Show that if \(G\) is a finite group acting transitively on a set \(X\) with at least two elements, then there exists \(g \in G\) which fixes no point of \(X\).
Let \(p\) be prime. For each abelian group \(K\) of order \(p^2\), how many subgroups \(H\leq \ZZ^{\times 3}\) are there with \(\ZZ^3/H \cong K\)?
Let \(\# G = pq\), with \(p,q\) distinct primes. Show that \(G\) has a nontrivial proper normal subgroup, and if \(p\not\equiv 1\mod q\) and \(q\not \equiv 1 \mod p\) then \(G\) is abelian.
Let \(G\) be a finite group and let \(p\) be the smallest prime dividing \(\# G\), and assume \(H\trianglelefteq G\) has order \(p\). Show that \(H \subset Z(G)\).
Let \(G\) be finite of even order and \(P\) a Sylow 2-subgroup. Assume \(P\) is cyclic and generated by \(x\). Show that the sign of the permutation of \(G\) given by left multiplication \(y\mapsto xy\) is \(-1\), and deduce that \(G\) has a nontrivial quotient of order 2.
Solution
1. A finite transitive action on at least two points has a fixed-point-free element.
For \(g\in G\), let \(X^g\) be its fixed-point set. Burnside’s orbit-counting lemma gives \[ 1=|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|, \] so \[ \sum_{g\in G}|X^g|=|G|. \] The identity fixes all of \(X\), and \(|X|\ge2\). If every \(g\in G\) fixed at least one point, then \[ \sum_{g\in G}|X^g| \ge |X|+(|G|-1) \ge |G|+1, \] a contradiction. Hence some \(g\) has \(X^g=\varnothing\).
2. The numbers of subgroups \(H\le\ZZ^3\) with quotient of order \(p^2\) are\[ \begin{array}{c|c} K & \#\{H:\ZZ^3/H\cong K\}\\ \hline \ZZ/p\ZZ\oplus\ZZ/p\ZZ & p^2+p+1,\\ \ZZ/p^2\ZZ & p^2(p^2+p+1). \end{array} \]
There are exactly two abelian groups of order \(p^2\).
For \(K\cong(\ZZ/p\ZZ)^2\), any such \(H\) contains \(p\ZZ^3\), and \(H/p\ZZ^3\) is a one-dimensional subspace of \[ \ZZ^3/p\ZZ^3\cong\FF_p^3. \] Conversely every line in \(\FF_p^3\) gives such an \(H\). The number of lines is \[ \frac{p^3-1}{p-1}=p^2+p+1. \]
For \(K\cong\ZZ/p^2\ZZ\), kernels are counted by surjections \[ \ZZ^3\twoheadrightarrow\ZZ/p^2\ZZ \] modulo postcomposition by automorphisms of the target. A homomorphism is determined by a triple in \((\ZZ/p^2\ZZ)^3\), and it is surjective exactly when the triple is not entirely divisible by \(p\). Thus there are \[ p^6-p^3 \] surjections. Two surjections have the same kernel exactly when they differ by an automorphism of \(\ZZ/p^2\ZZ\), and \[ |\operatorname{Aut}(\ZZ/p^2\ZZ)|=\varphi(p^2)=p(p-1). \] Therefore the number of kernels is \[ \frac{p^6-p^3}{p(p-1)} =p^2\frac{p^3-1}{p-1} =p^2(p^2+p+1). \]
3. Every group of order \(pq\) has a nontrivial proper normal subgroup; under the two stated congruence hypotheses it is abelian.
Interchange \(p\) and \(q\) if necessary so that \(p<q\). Let \(n_q\) be the number of Sylow \(q\)-subgroups. Then \[ n_q\mid p, \qquad n_q\equiv1\pmod q. \] Since \(p<q\), the only possibility is \(n_q=1\). Thus the Sylow \(q\)-subgroup \(Q\) is nontrivial, proper, and normal.
Now assume also \(q\not\equiv1\pmod p\); the other displayed congruence is automatic when \(p<q\). For the number \(n_p\) of Sylow \(p\)-subgroups, \[ n_p\mid q, \qquad n_p\equiv1\pmod p. \] Hence \(n_p=1\) or \(q\), and the congruence hypothesis excludes \(q\), so the Sylow \(p\)-subgroup \(P\) is also normal. Since \(P\cap Q=1\) and \(|P||Q|=|G|\), we have \(G=PQ\). For \(a\in P\) and \(b\in Q\), the commutator \(aba^{-1}b^{-1}\) lies in both \(P\) and \(Q\) because both are normal, hence is trivial. Thus \(P\) and \(Q\) commute elementwise and \(G\cong P\times Q\) is abelian.
4. If \(p\) is the smallest prime divisor of \(|G|\) and \(H\trianglelefteq G\) has order \(p\), then \(H\subseteq Z(G)\).
Since \(|H|=p\), \(H\cong C_p\), so \[ |\operatorname{Aut}(H)|=p-1. \] Normality of \(H\) gives the conjugation homomorphism \[ \theta:G\longrightarrow\operatorname{Aut}(H). \] The order of \(\operatorname{im}\theta\) divides both \(|G|\) and \(p-1\). Every prime divisor of \(p-1\) is smaller than \(p\), while \(p\) is the smallest prime divisor of \(|G|\). Hence \[ \gcd(|G|,p-1)=1, \] so \(\operatorname{im}\theta=1\). Thus every \(g\in G\) centralizes every \(h\in H\), proving \(H\subseteq Z(G)\).
5. If \(G\) has even order and its Sylow \(2\)-subgroup \(P=\langle x\rangle\) is cyclic, then left multiplication by \(x\) has sign \(-1\), and \(G\) has a quotient of order \(2\).
Write \[ |P|=2^r\quad(r\ge1), \qquad |G|=2^r m \] with \(m\) odd. In the left regular action of \(G\) on itself, the permutation \[ \lambda_x:G\to G, \qquad y\longmapsto xy, \] has orbits equal to the left action of \(\langle x\rangle\) on \(G\). Since \(x\) has order \(2^r\), every orbit is a cycle of length \(2^r\). There are \[ |G|/2^r=m \] such cycles. Each \(2^r\)-cycle has sign \[ (-1)^{2^r-1}=-1, \] and \(m\) is odd, so \[ \operatorname{sgn}(\lambda_x)=(-1)^m=-1. \]
The left regular representation \[ \lambda:G\longrightarrow S_G \] is a homomorphism, so \[ \varepsilon:=\operatorname{sgn}\circ\lambda:G\longrightarrow\{\pm1\} \] is a homomorphism. Since \(\varepsilon(x)=-1\), it is surjective. Therefore its kernel is normal of index \(2\), and \[ G/\ker\varepsilon\cong C_2. \]