Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 8

Interpret the functions as maps on $\mathbb{R}\setminus \{0\}$, so every reciprocal is defined. Let $e(x)=x$, $a(x)=-x$, $b(x)=1/x$ and $c(x)=-1/x$. With the left factor applied last, direct function composition gives the Cayley tableThe table proves closure and identifies $e$ as the identity. Every element is its own inverse. Associativity is inherited from composition of maps, so the set is a group. The three nonidentity elements are commuting involutions.

Its subgroups areThere are no others: a subgroup containing two distinct nonidentity elements also contains their product, the third, and hence is the whole group; a subgroup containing just one has order two.

The abstract dihedral group $D_2$ has commuting generators $R,m$ with $R^2=m^2=I$, for example a rectangle's half-turn and reflection in one axis. The assignment $I\mapsto e$, $R\mapsto a$, $m\mapsto b$, $Rm\mapsto c$ respects products and is bijective. ThusThis is the group generated by sign reversal and reciprocation, a realization of the Klein four-group. The nonzero domain is essential; treating these as functions defined at zero would not give the stated group.

For a subgroup $H$ of a finite group $G$, Lagrange's theorem statesThe left cosets partition the group and each has $|H|$ members, proving the formula. If $|G|=p$ is prime, the only divisors are one and $p$, so $H=\{e\}$ or $H=G$. Hence a group of prime order has no nontrivial proper subgroups.

The word “nontrivial” is necessary: $\{e\}$ is a proper subgroup of every prime-order group. Thus the source's shorter wording cannot literally exclude all proper subgroups.

The order of a group element $g$ is the least positive integer $q$ such that $g^q=e$, or infinity if there is no such integer. In a finite group such a $q$ exists, because two powers among $e,g,g^2,\dots$ must coincide.

The powers $e,g,\dots ,g^{q-1}$ are distinct: a repetition with exponents differing by less than $q$ would contradict its minimality. Division of any exponent by $q$ shows that these are all the elements of the cyclic subgroup $⟨g⟩$. Thus $|⟨g⟩|=q$. Applying Lagrange's theorem givesIn particular the identity has order one, and every nonidentity element has order greater than one. The divisibility assertion here concerns finite group order.

Let $|G|=p$ be prime and choose $g\ne e$. Its order divides $p$ by part (iii), and cannot be one. Hence its order is $p$, so its generated subgroup has the same number of elements as the whole group:This proves that a group of prime order is cyclic. In fact every nonidentity element is a generator. No commutativity assumption was needed; it follows from cyclicity.

For the intended finite nontrivial case, let $G=⟨g⟩$ have composite order $N$. Choose a prime divisor $p$ of $N$, with $1<p<N$. Thenis a nontrivial proper cyclic subgroup of order $p$. Indeed $(g^{N/p}{)}^p=e$, and no smaller positive power is the identity because $g$ has order $N$. Thus $1<|H|=p<N$.

If the source also allows an infinite cyclic group, $⟨g^2⟩$ is a proper cyclic subgroup: its powers have even exponents and do not include $g$. The trivial group, of order one, is an exception to the literal formulation “not of prime order”: it has no proper subgroup at all. The nontriviality assumption is therefore necessary for that general wording. These facts are instances of subgroups and quotients of a cyclic group.