Let \(D_8\) be the dihedral group with 8 elements. Show that every Sylow 2-subgroup of \(S_8\) contains a normal subgroup that is isomorphic to \(D_8 \times D_8\).
Solution
We first construct one Sylow subgroup with the required normal subgroup. All permutations below act on \(\{1,\ldots,8\}\) and fix unlisted points.
1. The subgroup \(H_1=\langle r,s\rangle\), where\(r=(1\,2\,3\,4)\) and \(s=(2\,4)\), is isomorphic to \(D_8\).
We have \(r^4=s^2=1\) and \[ srs=(1\,4\,3\,2)=r^{-1}. \] These relations put every word into the form \(r^i s^e\) with \(0\leq i<4\) and \(e\in\{0,1\}\). The four powers of \(r\) are distinct, and \(s\) is none of them: the two nontrivial odd powers are \(4\)-cycles and \(r^2=(1\,3)(2\,4)\). Therefore the two cosets \(\langle r\rangle\) and \(\langle r\rangle s\) are disjoint, so the eight normal forms are distinct. The actions of \(r\) and \(s\) on four cyclically ordered vertices are a quarter-turn and a reflection of a square. They generate its dihedral group of order \(8\).
2. There is a subgroup \(H\cong D_8\times D_8\) of order \(64\).
Let \[ t=(1\,5)(2\,6)(3\,7)(4\,8),\qquad H_2=tH_1t^{-1}. \] Then \(H_2=\langle(5\,6\,7\,8),(6\,8)\rangle\cong D_8\). The groups \(H_1,H_2\) have disjoint supports, so every element of one commutes with every element of the other. Their intersection is trivial: an element in both fixes each of the two four-point blocks pointwise. Thus multiplication induces an injective homomorphism \[ H_1\times H_2\longrightarrow S_8, \qquad (h_1,h_2)\longmapsto h_1h_2. \] Its image \(H=H_1H_2\) has order \(8\cdot8=64\) and the asserted direct-product structure.
3. The group \(P=\langle H,t\rangle\) is Sylow and \(H\lhd P\).
The involution \(t\) interchanges \(H_1\) and \(H_2\) under conjugation, so it normalizes their product \(H\). Also \(t\notin H\), because every element of \(H\) preserves each four-point block, whereas \(t\) interchanges them. Consequently \[ P=H\sqcup Ht,\qquad |P|=2|H|=128, \] and \(H\) is normal in \(P\): it is normalized by \(H\) and by \(t\), which together generate \(P\).
The exponent of \(2\) in \(8!\) is \(4+2+1=7\), counting one factor from every even integer, another from \(4\) and \(8\), and another from \(8\). Hence \(128=2^7\) is the full power of \(2\) dividing \(|S_8|=8!\), so \(P\) is a Sylow \(2\)-subgroup.
4. Every Sylow \(2\)-subgroup has the asserted normal subgroup.
Let \(P'\) be any Sylow \(2\)-subgroup of \(S_8\). Sylow conjugacy [@DF04] gives \(g\in S_8\) with \(P'=gPg^{-1}\). Then \(H'=gHg^{-1}\) is isomorphic to \(H\cong D_8\times D_8\). Conjugating the relation \(H\lhd P\) gives \(H'\lhd P'\), which proves the statement for the arbitrary chosen \(P'\).