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 \(\size G = pq\), 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 \(\size G\), and assume \(G\) has a normal subgroup of order \(p\). Show that \(H \subset Z(G)\).
Let \(G\) be finite and \(P\) a Sylow 2-subgroup. Assume \(P\) is cyclic and generated by \(x\). Show that the sign of the permutation of \(G\) corresponding to \(x\mapsto gx\) is 1, and deduce that \(G\) has a nontrivial quotient of order 2.
In the last part, the relevant permutation is \(y\mapsto xy\), and its sign is \(-1\), not \(1\); the argument needs \(|G|\) even, since for \(|G|\) odd the Sylow \(2\)-subgroup is trivial and \(G\) has no quotient of order \(2\).
Solution
1. A finite group acting transitively on a set \(X\) with \(|X|\ge2\) has an element with no fixed point.
For \(g\in G\) let \(X^g\) be its fixed-point set. By Burnside’s orbit-counting lemma, \(1=|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|\), so \(\sum_{g}|X^g|=|G|\). If every \(g\) had \(|X^g|\ge1\), then, since \(|X^e|=|X|\ge2\), \[\sum_{g\in G}|X^g|\ge 2+(|G|-1)=|G|+1,\] a contradiction.
2. The subgroups \(H\le\ZZ^3\) with \(\ZZ^3/H\cong K\) number \(p^2+p+1\) for \(K\cong(\ZZ/p\ZZ)^2\) and \(p^2(p^2+p+1)\) for \(K\cong\ZZ/p^2\ZZ\).
These are the two abelian groups of order \(p^2\). If \(\ZZ^3/H\cong(\ZZ/p\ZZ)^2\), then \(p\ZZ^3\subseteq H\) and \(H/p\ZZ^3\) is a line in \(\ZZ^3/p\ZZ^3\cong\FF_p^3\); conversely each line gives such an \(H\). There are \((p^3-1)/(p-1)=p^2+p+1\) lines.
If \(\ZZ^3/H\cong\ZZ/p^2\ZZ\), then \(H\) is the kernel of a surjection \(\ZZ^3\to\ZZ/p^2\ZZ\), and two surjections have the same kernel exactly when they differ by an automorphism of \(\ZZ/p^2\ZZ\). A homomorphism is a triple in \((\ZZ/p^2\ZZ)^3\), surjective exactly when some entry is a unit, so there are \(p^6-p^3\) surjections. Dividing by \(|\Aut(\ZZ/p^2\ZZ)|=p(p-1)\) gives \(p^2(p^2+p+1)\) kernels.
3. If \(|G|=pq\) with \(p<q\) primes, then the Sylow \(q\)-subgroup \(Q\) is normal; if moreover \(q\not\equiv1\pmod p\), then \(G\cong\ZZ/pq\ZZ\).
\(n_q\mid p\) and \(n_q\equiv1\pmod q\) with \(p<q\) force \(n_q=1\). If \(q\not\equiv1\pmod p\), then \(n_p\mid q\) and \(n_p\equiv1\pmod p\) force \(n_p=1\), so the Sylow \(p\)-subgroup \(P\) is normal too. Then \(P\cap Q=1\), \(G=PQ\), and for \(a\in P\), \(b\in Q\) the commutator \(aba^{-1}b^{-1}\) lies in \(P\cap Q=1\). So \(G\cong P\times Q\cong\ZZ/p\ZZ\times\ZZ/q\ZZ\cong\ZZ/pq\ZZ\).
4. If \(p\) is the smallest prime dividing \(|G|\) and \(H\trianglelefteq G\) has order \(p\), then \(H\subseteq Z(G)\).
Conjugation gives a homomorphism \(\theta\colon G\to\Aut(H)\cong(\ZZ/p\ZZ)^\times\), a group of order \(p-1\). \(|\operatorname{im}\theta|\) divides \(|G|\) and \(p-1\); every prime factor of \(p-1\) is less than \(p\), so \(\gcd(|G|,p-1)=1\) and \(\theta\) is trivial. Thus every element of \(G\) commutes with every element of \(H\).
5. If \(|G|=2^km\) with \(k\ge1\), \(m\) odd, and \(P=\langle x\rangle\) is a cyclic Sylow \(2\)-subgroup, then \(\lambda_x\colon y\mapsto xy\) has sign \(-1\) and \(G\) has a quotient of order \(2\).
The orbits of \(\langle x\rangle\) acting on \(G\) by left multiplication are the right cosets \(Py\), each a cycle of \(\lambda_x\) of length \(2^k\). There are \(m\) of them, each of sign \((-1)^{2^k-1}=-1\), so \(\operatorname{sgn}(\lambda_x)=(-1)^m=-1\). The left regular representation \(\lambda\colon G\to\operatorname{Sym}(G)\) is a homomorphism, so \(\operatorname{sgn}\circ\lambda\colon G\to\{\pm1\}\) is a homomorphism, surjective since it sends \(x\) to \(-1\). Its kernel has index \(2\).