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My professor has on our intranet uploaded some of his handwritten notes. I am a little worried about one of his statements and I'm suspecting that it contains a mistake.

"The dihedral group of degree 4 contains 4 different sylow 3-subgroups"

How is that possible?

The dihedral group of degree 4 ($D_{4}$) has order 8 and therefore every subgroup of $D_{4}$ need to have an order that divides 8 (according to lagrange) but a Sylow 3-subgroups has the order $3^{n}$ for some $1\leq n$, which makes the above statement false?

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    @bemyguest: For _alternating_ group of degree four the statement is true, however.. maybe that's what he meant.2012-03-26

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The statement is nonsense and has several mistakes in it.

Consider the dihedral group $D_{2n}$, where $3$ divides $n$. Now the group has a normal cyclic subgroup $H$ of order $n$ by definition. The subgroups of a cyclic normal subgroups are also normal, and thus the $3$-Sylow contained in $H$ is normal. But Sylow subgroups are conjugate so there can be only one $3$-Sylow. In particular there cannot be four. We can also generalize this a bit: any $p$-Sylow with $p \neq 2$ is normal in a dihedral group.

Thus the statement should talk about $2$-Sylows. But we cannot have four $2$-sylows, because $4 \not\equiv 1 \mod 2$. Also the dihedral group of degree $4$ also contains only one Sylow subgroup, which is the group itself.

My guess is that he meant the alternating group of degree four, which does have four $3$-Sylow subgroups.