Skip to main content

Section 1.4 Multiplication table

Subsection 1.4.1 The idea

So far we studied various examples of groups by giving explicit realizations of them. But in the end groups can be understood entirely abstractly, without referring to a particular realization of it. In this section we will focus on finite groups.

To specify a finite group all that we need to know is the number of elements in the set, and the result of multiplying these elements together. This is neatly encoded in the form of a multiplication table. Suppose that \(G\) is a finite group of order \(n\text{.}\) Then its multiplication table will be a square \(n \times n\) array with rows and columns labelled by elements of the group, and entries corresponding to the products of these elements:

\(\cdots\) \(g_j\) \(\cdots\)
\(\vdots\) \(\ddots\)
\(g_i\) \(\) \(g_i g_j\)
\(\vdots\) \(\ddots\)
Table 1.4.1. An example of a multiplication table

Subsection 1.4.2 The rearrangement theorem

Before we construct multiplication tables, an interesting fact to note is that each element of a group can only appear once and only once in each row, and same for each column. This is sometimes formally stated as the “rearrangement theorem”:

Suppose that two elements of \(g_k G\) are equal: \(g_k g_i = g_k g_j\) for \(i \neq j\text{.}\) Then multiply by \(g_k^{-1}\) on the left to get \(g_i = g_j\text{.}\) This is a contradiction, since \(i \neq j\text{.}\) Thus all elements of \(g_k G\) must be distinct. Since \(|g_k G| = n = |G|\text{,}\) it follows that all elements of \(G\) must appear once and only once in \(g_k G\text{.}\) Thus the sets \(G\) and \(g_k G\) are identical up to reordering of elements, hence the name of the theorem.

Subsection 1.4.3 Groups of order 2 and 3

With this under our belt let us start by constructing the multiplication table for a group of order two. Let us write \(G = \{e, a\}\text{,}\) where \(e\) is the identity element. We know that \(e^2 = e\text{,}\) \(e a = a\) and \(a e = a\text{.}\) Since each element can appear once and only once in each row and column of the multiplication table, it follows that we must have \(a^2 = e\text{,}\) to get:

\(e\) \(a\)
\(e\) \(e\) \(a\)
\(a\) \(a\) \(e\)
Table 1.4.3. The multiplication table for the group of order 2

Let us now move to groups with three elements \(G = \{e,a,b\}\text{.}\) We know by definition of the identity that

\(e\) \(a\) \(b\)
\(e\) \(e\) \(a\) \(b\)
\(a\) \(a\)
\(b\) \(b\)
Table 1.4.4. Constructing the multiplication table for the group of order 3

Further, the product \(a b\) cannot be equal to \(a\) (resp. \(b\)), since otherwise we could multiply by \(a^{-1}\) (resp. \(b^{-1}\)) on the left to conclude that \(b=e\) (resp. \(a=e\)). Thus we must have \(a b = e\text{.}\) We then complete the table using the requirement that each element appears once and only once in each row and column (this is like sudoku! :-) ), to conclude that

\(e\) \(a\) \(b\)
\(e\) \(e\) \(a\) \(b\)
\(a\) \(a\) \(b\) \(e\)
\(b\) \(b\) \(e\) \(a\)
Table 1.4.5. Multiplication table for the group of order 3

Again, this is the unique possibility. Hence there is a unique abstract group with three elements. Further, it is abelian, since the table is symmetric about its diagonal.

Going to higher order however we lose uniqueness. For groups with four elements, there are two distinct multiplication tables (corresponding to the groups \(\mathbb{Z}_4\) and \(\mathbb{Z}_2 \times \mathbb{Z}_2\text{,}\) see Example 1.3.4). And for higher order groups there are many possibilities. I will not construct multiplication tables for groups of higher order, but if you like this kind of thing feel free to do it yourself as an exercise. Try for instance the symmetric group \(S_3\text{,}\) which has order \(3! = 6\text{.}\) Lots of fun!

login