4.1 Basic definitions
Let G be a set. A binary operation on G is a function that takes as input a pair of elements of G and whose output is a single element of G: that is, a function \(G \times G \to G\). We usually write these binary operations in infix notation, so if the binary operation is called \(*\) we write \(g*h\) for the result of applying \(*\) to the pair (g,h) of elements of G.
Example 4.1
- \(+\) is a binary operation on the set of integers \(\mathbb{Z}\).
- \(-\) is a binary operation on the set of complex numbers \(\mathbb{C}\).
- \(-\) is not a binary operation on the set of strictly positive whole numbers \(\mathbb{N}\), because it doesn’t always output an element of \(\mathbb{N}\).
- Binary operations needn’t be some mathematical operation we’re familiar with, they can be something silly: defining \(a*b = 2\) for all \(a,b\) creates a perfectly good, if useless, binary operation on the real numbers \(\mathbb{R}\).
Definition 4.1 A group \((G, *)\) is a set G with a binary operation \(*\) which contains an element e such that
- (Identity axiom) For all \(g \in G\), we have \(e*g=g*e=g\).
- (Inverses axiom) For all \(g \in G\) there exists \(h \in G\) such that \(h*g=g*h=e\).
- (Associativity axiom) For all \(g,h,k \in G\) we have \((g*h)*k = g*(h*k)\).
Strictly speaking, to specify a group I have to tell you two things: the underlying set G and the binary operation \(*\), which we call the group operation. In practise people often do not specify the group operation when there is only one standard choice of group operation. For example, “the group \(\mathbb{Z}\)” refers to \((\mathbb{Z}, +)\) and “the symmetric group \(S_n\)” refers to \((S_n, \circ)\).
Sometimes people add an extra axiom to the ones above called the closure axiom which says that if \(g, h \in G\) then \(g*h\in G\). There is no need for us to include this because it is part of the definition of binary operation that \(g*h \in G\). It is important to remember that if you want to check if a certain set \(G\) under a certain operation \(*\) is a group, you have to check that \(*\) is genuinely a binary operation on \(G\) as well as checking that the three group axioms hold.
The element \(e\) referred to in the first group axiom is called an identity element for \(G\). We can show that in any group there is one and only one element of G which satisfies the the first axiom from Definition 4.1.
An element \(h \in G\) is called an inverse of \(g \in G\) if \(h*g=g*h=e\). Such an element is guaranteed to exists for every g by the second group action, we’ll show that in fact each g has only one inverse.
This proposition means that we are justified in talking about the inverse element of an element of a group. We usually write \(g^{-1}\) for the inverse of g, so that \[g*g^{-1} = g^{-1}*g = g.\]
Remark. There are only two ways to bracket a product of three elements \(a,b,c\) of a group: \[ a*(b*c) \,\,\, \text{ or } \,\,\, (a*b)*c \] and the associativity axiom tells you that these are equal, so that the product \(a*b*c\) is unambiguous. But the associativity axiom doesn’t tell you immediately that longer products are independent of how they are bracketed, for example, is \[ (a*b)*(c*d) \,\,\, \text{ equal to } \,\,\, ((a*b)*c)*d?\] In fact the answer is yes, for a product of any length: any bracketing you use to work out a product like \(g_1* \cdots * g_n\) gives the same result. You can prove this by induction: one such proof is given in the book (Green 1988) by J.A. Green in the suggested reading, or there is a proof at MSE 21581.
There is some special notation for what happens when you multiply a group element by itself some number of times.
Proposition 4.5 Let \(n,m \in \mathbb{Z}\) and let \((G, *)\) be a group. Then
- \(g^n * g^m = g^{n+m}.\)
- \((g^n)^m = g^{nm}\).
Proof.
- This result is clear if either n or m is zero, and follows directly from associativity of \((G,*)\) if n and m have the same sign. So suppose \(n>0\) and \(m<0\). The proof is by induction on n, and when \(n=1\), \[\begin{equation*} g* g^{m} = g* (\underbrace{g^{-1} * \cdots * g^{-1}}_{-m }) = (g*g^{-1}) * g^{m+1} = g^{m+1} \end{equation*}\] as required. Now for \(n>1\) we have \[\begin{equation*} g^n * g^m = g* g^{n-1} * g^m = g* g^{m+n-1} \end{equation*}\] by induction, and using either the base case (if \(m+n-1<0\)) or the comment at the start of the proof, this equals \(g^{m+n}\) as required. The case \(n<0\) and \(m>0\) is similar.
- For m positive or zero this follows immediately from the definition, so suppose \(m<0\). We do the case \(n>0\). The first part of this proposition implies \((g^n)^{-1}= g^{-n}\), so \[\begin{equation*} (g^n)^m = g^{-n} * \cdots * g^{-n} \end{equation*}\] with \(-m\) copies of \(g^{-n}\) appearing. This is the product of \(-mn\) copies of \(g^{-1}\), so equals \(g^{mn}\) by definition. The case when \(n<0\) is similar.
Very often we’ll write the product \(g*h\) of two group elements simply as \(gh\), especially when the group operation \(*\) is some kind of multiplication.
References
Green, J.A. 1988. Sets and Groups : A First Course in Algebra. Library of Mathematics. Springer Netherlands.