I have this permutation in $S_9$:
$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 8 & 7 & 6 & 9 & 4 & 2 & 3 & 5\end{pmatrix}$
I have obtained it as product of disjoint cyclics:
$(2837)(46)(59)$
I have found that each of them generate a subgroup:
$\left \langle (2837) \right \rangle = \left \{ (2837), (23)(78), (2738), (1) \right \} \\ \left \langle (46) \right \rangle = \left \{ (46), (1) \right \} \\ \left \langle (59) \right \rangle = \left \{ (59), (1) \right \}$
if it is correct, the order of $(2837)(46)(59)$ would be the $lcm$ between the orders of the subgroups of the above, hence $lcm(4,2,2) = 4$.
What I ask is: how can I obtain all the elements in $\left \langle (2837)(46)(59) \right \rangle$, once I have already been calculated the elements in the subgroups of the above?
I have tried to make a product between the elements $\left \langle (2837) \right \rangle \times \left \langle(46) \right \rangle \times \left \langle(59) \right \rangle = \left \{ (59)(46)(2837), (59)(46)(23)(78), \ldots \right \}$,
but it seems that it is not the corrrect way because I obtain more than $4$ elements of the order of $\left \langle (2837)(46)(59) \right \rangle$ calculated above through $lcm$.
Please, can you tell me what can I do? Thanks!
Best Answer
Using the $\times$ sign in group theory means a direct product, which is by no means related to the thing you're talking about. I guess you meant to write:
$$\langle (2837) \rangle\langle (46) \rangle\langle (59) \rangle = \{abc |a \in \langle (2837) \rangle, b \in \langle (46) \rangle, c \in \langle (59) \rangle\}$$
Unfortunately this isn't the way to go. In $\langle (2837)(46)(59) \rangle$ the elements are of the type $(2837)^k(46)^k(59)^k$, as cycles commute. On the otherside the elements in $\langle (2837) \rangle\langle (46) \rangle\langle (59) \rangle$ are of the type $(2837)^k(46)^m(59)^n$, where $k,m,n$ are not necessarily the same.
Anyway the best way to go is to calculate this by hand. In fact:
$$\langle(2837)(46)(59)\rangle = \{(2837)^0(46)^0(59)^0,(2837)^1(46)^1(59)^1, (2837)^2(46)^2(59)^2, (2837)^3(46)^3(59)^3\}$$
Now it's just a matter of calculation.