GroupsAQA A-Level Further Maths: Revision notes

A group $(G,*)$ is a set $G$ with a binary operation $*$ satisfying four axioms.

1. Closure: $a*b\in G$ for all $a,b\in G$.
2. Identity: there is $e\in G$ with $a*e=e*a=a$ for all $a$.
3. Inverses: for each $a$ there is $a^{-1}\in G$ with $a*a^{-1}=a^{-1}*a=e$.
4. Associativity: $(a*b)*c=a*(b*c)$ for all $a,b,c$. To show a set is not a group, find one axiom that fails. Example: $\mathbb{Z}$ under multiplication has closure, identity $1$ and associativity, but $2$ has no inverse in $\mathbb{Z}$. For a finite set, associativity of modular arithmetic is inherited from the integers, and may be quoted.

A Cayley table lists every product in a finite group. Closure holds if every entry is an element of the set. The identity is the element whose row and column repeat the headings. Inverses are found by locating the identity in each row. In a group, every element appears exactly once in each row and each column, so a repeated entry in a row means the set is not a group. Symmetry in the leading diagonal ($a*b=b*a$ for all pairs) shows the group is abelian.

The order of a group is the number of elements, written $|G|$. The order (or period) of an element $a$ is the smallest positive integer $n$ with $a^n=e$, where $a^n$ means $a*a*\dots*a$ ($n$ copies). The identity has order $1$. A subgroup of $G$ is a subset that is itself a group under the same operation: it contains $e$, is closed, and contains the inverse of each of its elements (associativity is inherited). The trivial subgroup is $\{e\}$; the proper subgroups are those not equal to $G$; the non-trivial subgroups are those other than $\{e\}$. So $\{e\}$ and $G$ are subgroups, but a proper non-trivial subgroup is neither of them.

A group is cyclic if one element $g$, a generator, has every element as a power of it: $G=\langle g\rangle=\{e,g,g^2,\dots\}$. A cyclic group of order $n$ has an element of order $n$. The integers modulo $n$ under addition, $\{0,1,\dots,n-1\}$, form a cyclic group generated by $1$. An abelian group has $a*b=b*a$ for all $a,b$. Every cyclic group is abelian, because powers of $g$ commute. The converse is false: $\{1,5,7,11\}$ under multiplication modulo $12$ is abelian, with every non-identity element of order $2$, so it has no element of order $4$ and is not cyclic (it is the Klein four-group).

Infinite groups include $(\mathbb{Z},+)$ and $(\mathbb{R},+)$. $(\mathbb{Z},\times)$ is not a group, since only $\pm1$ have inverses. Finite groups include $(\{0,\dots,n-1\},+_n)$ for every $n$, and $(\{1,\dots,p-1\},\times_p)$ for a prime $p$, for example $\{1,2,3,4,5,6\}$ under multiplication modulo $7$, which is cyclic with generator $3$. With a composite modulus, such as $\{1,\dots,5\}$ under multiplication modulo $6$, closure fails ($2\times3\equiv0$) so it is not a group; the elements coprime to $n$ do form a group, e.g. $\{1,5,7,11\}$ modulo $12$.

The symmetries of a regular $n$-sided polygon, under composition, form a group of order $2n$: $n$ rotations (including the identity) and $n$ reflections. Let $r$ be the rotation through $\frac{360^\circ}{n}$, so $r$ has order $n$ and the rotations form a cyclic subgroup $\{e,r,\dots,r^{n-1}\}$. If $s$ is a reflection then $s$ has order $2$ and $sr=r^{-1}s$. For an equilateral triangle the group has order $6$; for a square it has order $8$. For $n\ge3$ these groups are not abelian, because $r\neq r^{-1}$ and so $rs\neq sr$. Their proper non-trivial subgroups include the rotation subgroup and the subgroups $\{e,s\}$ generated by single reflections.