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 \((g*h)*k = g*(h*k)\).

So when I give you a group, I have to tell you two things: the underlying set G and the binary operation \(*\), which we call the group operation.

In fact there’s only one element of G which satisfies the condition of the first axiom from Definition 4.1.

Proposition 4.1 Suppose \((G, *)\) is a group. Then there is a unique element e of G such that \(e*g=g*e=g\) for all \(g \in G\).
Proof. Suppose that \(e, e' \in G\) and for any \(g\in G\) we have \[\begin{align} \tag{4.1} e*g & =g*e=g \\ \tag{4.2} e' *g &= g*e' = g \end{align}\] Then putting \(g=e\) in (4.2) we get \(e = e *e'\) and putting \(g=e'\) in (4.1) we get \(e*e'=e'\). So \(e = e*e' = e'\).

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.

Proposition 4.2 Suppose \((G,*)\) is a group, \(g \in G\). Then g has only one inverse element.
Proof. Suppose h and k are inverses of g, so that in particular \(h*g=e\) and \(g*k=e\). Then \((h*g)*k = e*k = k\), but \(h*(g*k) = h*e = h\). But the associativity law tells us \((h*g)*k=h*(g*k)\), which says \(k=h\).

So 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 \[ g*g^{-1} = g^{-1}*g = g.\]

Proposition 4.3 Let \((G,*)\) be a group and let \(g \in G\). Then \((g^{-1})^{-1} = g\), that is, the inverse of \(g^{-1}\) is g.
Proof. To show that g is the inverse of \(g^{-1}\) we have to check that \(g* g^{-1}=g^{-1} *g = e\). But this is true because \(g^{-1}\) is the inverse of 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.

Proposition 4.4 Let \((G,*)\) be a group and \(g,h \in G\). Then \((g*h)^{-1} = h^{-1}* g^{-1}\).
Proof. We are going to start by showing that \((h^{-1}*g^{-1})*(g*h) = e\). We are free to bracket the product on the left in whichever way we like as discussed in the previous remark, so \[\begin{equation} \tag{4.3} (h^{-1}*g^{-1})*(g*h) = h^{-1}*(g^{-1}*g)*h = h^{-1} * e * h = h^{-1}*h = e.\end{equation}\] If \(x*y=e\) then multiplying on the right by \(y^{-1}\) gives \(x*y*y^{-1}=e*y^{-1}\) so \(x = y^{-1}\). Applying that to (4.3) gives that \(h^{-1}*g^{-1} = (g*h)^{-1}\) as required.
There is some special notation for what happens when you multiply a group element by itself some number of times.
Definition 4.2 Let \(n \in \mathbb{Z}\), let \((G,*)\) be a group and let \(g \in G\). Then we define \(g^n\) as follows: \[ g^n = \left\{ \begin{array}{ll} g * g* \cdots * g & n>0 \\ g^{-1} * g^{-1} * \cdots * g^{-1} & n<0 \\ e & n=0 \end{array} \right. \] where in the first case there are n copies of g in the product and in the second there are \(-n\) copies of \(g^{-1}\), so that \(g^{n}=(g^{-1})^{-n}\).
These exponents behave exactly how you would expect them to.

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.

Definition 4.3 Let \((G,*)\) be a group. If G is a finite set with exactly n different elements we write \(|G|=n\) and say G has order n, and that G is a finite group. Otherwise we say that \((G,*)\) is an infinite group.

References

Green, J.A. 1988. Sets and Groups : A First Course in Algebra. Library of Mathematics. Springer Netherlands.