How many Sylow \(2\)-subgroups does the dihedral group \(D_n\) of order \(2n\) have when \(n\) is odd?
Solution
Use the presentation \[ D_n = \langle r,s\mid r^n=s^2=1,\ srs=r^{-1}\rangle. \]
1. Every Sylow \(2\)-subgroup of \(D_n\) has order \(2\).
Since \(n\) is odd, \[ \abs{D_n}=2n \] is divisible by \(2\) but not by \(4\). Hence the largest power of \(2\) dividing the group order is \(2\).
2. For every \(k=0,1,\ldots,n-1\), the element \(sr^k\) has order \(2\).
The relation \(srs=r^{-1}\) implies \[ sr^ks=r^{-k}. \] Therefore \[ (sr^k)^2 = sr^ksr^k = r^{-k}r^k =1. \] The element \(sr^k\) is not the identity because it lies outside the rotation subgroup \(\langle r\rangle\).
3. The subgroups\[ \langle sr^k\rangle, \qquad 0\le k<n, \]are \(n\) distinct Sylow \(2\)-subgroups.
By steps <1>1–<1>2, each subgroup \(\langle sr^k\rangle\) has the required Sylow order \(2\). The elements \(sr^k\) are pairwise distinct, and a subgroup of order \(2\) has exactly one nonidentity element. Hence the generated subgroups are pairwise distinct.
4. There are no other Sylow \(2\)-subgroups.
Every element of \(D_n\) is either a rotation \(r^k\) or a reflection \(sr^k\). Since \(\langle r\rangle\) has odd order \(n\), no nonidentity rotation has order \(2\). Therefore every subgroup of order \(2\) is generated by one of the reflections listed in step 3.
5. Hence the number of Sylow \(2\)-subgroups is\[ \boxed{n}. \]
Step 3 produces \(n\) such subgroups, and step 4 proves the list is exhaustive.
6. Q.E.D.
Step 5 gives the requested number.