I'm reading some notes concerning Sylow subgroups of S4:(adsbygoogle = window.adsbygoogle || []).push({});

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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# Sylow subgroups of S4

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