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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    $\begingroup$ You are correct. There is a mistake/typo somewhere in the notes. $D_4$ (the dehidral group of order $8$ = symmetries of a square) has only 1 Sylow subgroup, namely the group itself. $D_4$ is its own Sylow 2-subgroup. :) $\endgroup$ – Bill Cook Mar 26 '12 at 13:06
  • $\begingroup$ It could be the dihedral group of order $2^n3$ for any $n\geq 2$, as in each case $4\equiv 1\text{ mod }3$ and $4\mid 2^n$. So...he could be talking about $D_{6}$...but these groups only have one copy of $C_3$...($(a^ib)^3=a^ibb^{-1}a^ia^ib=a^{3i}b\neq 1$ so our only choice for an element or order $3$ is $a^{\pm i}$, $i=2^{n-1}$). $\endgroup$ – user1729 Mar 26 '12 at 13:26
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    $\begingroup$ @bemyguest: For alternating group of degree four the statement is true, however.. maybe that's what he meant. $\endgroup$ – spin Mar 26 '12 at 14:59

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.


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