Theorem: The intersection of subgroups \(H_1, H_2, ....\) is a subgroup of each of \(H_1, H_2, ...\)

We say the elements \(g_1,...,g_m\) are independent if none of them can be expressed in terms of the others, that is, \(g_i \notin \langle g_1,...,g_{i-1},g_{i+1},...,g_m \rangle\). Clearly every finite group has at least one set of independent generators. Independent elements can have relations between them, e.g. if \(a,b\) are independent then we may have \((a b)^2 = 1\) for example. Such a relation is called a defining relation.

Given any two groups \(G, H\) we may form their direct product \(G\times H\), whose elements are pairs \((g,h)\) with \(g\in G, h\in H\), and the group operation applies coordinatewise. The direct product of abelian groups is abelian.

Suppose every element of a group \(F\) has the form \(g h\) where \(g\in G, h\in H\) for some subgroups \(G, H\) of \(F\), and furthermore, suppose every element of \(G\) commutes with every element of \(H\) and \(G \cap H = \{1\}\). Then \(F \cong G\times H\).

It is clear how to generalize this to define the direct product to \(k\) groups.

Example: \(\mathbb{Z}_{15}^* \cong \mathbb{Z}_4 \times \mathbb{Z}_2\).

Ben Lynn 💡