Sylow subgroups of S4

1. Mar 26, 2012

will8655

I'm reading some notes concerning Sylow subgroups of S4:

Since #S4 = 24 = 3 . 2^3 from Sylow's Theorems we know there are either 1 or 4 Sylow 3-subgroups and either 1 or 3 Sylow 2-subgroups.

The question I have regards how we narrow this down:

My notes argue that since S4 has 9 elements of order 2, they cannot all fit in a subgroup of order 8 and so we must have 3 Sylow 2-subgroups. Similarly we must have 4 Sylow 3-subgroups.

I don't quite follow this reasoning, why must all elements of order 2 lie in a Sylow 2-subgroup? For instance, why couldn't we have some of them lying in a subgroup of order 6?

I know you can argue that if there is only one Sylow subgroup then it must be normal and hence a union of conjugacy classes and get a contradiction that way, but I'd like to understand the logic behind the above reasoning.

Thanks
Will

2. Mar 27, 2012

morphism

The given argument is simply using the fact that an element of order 2 must belong to a Sylow 2-subgroup. (Because such an element generates a 2-group, which must be contained in a maximal 2-group, i.e. a Sylow 2-subgroup.) It could still belong to a subgroup of order 6, of course.

3. Mar 29, 2012

will8655

Ah I see. Thanks for clearing that up.