Understanding Sylow Subgroups of S4: The Logic Behind Grouping Elements

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The discussion centers on the Sylow subgroups of S4, specifically the number of Sylow 2-subgroups and Sylow 3-subgroups. It is established that S4 has 9 elements of order 2, which cannot all fit into a single subgroup of order 8, leading to the conclusion that there must be 3 Sylow 2-subgroups. The reasoning is clarified that each element of order 2 must belong to a Sylow 2-subgroup, as they generate 2-groups contained within maximal 2-groups. The possibility of these elements belonging to a subgroup of order 6 is acknowledged, but the focus remains on the necessity of their inclusion in Sylow 2-subgroups. Understanding this logic is essential for grasping the structure of S4's subgroups.
will8655
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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
 
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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.
 
Ah I see. Thanks for clearing that up.
 
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