2.5 Subgroup generated by a set
In this section, let’s let \(G\) be a group with identity element \(e\).
Suppose \((H_i, i\in I)\) is a family of subgroups of \(G\). Then \(H:=\cap _{i\in I} H_i\) is a subgroup of \(G\).
Since \(H_i\leq G\) for each \(i\), \(e\in H_i\) for each \(i\). Therefore, \(e\in H\). Suppose \(x,y\in H\). Then \(xy^{-1}\in H_i\) for all \(i\). Therefore \(xy^{-1}\in H\).
The set \(I\) in Theorem 2.36 need not be finite. For example, for each positive integer \(n\) set \(H_n := n\mathbb {Z} = \{ nx: x\in \mathbb {Z}\} \). Then \(H_n \leq \mathbb {Z}\) for all \(n\in \mathbb {P}\). It’s easy to see that
the trivial subgroup of \(\mathbb {Z}\).
Suppose \(G\) is a group and \(S\) is a subset of \(G\). The subgroup \(\langle S\rangle \) of \(G\) generated by \(S\) is the intersection of all subgroups of \(G\) containing \(S\).
If \(S=\{ g_1,\ldots , g_k\} \), we abuse notation and write \(\langle g_1,\ldots , g_k\rangle \) for \(\langle S\rangle \), which is said to be generated by the elements \(g_1,\ldots g_k\). A subgroup of \(G\) is called cyclic if it can be generated by a single element.
Suppose \(S\) is a subset of a group \(G\) with identity element \(1\). Set \(T = S \cup S^{-1}\cup \{ 1\} \), and set \(H = \cup _{n\geq 0} T^n\). Then \(H=\langle S\rangle \).
First, let’s show that \(H\) is a subgroup of \(G\). Clearly, \(e\in H\). Suppose \(x=g_1\ldots g_r\) and \(y=h_1\ldots h_s\) are in \(H\). Then \(xy^{-1}= g_1\ldots g_r h_s^{-1}h_{s-1}^{-1}\ldots h_1^{-1}\) is of the same form. It follows that \(H\leq G\). Clearly, \(S\subset H\). So, since \(\langle S\rangle \) is the intersection of all subgroups of \(G\) containing \(S\), \(\langle S\rangle \leq H\).
Suppose \(K\) is a subgroup of \(G\) containing \(S\). Then any element \(g\) of the form \(t_1\ldots t_r\) with \(t_i\in T\) is in \(K\). Therefore any such element is in \(\langle S\rangle \). So \(H\leq \langle S\rangle \). Therefore \(H=\langle S\rangle \).