Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 7D (Groups)

Cayley theorem states that every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. For $g\in G$, defineEach $L_g$ is a permutation, and $L_g{L}_h=L_{gh}$, so $g\mapsto L_g$ is a homomorphism. If $L_g$ is the identity permutation, then $g=L_g(1)=1$; hence the homomorphism is injective. For finite $G$ of order $n$, this embeds $G$ in $S_n$.

The Cauchy theorem for groups says that if a prime $p$ divides $|G|$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. Cyclic rotation acts on $X$: if the product is one, thenEvery orbit has size one or $p$. The fixed points are exactly the constant tuples $(x,\dots ,x)$ satisfying $x^p=1$. Since $|X|\equiv 0(\mathrm{mod}p)$, the number of fixed points is divisible by $p$. The identity supplies one, so there is a nonidentity $x$ with $x^p=1$. Its order divides the prime $p$ and is not one, hence is $p$.

Let $n=p^k$ and take $G=C_{p^k}$. If it embedded in $S_{n-1}$, that symmetric group would contain a permutation of order $p^k$. The order of a permutation is the least common multiple of its cycle lengths. For this least common multiple to be divisible by $p^k$, one cycle length must itself be divisible by $p^k$, requiring at least $p^k=n$ points. This is impossible in $S_{n-1}$. Thus $C_n$ is the required group.

Because $n$ is not a prime power, distinct primes $p<q$ divide $n$. By Cauchy's theorem, $G$ has subgroups $H,K$ of orders $p,q$. Combine the two coset actions to obtainThe kernel of the first action is contained in $H$, and that of the second is contained in $K$. Their intersection is trivial because $H\cap K=\{1\}$, so the combined action is faithful.

Since $p\ge 2$, $q\ge 3$, and $n\ge 6$,Adding fixed points embeds this symmetric group into $S_{n-1}$. Thus every group of non-prime-power order $n$ is a subgroup of $S_{n-1}$, as summarized by symmetric-group embedding at one less than the group order.