Determine if exists a subgroup of order $3$ of $H=\langle\sigma^{8440}\rangle$

Question

Consider the following permutation of $S_{13}$.

$\sigma=(1\;3\;13\;5\;11\;8)(2\;10\;4\;6\;12\;7\;9)$

Determine if exists a subgroup of order 3 of $H=\langle\sigma^{8440}\rangle$. If yes, exhibit it if no say why.

My attempt by using some properties of cyclic groups.

First of all, I calculated the order of $\sigma$ which is $ \operatorname{lcm}(6,7)=42$. Then the order of $\sigma^{8440}$ which is $\frac{42}{\gcd(8440,42)}$. Since $\gcd(8440,42) = 2$ then $|H|=o(\sigma^{8440}) = 21$ (since we are working with cyclic groups). Finally, since $3\mid21$, there exists a subgroup of $H$ of order $3$. I call it $G$ and it is generated by $\langle\sigma^{\frac{21}{3}}\rangle=\langle\sigma^{7}\rangle$.

Is my attempt correct?

Answer 1

You make a mistake at the very end of your argument; $\sigma^7$ is not an element of $H$. Because $8440\equiv40\pmod{42}$ you have $$H=\langle\sigma^{8440}\rangle=\langle\sigma^{40}\rangle=\langle\sigma^{-2}\rangle=\langle\sigma^2\rangle.$$ The subgroup $H\subset\langle\sigma\rangle$ is the subgroup of even powers of $\sigma$. I'll leave it to you (for now) to determine the subgroup of order $3$.

Besides this, you are entirely correct. Depending on how complete you want to be, you could

• note that $\sigma$ is given as a product of disjoint cycles, and hence its order is $\operatorname{lcm}(6,7)$.
• make explicit what the elements of the subgroup of order $3$ are, i.e. write out the appropriate powers of $\sigma$ in cycle notation.