Find a cylic subgroup of order $6$ inside $S_{5}$ . Find all the abelian subgroups of $S_{5}$ .

Question

I am not sure how to approach this question. I know a subgroup of order six would be $(123)(45)$, but I'm not sure how this would be related to abelian groups. I imagine every transposition is abelian.

Answer 1

A subgroup of order six are isomorphic to $\mathbb{Z}_{6}$, or $S_{3}$, but this last one isn't of course cyclic (it isn't abelian).

If we want to find a subgroup isomorphic to $\mathbb{Z}_{6}$ we just need to find a generator, which means an element of order $6$. Since in $S_{5}$ can't exist $6$-cycles the only way we can build an element of order $6$ is as a product of a $3$ cycle and a trasposition, for the sake of the example $\langle \sigma \rangle$, where $\sigma = (1 2 3)(45)$ works. Notice that you can even count them all, just by doing cases.

As far as concern the abelian groups, we notice that the only abelian groups we can find $S_{5}$ are determined by the cyclic structure obtainable from the "digits" we are permuting, which are $1,2,3,4,5$.

So of course we have the identity, $\mathbb{Z}_{2}$ given by the traspositions or the doble trasposition, as for example $(12)(34)$, $\mathbb{Z}_{3}$ given only by $3$-cycles, $\mathbb{Z}_{4}$ given by $4$-cycles or $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$, given by the subgroups generated by disjoint tranpositions; $\mathbb{Z}_{5}$ generated by $5$-cycles and in the end as we already saw $\mathbb{Z}_{6}$.

We can stop here since an abelian group of higher order would be : cyclic or not cyclic.

In the first case we would need an element of order stricly greater that $6$, which is not possibile since the greatest order you can build with just $5$ digits is $6$. Analogously in the second case you see there is no "space" to create a different abelian subgroup. Think for example to the minimum costruction you would like to have, for example $(\mathbb{Z}_{2})^{3}$, can you see why you can't find it ? The reasoning can be mimed to exclude the others abelian groups.