# Normalizer of a Sylow 2-subgroups of dihedral groups

I can't solve the following exercise which is the last exercise in page 146 of Dummi & Foote's Abstract Algebra:

Let $2n=2^ak$ where $k$ is odd. Prove that the number of Sylow 2-subgroups of $D_{2n}$ is $k$. [Prove that if $P\in Syl_2(D_{2n})$ then $N_{D_{2n}}(P)=P.$]

Here $D_{2n}=<r,s\mid r^n=s^2=1,\,rs=sr^{-1}>$ is the dihedral group of order $2n$ and $Syl_2(D_{2n})$ is the set of all Sylow 2-subgroups of $D_{2n}$.

It is easy to solve the exercise if we prove that $N_{D_{2n}}(P)=P$ and that's what I can't prove. Could you give me some hints?

My attempt

I got the following results. They might be useless, but who knows?

1) If $a=1$ then Sylow $2$-subgroups of $D_{2n}$ have order $2$ and so their number is the number of elements of $D_{2n}$ of order $2$ which is $n$ because the set of elements of order $2$ of $D_{2n}$ is $\{sr^i:i\in\{0,\,\dots,\,n-1\}\}$. So we can assume that $a>1$.

2) Let $P\in Syl_2(D_{2n})$ and suppose $a>1$ (from 1). Then $P$ must have an element of order $2$ in $\{sr^i:i\in\{0,\,\dots,\,n-1\}\}$ . In fact, we know that all cyclic subgroups of order larger than $2$ of a dihedral group are generated by a power of $r$, so if $P$ is cyclic then $\exists m\in\mathbb{Z}^+,\,P=<r^m>$ and $m$ is a divisor of $n$. Then we have:$|r^m|=\frac{2^{a-1}k}{m}=2^a$ and so $k=2m$ which is false because $k$ is odd. Thus $P$ isn't cyclic and so $P$ is generated by at least two elements ($P$ is finite so it has a finite number of generators and let $S$ be a set of generators of $P$ so that $P=<S>$). If all elements of $S$ have order larger than $2$ they must be powers of $r$ and so $S\subset <R>$ thus $P\le <r>$ which is impossible, otherwise $P$ would be cyclic. Thus at least one element in $S$ has order $2$. But we know that there exists a unique element of $<r>$ which order is $2$. Thus we have two possible cases:

• Case 1: There is an element of order larger than $2$ in $S$.

This element must be a power of $r$. If all other elements of $S$ are powers of $r$, then $P\le <r>$ and we get the same contradiction as before.

• Case 2: All elements of $S$ have order $2$.

There are at least two distinct ones otherwise $P$ would be cyclic. They can't both be powers of $r$ since only one power of $r$ has order $2$ and so one isn't a power of $2$.

• Conclusion: In both cases, there's at least one element of $S$ which order is $2$ and isn't a power of $r$, so $\{sr^i:i\in\{0,\,\dots,\,n-1\}\}\cap P\neq\emptyset$

So here's all I the informations I could get (that might be useless and the exercise might not need any of them ^^). Could you please give me some hints?

• There are $n=2^{a-1}k$ elements of $D_{2n}$ lying outside of the cyclic subgroup $\langle r \rangle$, all of order $2$. (These are usually called reflections.) Each such reflection is contained in at least one Sylow $2$-subgroup. A Sylow $2$-subgroup has order $2^a$ and contains $2^{a-1}$ reflections. So there must be at least $k$ Sylow $2$-subgroups - but that's the largest possible number. – Derek Holt Aug 26 '15 at 18:39
• @DerekHolt Thank you very much for your help! (you really should write this as an answer not a comment!). Yesterday night I started working on your hints and I'll work again on that today. I'll tell you whenever I prove a point of your hints or if I'm stuck. – Scientifica Aug 27 '15 at 10:03
• @DerekHolt Thank you very much! In the case where $a>1$ as I was working on the first hint, I only managed to prove that a Sylow 2-subgroup of $D_{2n}$ must contain a rotation and half of all reflections are contained in a Sylow 2-subgroup using conjugacy classes of $D_{2n}$ (Sylow 2-subgoups are conjugate). I was going to ask your help again, but looking at the second hint: the number of reflections of a Sylow 2-subgroup is $2^{a-1}=\frac{2^a}{2}$, thinking about a Dihedral subgroup would be good. In fact, $<s,r^k>\cong D_{2^a}$ is a Sylow 2-subgroup of $D_{2n}$... – Scientifica Aug 27 '15 at 12:21
• @DerekHolt ... and all Sylow 2-subgroups are isomorphic to it since they are conjugate. This shows that every Sylow 2-subgroup has $2^{a-1}$ reflections. Using conjugacy classes of $D_{2n}$ when $n$ is odd ($a>1$), we know that all reflections $sr^i$ where $i$ is odd are contained in a Sylow 2-subgroup because they are conjugate to $sr^k\in <s,r^k>$ ($k$ is odd). We get the same thing for the reflections $sr^i$ when $i$ is even because $sr^{2k}\in <s,r^k>$. Then the conclusion follows as you showed in the comment... – Scientifica Aug 27 '15 at 12:27
• @DerekHolt So you should post your comment as an answer ;) – Scientifica Aug 27 '15 at 12:28

There are $n=2^{a−1}k$ elements of $D_{2n}$ lying outside of the cyclic subgroup $\langle r \rangle$, all of order 2. (These are usually called reflections.) Each such reflection is contained in at least one Sylow $2$-subgroup. A Sylow $2$-subgroup has order $2^a$ and contains $2^{a-1}$ reflections. So there must be at least $k$ Sylow $2$2-subgroups.

But a Sylow $2$-subgroup has index $k$, so there at most $k$ Sylow $2$-subgroups. Hence there are exactly $k$, and each reflection is contained in exactly one Sylow $2$-subgroup.

• @Derek Holt, why must a Sylow 2-group contain $2^{a-1}$ reflections? Why can't it contain some nontrivial powers of $r$? – The Substitute Jun 1 '16 at 7:01
• It does contain powers of $r$. Half of its elements are powers of $r$ (called rotations) and the other half are reflections. – Derek Holt Jun 1 '16 at 9:14

Your step (1) looks good; after that, I would look at a factor group:

Assume that $a>1$ and let $N:=\langle r\rangle$ and $P\in Syl_2(D_{2n})$. Then $N$ is cyclic and therefore abelian, and $Q:=P\cap N$ has index $2$ in $P$, so $Q$ is normalized by $\langle P, N\rangle = D_{2n}$. By looking at the relations in the definition of $D_{2n}$, it's easy to see that $D_{2n}/Q\simeq D_{2k}$, and so the claim follows from (1).

• Thank you for your answer. Unfortunately, $r^n=1$ so $N=1$. – Scientifica Aug 27 '15 at 12:40
• Oops, sorry; what I meant was $N := \langle r\rangle$. (Fixing it in my answer). – jpvee Aug 27 '15 at 12:56
• No worries :D. $Q\le N$ so $\exists q\in\mathbb{Z}^+,\,Q=<r^q>$ and $q\mid n$ (I can show that $P$ contains nonidentity powers of $r$ using step (2)). . We have then $|r^q|=\frac{n}{q}=\frac{2^{a-1}k}{q}$. But by Lagrange's theorem: $Q\le P\Rightarrow |r^q|=2^\beta$ for some $\beta\in\{ 1,\,\dots,\,a\}$. Therefore $q$ must be a multiple of $k$ and so $Q\le <r^k>$. Now can you please give me a hint to prove that $Q=<r^k>$? (so that we have $|P:Q|=2$ and $D_{2n}/Q\cong D_{2k}$) I can't see any other method than the one in the comments of my question that uses the fact that $P\cong <s,r^k>$... – Scientifica Aug 27 '15 at 13:30
• .. but still using this fact proves that $Q=<r^k>$. If there isn't another method it's not a problem. Afterall that fact isn't "all" the method given by Derek Holt. So then we get $D_{2n}/Q\cong D_{2k}$ and the number of Sylow 2-subgroups of $D_{2k}$ is $k$ from (1) so the number of Sylow 2-subgroups of $D_{2n}/Q$ is $k$. Unfortunately, I can't conclude from this that the number of Sylow 2-subgroups of $D_{2n}$ is $k$ (I thought on the Fourth Isomorphism Theorem, also called Lattice Isomorphism Theorem). Could you please help me? – Scientifica Aug 27 '15 at 13:36
• Oh! It's ok now I see I see. Let $\bar{P}\in Syl_2(D_{2n}/Q)$. Thus $|\bar{P}|=2$ because $|D_{2n}/Q|=|D_{2k}|=2k$ and $k$ is odd. By the Four Isomorphism Theorem, we can find $P\le D_{2n}$ such as $P/Q=\bar{P}$. Since $|Q|=2^{a-1}$, $P$ is indeed a Sylow 2-subgroup of $D_{2n}$. I'll think later about the fact that distinct Sylow 2-subgroups of $D_{2n}$ gives distinct Sylow 2-subgroups in $D_{2n}/Q$. – Scientifica Aug 27 '15 at 13:43

Just note that each $$P$$ can separate into two part $$A,B$$, which contains elements of the form $$r^i$$ and $$sr^j$$ respectively.

Soppose $$r^i\in P$$ has the least power, then all element of the form $$r^j\in P$$ must have $$i\vert j$$. As your step (2), $$i=sk$$ where $$2\nmid s$$, which implies that all $$P\in Syl_2(D_{2n})$$ has the same part $$A=$$.

So we may find $$k$$ different $$B$$ to make $$P$$ be a subgroup. It's easy to check that $$P_i=\{A,B_i\}$$ where $$B_i=\{sr^{k+i},sr^{2k+i},\cdots,sr^{2^{\alpha-1}k+i}\}$$ satisfies this property.

Since $$n_2|k$$ and we have already find $$k$$ Sylow $$2$$-subgroups, we have $$n_2=k$$.