Let $S_6$ be the symmetric group of a set of six elements. Let $\sigma=(1,3,4,6)$ and $\tau=(1,3,4,6)(2,5)$ be two cycles in $S_6$.
(A) Determine the order of the two permutations and the order of $G:=\langle\sigma,\tau\rangle$.
(B) Determine all the non trivial possible decompositions of $G$ in a direct product of its subgroups.
I wrote that $o(\sigma)=4$ and $o(\tau)=lcm(4,2)=4$. Since $(1,3,4,6)\cdot(1,3,4,6)\cdot(2,5)=(1,3,4,6)^2\cdot(2,5)=$
$(1,3,4,6)\cdot(2,5)\cdot(1,3,4,6)\implies$ $\sigma$ and $\tau$ are permutable, hence $\langle\sigma\rangle\langle\tau\rangle$ has elements of form $\sigma^h\tau^k$.
This means that $\langle\sigma\rangle\langle\tau\rangle=\langle\sigma,\tau\rangle\le G$ (I think that in general $\langle\sigma,\tau\rangle$ has the same form of $\langle\langle\sigma\rangle\langle\tau\rangle\rangle$).
$$|\langle\sigma,\tau\rangle|=\dfrac{|\langle\sigma\rangle|\cdot|\langle\tau\rangle|}{|\langle\tau\rangle\cap\langle\sigma\rangle|}=\dfrac{4\cdot4}{2}=8.$$
I state this because $\langle\sigma\rangle=\{id,(1,3,4,6),(1,4)(3,6),(1,6,4,3)\}$ and $\langle\tau\rangle=\{id,(1,3,4,6)(2,5),(1,4)(3,6),(1,6,4,3)(2,5)\}\implies$ the intersection contains $id$ and $(1,4)(3,6).$
I think that in order to classify the type of decomposition of $G$ I have to look at the normal subgroups of $G$ whose intersection is trivial but I always find an element in the intersection… What did I miss?
Best Answer
The subgroup generated by $\tau$ is cyclic of order $4$; the subgroup generated by $\sigma$ is cyclic of order $4$; you correctly calculate (since $\tau$ and $\sigma$ commute) that $\langle \sigma,\tau\rangle=\langle\sigma\rangle\langle\tau\rangle$, and that the order is $8$.
This is an abelian group. As such, it must be (abstractly) isomorphic to either $C_8$, $C_4\times C_2$, or $C_2\times C_2\times C_2$. The latter has no subgroups of order $4$, so that can't be it. The first has a unique subgroup of order $4$, so that can't be it, either. So that means that $G$ is abstractly isomorphic to $C_4\times C_2$. As such, it definitely has direct product decompositions.
Being abelian, every subgroup is normal. So you are just looking for subgroups with trivial intersection, one of order $4$ and one of order $2$ (those are really the only possibilities, because of the structure theorem for finite groups).
For instance, can we find a $2$-element subgroup $K$ such that $G=\langle \tau\rangle \times K$? Note that $(2,5)=\tau\sigma^{-1}\in G$. What is the intersection of $\langle (2,5)\rangle$ with $\langle \tau\rangle$? With $\langle \sigma\rangle$? Can we find other subgroups of order $4$ and decompositions?
If you find it difficult to it directly with the permutations, find an explicit isomorphism to $C_4\times C_2$ and work in the latter, then translating back to $G$.