Prove that if \(G\) is a finite group and \(H<G\) is a proper subgroup, then \(G\) cannot be written as a union of conjugates of \(H\).
Use this to prove that if \(G=\Sym(X)\) is the group of permutations of a finite set \(X\) with \(|X|=n>1\), then there exists \(g\in G\) with no fixed point in \(X\).
Solution
1. Let \(G\) act transitively on the coset space \(X=G/H\) by left multiplication. Then an element \(g\in G\) fixes some coset if and only if it lies in a conjugate of \(H\).
A coset \(aH\) is fixed by \(g\) exactly when \[ gaH=aH, \] which is equivalent to \[ a^{-1}ga\in H. \] Equivalently, \[ g\in aHa^{-1}. \] Thus the elements lying in at least one conjugate of \(H\) are exactly the elements having at least one fixed point in the coset action.
2. Every finite transitive action on a set with more than one point has a fixed-point-free element.
Let a finite group \(G\) act transitively on a finite set \(X\) with \(|X|>1\). For \(g\in G\), write \(X^g\) for the 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 contributes \(|X|>1\). If every nonidentity element fixed at least one point, then \[ \sum_{g\in G}|X^g| \ge |X|+(|G|-1)>|G|, \] a contradiction. Hence some element has no fixed points.
3. A finite group is not the union of the conjugates of a proper subgroup.
If \(H<G\), then the coset space \(G/H\) has more than one point and the action is transitive. By 2, some \(g\in G\) fixes no coset. By 1, this \(g\) belongs to no conjugate of \(H\). Therefore \[ \bigcup_{a\in G}aHa^{-1}\ne G. \]
4. For every finite set \(X\) with \(|X|=n>1\), the symmetric group \(\Sym(X)\) contains a derangement.
Fix \(x\in X\) and let \[ H=\operatorname{Stab}_{\Sym(X)}(x). \] This is a proper subgroup. Since \(\Sym(X)\) acts transitively on \(X\), the conjugates of \(H\) are exactly the point stabilizers \[ \operatorname{Stab}_{\Sym(X)}(y),\qquad y\in X. \] By 3, choose \[ g\notin\bigcup_{y\in X}\operatorname{Stab}(y). \] Then \(g\) fixes no point of \(X\), so \(g\) is a derangement.