Subgroups

Consider the set of nonzero real numbers, \({\mathbb R}^*\), with the group operation of multiplication. \(\mathbb R^*\) the multiplicative group of real numbers The identity of this group is \(1\) and the inverse of any element \(a \in {\mathbb R}^*\) is just \(1/a\). We will show that \[\begin{aligned}\end{aligned}\] is a subgroup of \({\mathbb R}^*\). \(\mathbb Q^*\) the multiplicative group of rational numbers The identity of \({\mathbb R}^*\) is \(1\); however, \(1 = 1/1\) is the quotient of two nonzero integers. Hence, the identity of \({\mathbb R}^*\) is in \({\mathbb Q}^*\). Given two elements in \({\mathbb Q}^*\), say \(p/q\) and \(r/s\), their product \(pr/qs\) is also in \({\mathbb Q}^*\). The inverse of any element \(p/q \in {\mathbb Q}^*\) is again in \({\mathbb Q}^*\) since \((p/q)^{-1} = q/p\). Since multiplication in \({\mathbb R}^*\) is associative, multiplication in \({\mathbb Q}^*\) is associative.

Recall that \({\mathbb C}^{\ast}\) is the multiplicative group of nonzero complex numbers. Let \(H = \{ 1, -1, i, -i \}\). Then \(H\) is a subgroup of \({\mathbb C}^{\ast}\). It is quite easy to verify that \(H\) is a group under multiplication and that \(H \subset {\mathbb C}^{\ast}\).