Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 5D (Groups)

No such nonabelian group exists. Every element, including the identity, satisfies $g^2=1$, so $g^{-1}=g$. For any $a,b$,Therefore the group is an abelian group. This proves that a group of exponent two is abelian without any finiteness assumption.

An example is the unitriangular group of degree three over F3, consisting ofThis is an upper unitriangular group of order $3^3=27$. Direct matrix multiplication givesThe formula gives closure, identity $M(0,0,0)$, and inverse $M(-a,-b,ab-c)$; associativity follows from matrix multiplication. The two elements $M(1,0,0)$ and $M(0,1,0)$ do not commute: their products have respectively $c=1$ and $c=0$.

Every element has the form $I+N$ with $N^3=0$. Since the finite field ${\mathbb{F}}_3$ has characteristic three, the binomial expansion givesHence every nonidentity element has order exactly three. This is a nonabelian finite group with exponent of a finite group equal to three.

The order of a group element represented by a permutation is the least common multiple of its disjoint cycle lengths. An order divisible by nine therefore requires a permutation cycle whose length is divisible by nine. In $S_9$ that must be a nine-cycle, which uses every letter and has order nine, not eighteen. There is no element of $S_9$ of order eighteen.

Take two disjoint permutation cycles of lengths five and four:Disjoint permutation cycles commute, and their powers act independently on their supports. Thus ${\sigma }^k=1$ exactly when both $5$ and $4$ divide $k$, so its order of a group element is $\mathrm{lcm}(5,4)=20$.

No such finite group exists. Cayley theorem embeds every finite group $G$ of order $n$ into $S_n$: the left-regular group action sends $g$ to the permutation $x\mapsto gx$, and this is faithful because the image of the identity element is $g$.

To make every image permutation even, let $t=(n+1n+2)$, with support disjoint from the first $n$ letters. For $\sigma \in S_n$, let $e(\sigma )=0$ for an even permutation and $e(\sigma )=1$ for an odd permutation, and setThe sign homomorphism gives $e(\sigma \tau )\equiv e(\sigma )+e(\tau )(\mathrm{mod}2)$, and $t$ commutes with all permutations of the first $n$ letters. Thus $\iota$ is a group homomorphism. It is injective by restriction to those letters, and its sign is $(-1{)}^{e(\sigma )}(-1{)}^{e(\sigma )}=1$. ConsequentlyThis is the alternating-group embedding of a finite group.