Solution

Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 19 / 4 / Solution

Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 38 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 1 / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 2 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 4 / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 5 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 50 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 1D / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 13F / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 17 / 2 / Solution

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{wiki=Group_(mathematics)}

A <group> packages reversible operations together with a rule for composing them. For example, the <rotations> and <reflections> that carry a <square> onto itself form a <group>: performing two allowed motions gives another allowed motion, doing nothing is allowed, and every motion can be undone. The elements are the motions themselves, rather than the vertices they move. Labels in the figure track those vertices; they are markers, rather than extra decorations that the motions must preserve.

\Image[/square-symmetry-composition.png]
{title=Square symmetries composed in opposite orders}
{description=The <rotation> $r$ is counterclockwise through $90^\circ$, and the <reflection> $s$ is in the vertical axis. Products act rightmost first: $rs=r\circ s$. The red vertex starts at $(1,1)$ and ends at $(-1,-1)$ under $rs$, but at $(1,1)$ under $sr$. Both motions preserve the unmarked <square>, while their <permutations> of its vertices differ.}

Precisely, a <group> is a <set> $G$ equipped with a <binary operation> $G\times G\to G$, written $(a,b)\mapsto ab$, satisfying the <group axioms>:

* Closure: $ab\in G$ for all $a,b\in G$. This is already part of specifying the operation's target as $G$.
* <Associativity>: $(ab)c=a(bc)$ for all $a,b,c\in G$.
* An <identity element> $e\in G$ satisfies $ea=ae=a$ for every $a\in G$.
* Every $a\in G$ has an <inverse element> $a^{-1}\in G$ with $aa^{-1}=a^{-1}a=e$.

The <group operation> need not be ordinary numerical multiplication. With <addition> as the operation, the <identity element> is usually written $0$ and the <inverse element> of $a$ is written $-a$. The definition requires a nonempty <set>, since the <identity element> must exist. See https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/faeeeaf01f838584ce2172cfd85cf74d_MIT18_703S13_pra_l_1.pdf[McKernan's MIT lecture 1, Definition 1.1] for the axioms.

Why insist on <associativity>? It makes a string of operations unambiguous without specifying parentheses. For <function composition>, both $(fg)h$ and $f(gh)$ send $x$ to $f(g(h(x)))$. This fixes the order of execution while allowing regrouping. <Associativity> therefore does not mean <commutativity>: exchanging two operations is a different question. Ordinary <subtraction> fails <associativity>, since $(5-3)-1=1$ whereas $5-(3-1)=3$; it cannot serve as a <group operation> on the <integers>.

An <inverse element> makes reversal exact and equations solvable. In any <group>, $ax=b$ has the unique solution $x=a^{-1}b$, since $a^{-1}(ax)=(a^{-1}a)x=x$. Similarly, $xa=b$ has solution $x=ba^{-1}$. The side matters in a <non-abelian group>. Undoing a composite operation requires reversing its order:
$$
(ab)^{-1}=b^{-1}a^{-1},\qquad
(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=e.
$$
The analogous product in the other order also equals the <identity element>. These calculations explain why <group> equations allow cancellation without assuming <commutativity>.

For the <square> with vertices $(\pm1,\pm1)$, choose the <rotation> and <reflection>
$$
r(x,y)=(-y,x),\qquad s(x,y)=(-x,y).
$$
Throughout this example, multiplication is <function composition> with the rightmost operation performed first: $(gh)(v)=g(h(v))$. Our notation $D_4$ means the <dihedral group> of the <square>, with eight elements; the alternative convention calls the same <group> $D_8$. Its elements are
$$
D_4=\{e,r,r^2,r^3,s,rs,r^2s,r^3s\},\qquad
r^4=e,\quad s^2=e,\quad srs=r^{-1}.
$$
Here $r^{-1}=r^3$ is clockwise <rotation> through $90^\circ$, while each <reflection> is its own <inverse element>. There are exactly eight motions: the image of one vertex has four choices, and the image of an adjacent vertex has two choices; together these determine the motion. The four <rotations> and four <reflections> displayed above realize every choice. The vertex <permutations> give a concrete <group action>. The general polygon construction is discussed in <Group Theory (v4.01) by James S. Milne> (section 1.17).

Computing the two products rather than merely comparing their names gives
$$
(rs)(x,y)=r(-x,y)=(-y,-x),\qquad
(sr)(x,y)=s(-y,x)=(y,x).
$$
Thus $(rs)(1,1)=(-1,-1)$ but $(sr)(1,1)=(1,1)$, proving $rs\ne sr$. In fact $rs$ is <reflection> in the line $y=-x$, whereas $sr$ is <reflection> in $y=x$. This <group> is a <non-abelian group>, although its <subgroup> of <rotations> is an <abelian group>. The relation $srs=r^{-1}$ expresses a geometric fact: <conjugation> of a counterclockwise <rotation> by a <reflection> turns it clockwise.

A contrasting example is the <cyclic group> $\mathbb Z/4\mathbb Z$ under <addition> in <modular arithmetic>. Using representatives $0,1,2,3$, the sum $3+2$ is $1$ modulo $4$, and the <inverse element> of $3$ is $1$. Repeated <addition> of $1$ runs through $0,1,2,3,0$. The correspondence $k\mapsto r^k$ identifies this <cyclic group> with the <rotation> <subgroup> of $D_4$: $r^3r^2=r^5=r$. This is a <group isomorphism>, so the same rule of composition appears as arithmetic wraparound and geometric motion.

By comparison, $(\mathbb Z,+)$ is an <infinite cyclic group>. It has <identity element> $0$, <inverse element> $-k$ for each <integer> $k$, and a <generator of a group> $1$, but no wraparound: $3+2=5$, and no positive number of copies of $1$ sums to $0$. Both examples are <abelian groups>. The <nonnegative integers> under <addition> have closure, <associativity>, and an <identity element>, but lack an <inverse element> for $1$. Reversibility is precisely what fails.

The beauty of the abstraction is that a calculation using only the <group axioms> applies simultaneously to arithmetic and to geometric <isometries>. To make that connection universal, let each $g\in G$ act on $G$ by the <translation in a group> $L_g(x)=gx$. Each $L_g$ is a <bijection>, with inverse $L_{g^{-1}}$, and $L_gL_h=L_{gh}$. Distinct elements give distinct maps because $L_g(e)=g$. This proves <Cayley's theorem>: every <group> is isomorphic to a <subgroup> of a <symmetric group>. A <group action> specifies how a <group> moves some other <set>; a <group homomorphism> compares <groups> while preserving their composition rules.

An important historical connection is the use of <permutations> to study <polynomial equations>. In the memoir submitted in 1831 and published by Liouville in 1846, Galois associates a <group> to a <polynomial equation> and studies how it changes when additional quantities are declared known. Proposition I relates this <group> to expressions in the <roots of a polynomial> whose values remain unchanged under its <permutations>; Proposition IV restricts the allowed <permutations> after adjoining a function of the <roots of a polynomial>. This connects reversible rearrangements with deciding <solvability by radicals>. See https://www.numdam.org/item/JMPA_1846_1_11__381_0.pdf[Galois's original memoir, pp. 417–433, especially Propositions I and IV; Liouville's introduction, p. 382].

Group

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