Let \(G\) be a group with \(|G| = p(p+1)\), where \(p\) is an odd prime and \((p+1) = 2^n\) is a power of 2. Suppose further that \(G\) does not have a normal Sylow \(p\)-subgroup.
Show that \(G\) has a normal subgroup \(H\) of order \(p+1\). Show that \(G\) is isomorphic to a semidirect product \(H \rtimes \mathbb{Z}_p\).
Prove that the subgroup \(H\) in part (a) is abelian.
Hint: Given a nonidentity element \(x \in H\), consider the possible size of the centralizer subgroup \(C_G(x)\).
Solution
Write \[ |G|=p(p+1)=p2^n. \]
1. The number of Sylow \(p\)-subgroups is \(p+1\).
Let \(n_p\) denote the number of Sylow \(p\)-subgroups. By Sylow’s theorem, \[ n_p\mid p+1 \qquad\text{and}\qquad n_p\equiv1\pmod p. \] Since \(G\) has no normal Sylow \(p\)-subgroup, \(n_p\neq1\). Every divisor of \(p+1\) lies between \(1\) and \(p+1\), and the only integers in that interval congruent to \(1\) modulo \(p\) are \(1\) and \(p+1\). Hence \[ n_p=p+1. \]
2. A Sylow \(2\)-subgroup \(H\) is unique, hence normal, and has order \(p+1\).
Each Sylow \(p\)-subgroup has order \(p\). Distinct such subgroups intersect only in the identity, since an intersection of two groups of prime order is either trivial or the whole subgroup. By 1, the Sylow \(p\)-subgroups therefore contain \[ (p+1)(p-1)=p^2-1 \] distinct nonidentity elements.
The group \(G\) has \[ p(p+1)=p^2+p \] elements altogether. After removing those \(p^2-1\) nonidentity \(p\)-elements, exactly \[ (p^2+p)-(p^2-1)=p+1 \] elements remain, including the identity. Thus exactly \(p\) nonidentity elements of \(G\) do not lie in a Sylow \(p\)-subgroup.
Let \(H\) be any Sylow \(2\)-subgroup. Then \[ |H|=2^n=p+1, \] so \(H\) has exactly \(p\) nonidentity elements. None of them can lie in a subgroup of order \(p\), because a \(2\)-group and a group of order \(p\) have trivial intersection. Hence the \(p\) nonidentity elements of \(H\) are precisely all of the nonidentity elements left over above. This description is independent of the choice of Sylow \(2\)-subgroup, so every Sylow \(2\)-subgroup has the same elements. Therefore \(H\) is unique and hence normal in \(G\).
3. One has\[ G\cong H\rtimes\mathbb Z_p. \]
Let \(P\) be a Sylow \(p\)-subgroup. Then \(|P|=p\), so \[ P\cong\mathbb Z_p. \] Since \(|H|=p+1\) and \(|P|=p\) are coprime, \[ H\cap P=1. \] Because \(H\triangleleft G\), the product \(HP\) is a subgroup, and \[ |HP|=\frac{|H||P|}{|H\cap P|}=p(p+1)=|G|. \] Thus \(G=HP\). Therefore \(G\) is the internal semidirect product \[ G=H\rtimes P\cong H\rtimes\mathbb Z_p. \]
4. If \(1\neq x\in H\), then \(p\) does not divide \(|C_G(x)|\).
Suppose instead that \(p\mid |C_G(x)|\). By Cauchy’s theorem, \(C_G(x)\) contains an element \(y\) of order \(p\). Let \[ P:=\langle y\rangle. \] Then \(x\) centralizes \(P\), hence normalizes \(P\), so \[ x\in N_G(P). \] By 1, \[ [G:N_G(P)]=n_p=p+1, \] and therefore \[ |N_G(P)|=p. \] Since \(P\subseteq N_G(P)\) also has order \(p\), one has \[ N_G(P)=P. \] Thus \(x\in P\). But \(x\in H\) and \(H\cap P=1\), contradicting \(x\neq1\). Hence \(p\nmid |C_G(x)|\).
5. The subgroup \(H\) is abelian.
Fix \(1\neq x\in H\). Because \(H\triangleleft G\), every conjugate of \(x\) lies in \(H\). Hence the conjugacy class of \(x\) has at most \[ |H|-1=p \] elements. By 4, the centralizer \(C_G(x)\) has order a power of \(2\), say \(2^k\) with \(0\le k\le n\). The orbit-stabilizer formula for conjugation gives \[ |x^G|=[G:C_G(x)] =\frac{p2^n}{2^k} =p2^{n-k}. \] Since \(|x^G|\le p\), necessarily \(k=n\). Thus \[ |C_G(x)|=2^n=|H|. \] The centralizer is a \(2\)-subgroup, so by uniqueness of the Sylow \(2\)-subgroup from 2, \[ C_G(x)=H. \] Therefore \(x\) commutes with every element of \(H\). Since this holds for every nonidentity \(x\in H\) (and trivially for \(x=1\)), \[ H=Z(H), \] so \(H\) is abelian.