $D_4\le S_4$. To take advantage of this prove the identity $$\sigma\beta\sigma^{-1}=\beta^{-1}.$$
• So now you should be able to prove that $\sigma$ and $\beta$ generate a dihedral group $D_{12}$ of order $24$.
• Show that $\langle\sigma,\beta\rangle=\langle\sigma,\tau\rangle$.