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I am trying to describe the Sylow $2$-subgroups of an arbitrary dihedral group $D_n$ of order $2n$.

In the case that $n$ is odd, $2$ is the highest power dividing $2n$, so that all Sylow $2$-subgroups have order $2$, and it is fairly easy to describe them.

However, if $n$ isn't odd, we may factor a power of $2$ out and write $|D_n|=2^{k}m$ for some odd integer $m$.

There is a proof that there exist precisely $m$ Sylow $2$-subgroups, but it does not provide an explicit description of such subgroups in the case $n$ is odd.

Additionally, someone has asked a similar question in the past, and they claim to give a description of the Sylow $2$-subroups in the case $n$ is odd, but I can not find a source or an explanation. The question is here.

Can anyone provide a description of how one may determine precisely the Sylow $2$-subgroups for the case $n$ is odd?

Thank you.

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KyleP
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1 Answers1

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This simple lemma may be useful:

Intersection of a normal subgroup $N$ with Sylow-$p$ subgroup of $G$ is Sylow-$p$ subgroup of $N$.

In dihedral group of order $2n$, there is a natural normal subgroup - cyclic group of order $n$, consisting of rotations. In cyclic groups, its much easier and trivial to say what could be the Sylow-$2$ subgroup of it? Let $H$ be the Sylow-$2$ subgroup of this cyclic subgroup. Let $t$ be any element of order $2$ which is outside cyclic subgroup of order $n$; such element is reflection. then $\langle H,t\rangle =H\cup Ht$ is a Sylow-$2$ subgroup of dihedral group.

Why "a"? For example, consider dihedral group of order $12$. It has a cyclic subgroup of order $6$; what is Sylow-$2$ subgroup in that? It is $H\cong \mathbb{Z}_2$. To this $H$ we attach an element of order $2$ outside cyclic group oforder $6$, say $t$. Then $H\cup Ht$ is Sylow-$2$ subgroup. In this Sylow-$2$ subgroup, the elements of $Ht$ are reflections; there are $|Ht|=|H|=2$ reflections in one Sylow-$2$ subgroup. In $G$ there are $6$ reflections, and in Sylow-$2$ subgroup, there come $2$ reflections. So we should have at least three Sylow-$2$ subgroups; and in fact, we have exactly $3$ Sylow-$2$ subgroups here.

Riccardo
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Groups
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  • Why is $\left \langle H,t \right \rangle = H \cup Ht\ ?$ It is clear that $H \cup Ht \subseteq \left \langle H,t \right \rangle.$ So if it can be shown that $H \cup Ht$ is a group then we are through. But why is it actually the case? Also I don't understand where you have applied the lemma you have stated at the very beginning of your proof. – Anacardium Oct 08 '20 at 06:07