8.2 Formal definition
We want to formalize the structure of symmetries of the kind we looked at in the last section. First of all noticed that in all cases we looked at combining two permutations/symmetries to get a third. We can formalize this as follows.
The word “binary” refers to the fact that the function takes two inputs \(x\) and \(y\) from \(G\). (Technically speaking, it takes one input from \(G\times G\).)
We’ll usually write \(x\star y\) instead of \(\star(x,y)\).
Let \(X\) be a non-empty set. Composition \(\circ\) is a binary operation on the set \(G\) of permutations of a set \(X\).
Addition \(+\) is a binary operation on the set \(\mathbb{R}\) (or on \(\mathbb{Z}\), or on \(\mathbb{Q}\)).
Multiplication and subtraction are also binary operations on \(\mathbb{R}\), \(\mathbb{Q}\) or \(\mathbb{Z}\).
Intersection and unions are binary operations on the set of subsets of \(\mathbb{R}\).
Note that in the definition of a binary operation, the function \(\star\) maps to \(G\), so if we have a definition of \(x\star y\) so that \(x\star y\) is not always in \(G\), then this is not a binary operation on \(G\) (we say that \(G\) is not closed under \(\star\)). Also, the domain of \(\star\) is \(G\times G\), so \(x\star y\) needs to be defined for all pairs of elements \(x,y\).
Subtraction is not a binary operation on \(\mathbb{Z}_+\) since, for example, \(4-7=-3\notin \mathbb{Z}_+\). So \(\mathbb{Z}_+\) is not closed under subtraction (it is closed under addition).
Division is not a binary operation on \(\mathbb{Z}\) since, for example, \(1/2\notin \mathbb{Z}\).
Division is not a binary operation on \(\mathbb{Q}\) since \(x/y\) is not defined when \(y=0\). (But division is a binary operation on the set \(\mathbb{Q}\setminus\{0\}\) of non-zero rational numbers).
For a general binary operation, the order of the elements \(x,y\) matters: \(x\star y\) is not necessarily equal to \(y\star x\).
Some examples and non-example of commutative binary operations
Addition and multiplication are commutative binary operations on \(\mathbb{R}\) (axioms (A3) and (A9) ).
Subtraction is not commutative on \(\mathbb{R}\) since, for example, \(2-1=1\) but \(1-2=-1\).
Composition is not a commutative operation on the set of permutations of the set \(\{1,2,3\}\).
Bearing in mind the analogies we drew between some of the axioms of \(\mathbb{R}\) and the compositions of permutations/symmetries, we’ll now define a group.
A group \((G,\star)\) is a set \(G\) together with a binary operation \(\star:G\times G\to G\) satisfying the following properties (or “axioms”):
(Associativity) For all \(x,y,z\in G\), \[ (x\star y)\star z=x\star(y\star z). \]
(Existence of an identity element) There is an element \(e\in G\) (called the identity element of the group) such that, for every \(x\in G\), \[x\star e=x=e\star x.\]
(Existence of inverses) For every \(x\in G\), there is an element \(x^{-1}\in G\) (called the inverse of \(x\)) such that \[x\star x^{-1}=e=x^{-1}\star x.\]
Strictly speaking, the group consists of both the set \(G\) and the operation \(\star\), but we’ll often talk about “the group \(G\)” if it’s clear what operation we mean, or say “\(G\) is a group under the operation \(\star\)”. But the same set \(G\) can have several different group operations, so we need to be careful.
If \(X\) is a set, \(S(X)\) is the set of all permutations of \(X\) (i.e., bijective functions \(X\to X\)), and \(f\circ g\) is the composition of bijections \(f\) and \(g\), then \(\left(S(X),\circ\right)\) is a group.
Note that in this example, there are two sets involved (\(X\) and the set \(S(X)\) of permutations). It is the set \(S(X)\) that is the group, not \(X\) (we haven’t defined a binary operation on \(X\)).
The set of all symmetries of a regular \(n\)-sided polygon is a group under composition, as is the set of all symmetries of a cube, or a Rubik’s cube.
These examples, and similar ones, of symmetry groups, are the motivation for the definition of a group, but there are some other familiar examples of sets of numbers with arithmetical operations that fit the definition.
Here we explore some sets and associated operations and check if they are a group or not.
We have \((\mathbb{R},+)\) is a group. [by axioms (A2),(A4),(A5)]. Similarly \((\mathbb{Z},+)\) and \((\mathbb{Q},+)\) are groups.
The set \(\mathbb{Z}_+\) is not a group under addition, since it doesn’t have an identity element [\(0\notin \mathbb{Z}_+\)]. Note \(\mathbb{Z}_{\geq 0}\) is still not a group (under addition), as although it has the identity element \(0\), no integer \(n>0\) has an inverse (since for any \(n\in \mathbb{Z}_+\), \(-n \notin \mathbb{Z}_{\geq 0}\)).
We have \((\mathbb{R}\setminus\{0\},\cdot)\) is a group [by axioms (A8),(A10) and (A11)]. Similarly \((\mathbb{Q}\setminus\{0\},\cdot)\) is a group.
We have \((\mathbb{R},\cdot)\) is not a group, since \(0\) does not have an inverse.
We have \(\mathbb{R}\) is not a group under subtraction, since associativity fails: \((x-y)-z=x-y-z\), but \(x-(y-z)=x-y+z\), and so these are different whenever \(z\neq 0\).
Matrices are another source of examples.
Let \(M_n(\mathbb{R})\) be the set of \(n\times n\) matrices with real entries. Then \(M_n(\mathbb{R})\) is a group under (matrix) addition.
\(M_n(\mathbb{R})\) is not a group under matrix multiplication, since not every matrix has an inverse. However, the set of invertible \(n\times n\) matrices is a group under multiplication.
We’ve stressed that \(x\star y\) and \(y\star x\) are typically different in symmetry groups. But in the examples coming from addition and multiplication of integers, they are the same.
The word “abelian” was introduced in the 1880s by German mathematician Heinrich Weber (1842 - 1913). He derived the name from the Norwegian mathematician Niels Henrik Abel (1802 - 1829) who studied the permutation of solutions of polynomials. Very loosely speaking he showed in 1824 that for a given polynomial if the permutation of the roots form an Abelian group then the polynomial can be solved in radicals (i.e. involving \(+,-,\cdot,\div\) and \(n\)-th roots only). You can learn more about this in the fourth year course Galois Theory (named after French mathematician Évariste Galois, 1811 - 1832).
Even in a non-abelian group, \(x\star y=y\star x\) may hold for some elements \(x\) and \(y\), in which case we say that \(x\) and \(y\) commute. For example, the identity commutes with every other element. However for the group to be abelian it must hold for all elements.
The group \(S(X)\) of permutations of a set \(X\) is non-abelian if \(X\) has at least three elements, since if \(x,y,z\in X\) are three distinct elements, \(f\) is the permutation that swaps \(x\) and \(y\) (and fixing all other elements), \(g\) is the permutation that swaps \(y\) and \(z\), then \(f\circ g\neq g\circ f\).
More generally, symmetry groups are typically non-abelian (although sometimes they are). In particular, the symmetry group of a regular \(n\)-sided polygon (where \(n\geq 3\)) is non-abelian.
\((\mathbb{R},+)\) is abelian by axiom (A3).
\((\mathbb{R}\setminus\{0\},\cdot)\) is abelian by axiom (A9).
\(M_n(\mathbb{R})\) is an abelian group under matrix addition.
If \(n\geq 2\), then the set of invertible \(n\times n\) real matrices is a non-abelian group under matrix multiplication, since for example \[\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} =\begin{pmatrix}1&1\\1&0\end{pmatrix}\] but \[\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} =\begin{pmatrix}0&1\\1&1\end{pmatrix}.\]
Often, especially when we’re dealing with abstract properties of general groups, we’ll simplify the notation by writing \(xy\) instead of \(x\star y\), as though we’re multiplying. In this case we’ll say, for example, “Let \(G\) be a multiplicatively-written group”.
Note that this is purely a matter of the notation we use for the group operation: any group can be “multiplicatively-written” if we choose to use that notation, so if we prove anything about a general multiplicatively-written group, it will apply to all groups, no matter what the group operation is.
Of course, if we’re looking at something like the group of real numbers under addition, it would be incredibly confusing to write this multiplicatively, so in cases like that, where multiplication already has some other meaning, or where there’s already another standard notation for the group operation, we’ll tend not to use multiplicative notation.
Notice that in a multiplicatively-written group, the associativity axiom says \[(xy)z=x(yz),\] and from this it follows easily (by induction, for example) that any product such as \(wxyz\) has a unique meaning: we could bracket it as \((wx)(yz)\) or \((w(xy))z\) or \(w((xy)z)\) or any of the other possible ways, and because of associativity they would all give the same element.