Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 8

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.

The cyclic subgroup $A=⟨a⟩$ has order four and index two. Every index-two subgroup is normal: its complement is both the only nontrivial left coset and the only nontrivial right coset. Thus conjugation by $b$ is an automorphism of $A$. It must send its generator $a$ to another element of order four, namely $a$ or $a^3$. Since $b^{-1}=b$,equivalently $ab=ba$ or $ab=b{a}^3$, as required. Also $b\notin A$, since otherwise $a,b$ could generate at most four elements.

In the commuting case, every word reduces uniquely to $a^j{b}^{\epsilon }$ with $0\le j<4$ and $\epsilon \in \{0,1\}$. The eight expressions are distinct because $A$ and $Ab$ are disjoint. Multiplication is addition of the two exponents modulo four and two, respectively. ThereforeThis gives the multiplication table independently of any other properties of $G$, proving uniqueness up to group isomorphism.

In the inversion case, the relation $ba=a^{-1}b$ gives the same eight unique normal forms, now with multiplicationwith exponents reduced modulo four and two. Thus this case also has a uniquely determined group structure. A faithful geometric realization is the dihedral group of a square, with $a$ a quarter-turn and $b$ a reflection in a line through opposite edge midpoints. For example, on the plane,These preserve the square with vertices $(\pm 1,\pm 1)$ and satisfy $a^4=b^2=I$, $bab=a^{-1}$. They generate eight distinct transformations, so

No. The quaternion groupis another non-abelian group of order eight: for example $ij=k$ while $ji=-k$. Its multiplication comes from the associative quaternion algebra, so it does define a group.

It is not isomorphic to the dihedral group of the square. In $Q_8$, only $-1$ has order two; the six elements $\pm i,\pm j,\pm k$ have order four. The square's dihedral group has five elements of order two: four reflections and the half-turn. Group isomorphisms preserve element orders, so these groups are distinct. The hypothesis in part (c) excludes $Q_8$, since its only element of order two is already in each cyclic subgroup of order four.