Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 8

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

In a group of order six, element orders divide six. If there were no element of order three or six, every nonidentity element would have order two, making the group abelian by the preceding argument. Choose $a\ne e$ and $b\notin \{e,a\}$. Then $\{e,a,b,ab\}$ is a subgroup of order four: its elements are distinct, it is closed because $a,b$ commute and square to $e$, and each element is its own inverse. This contradicts Lagrange's theorem, since four does not divide six. Hence an element of order three or six exists.

Suppose $x,y$ have order three but $x\ne y,y^2$. Their cyclic subgroups have trivial intersection: any common nonidentity element would generate both subgroups and force $x=y$ or $y^2$. The nine products $x^i{y}^j$, $0\le i,j<3$, are distinct. Indeed $x^i{y}^j=x^k{y}^{\ell }$ implies $x^{i-k}=y^{\ell -j}$ in that intersection, so $i=k$ and $j=\ell$ modulo three. Nine distinct elements cannot fit in a group of order six, proving the claim without assuming commutativity.

If an element has order six, it generates $G$, so $G\cong C_6$. Otherwise choose an element $x$ of order three. The preceding result excludes any further order-three elements outside $\{e,x,x^2\}$, and there are no elements of order six by assumption. Each of the other three elements must therefore have order two. ThusThe noncyclic case is the symmetric group on three points, as the classification of a group of order six further identifies.