Analyzing Cyclic Subgroups in S6

[Kate2010]

Homework Statement

I have to determine whether some groups are cyclic. The first is the subgroup of S₆ generated by (1 2 3)(4 5 6) and (1 2)(2 5)(3 6)

Homework Equations

Lagrange's Theorem?

The Attempt at a Solution

I don't really know how to tackle this problem. I have only posted (i) as I'm hoping that after I get to grips with how to go about answering it, I will be able to figure the rest out on my own. The question advises to use general facts rather than elaborate calculations.

I think o[(1 2 3)(4 5 6)] = 3 and o[(1 2)(3 4)(5 6)] = 2, therefore would the order of the subgroup generated by them be 6? I may have completely made that up. I don't think I even know what it means by the subgroup generated by them.

Also, I'm unsure of what general facts about being cyclic I'm meant to be using.

Sorry about all the confusion in this post. I'm currently feeling very out of my depth with this module. Thanks :)

 

[eok20]

You say the subgroup is generated by (1 2 3)(4 5 6) and (1 2)(2 5)(3 6) but then you compute the order (correctly) of (1 2)(3 4)(5 6), which has a different order than (1 2)(2 5)(3 6).

Some general thoughts: an element of order 2 and an element of order 3 may not generate a group of order 6. However, by Lagrange's theorem the group they generate will have to be divisible by 2 and 3 (and therefore 6). One thing you know about cyclic groups is that they are abelian. So if the generators don't commute then the group can't be cyclic. The only other way I can think of is to find all the elements of the group and see if there is one that has the same order as the group (which would imply that that element generates the group, i.e. the group is cyclic).

 

[Kate2010]

Thanks, so I think that group is non-abelian hence non-cyclic.

If a group (again a subset of S6) is generated by (1 2 3) and (4 5 6), then I think it contains e, (1 2 3), (1 3 2), (4 5 6), (4 6 5)

Does it also contain, for example, (1 2 3)(4 5 6) and the other compositions of the above cycles that are disjoint? If so, then there would be an extra 4 elements, the order of G would be 9, yet the order of all elements is either 3 or 6, so it's non-cyclic?

 

[eok20]

If a group (again a subset of S6) is generated by (1 2 3) and (4 5 6), then I think it contains e, (1 2 3), (1 3 2), (4 5 6), (4 6 5)

Does it also contain, for example, (1 2 3)(4 5 6) and the other compositions of the above cycles that are disjoint? If so, then there would be an extra 4 elements, the order of G would be 9, yet the order of all elements is either 3 or 6, so it's non-cyclic?

Yes, it also must contain elements like (1 2 3)(4 5 6) since it has to be closed under the operation.