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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Derek Holt Aug 26 '15 at 18:39
  • $\begingroup$ @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. $\endgroup$ – Scientifica Aug 27 '15 at 10:03
  • $\begingroup$ @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 $D_{2n}$... $\endgroup$ – Scientifica Aug 27 '15 at 12:21
  • $\begingroup$ @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... $\endgroup$ – Scientifica Aug 27 '15 at 12:27
  • $\begingroup$ @DerekHolt So you should post your comment as an answer ;) $\endgroup$ – 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.

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

  • $\begingroup$ Thank you for your answer. Unfortunately, $r^n=1$ so $N=1$. $\endgroup$ – Scientifica Aug 27 '15 at 12:40
  • $\begingroup$ Oops, sorry; what I meant was $N := \langle r\rangle$. (Fixing it in my answer). $\endgroup$ – jpvee Aug 27 '15 at 12:56
  • $\begingroup$ 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 $P\cong <s,r^k>$... $\endgroup$ – Scientifica Aug 27 '15 at 13:30
  • $\begingroup$ .. 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? $\endgroup$ – Scientifica Aug 27 '15 at 13:36
  • $\begingroup$ 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 $D_{2n}/Q$. $\endgroup$ – 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 $A=<r^k>$.

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 $n_2=k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.