Is $G$ isomorphic to the dihedral group $D_{10}$?

Question

Let $G\le S_6$ be the subgroup generated by the permutations $\sigma=(12356)$ and $\tau=(26)(35)$.

I'm asked to determine: (a) the order of $G$ and the period of each element; (b) if $G$ is isomorphic to a dihedral group.

By calculation I've found $G=\{\operatorname{Id}, (12)(36), (13)(56), (15)(23), (16)(25), (26)(35), (12356), (13625), (15263), (16532) \}$ so $|G|=10$ and the period of each element is clear. Could I have done this part without finding $G$ explicitly? In particular, could I have deduced the order from the lengths of $\sigma$ and $\tau$? I think then I could complete this part: then Cauchy's theorem would imply the existence of at least an element of period $2$ and one of period $5$, and since for example there must be elements whose decomposition contains a 2-cycle that sends $1$ to each one of the other elements and itself, that's 5 elements with period $2$; with an analogous reasoning I think one can conclude there must be 4 elements with period $5$.

For part (b), I noticed that $|D_{10}|=10=|G|,$ but how can I prove that they are isomorphic?

Answer 1

An easy computation shows that $\tau\sigma\tau=\sigma^{-1}$. Since the orders of $\tau$ and $\sigma$ are $2$ and $5$ respectively, this shows that $G$ is a quotient of $D_{10}$. On the other hand, since $G$ has an element of order $2$ and an element of order $5$, its order must be at least $10$. Therefore, $G$ and $D_{10}$ are isomorphic.

Answer 2

Just for part (b): If you already know that $|G|=10$, then you can forget everything else you know about $G$! Instead, apply the principle that there isn't much diversity among groups of semiprime order. In this case, $|G|=10$ means that $G$ is either dihedral or cyclic. And $G$ can't be cyclic because there is no element of order $10$ in $S_6$.