Proposition 8

If \(\mathcal{A}\) is any nonempty collection of subgroups of \(G\), then the intersection of all members of \(\mathcal{A}\) is also a subgroup of \(G\).

Proof

Let

$$ K = \bigcap_{H \in \mathcal{A}} H. $$

Each \(H \in \mathcal{A}\) is a group, so \(1\) belongs to each \(H \in \mathcal{A}\). Therefore, \(K \ne \emptyset\). Suppose \(a, b \in K\). Then \(a, b \in H\) for every \(H\). Since every \(H\) is a group, \(ab^{-1} \in H\) for every \(H\). Hence, \(ab^{-1} \in K\).

By the Subgroup Criterion, \(K\) is a subgroup of \(G\).

$$\tag*{$\blacksquare$}$$

Subgroup generated by a subset

If \(A\) is any subset of the group \(G\), define

$$ \langle A \rangle = \bigcap_{\substack{A \subseteq H \\ H \le G}} H. $$

This is called the subgroup of \(G\) generated by \(A\). We write

• \(\langle a_1, a_2, …, a_n \rangle\) instead of \(\langle \{a_1, a_2, …, a_n\} \rangle\) for \(A = \{a_1, a_2, …, a_n\}\)
• \(\langle A, B \rangle\) instead of \(\langle A \cup B \rangle\) if \(A\) and \(B\) are two subsets of \(G\).

Proposition 9

Let \(G\) be a group and \(A\) be any subset of \(G\). Let \(\overline A\) be the set of all finite products of elements of \(A\) and inverses of elements of \(A\). More precisely,

$$ \overline A = \{a_1^{\epsilon_1}a_2^{\epsilon_2}…a_n^{\epsilon_n} \mid n \in \mathbb{Z}, n \ge 0 \text{ and } a_i \in A, \epsilon_i = \pm 1 \text{ for each }i\} $$

where \(\overline A = \{1\}\) if \(A = \emptyset\). Then \(\overline A = \langle A \rangle\).

Proof

We first prove \(\overline A\) is a subgroup. By definition of \(\overline A\), \(\overline A\) is always nonempty, even if \(A\) is empty. Suppose \(a, b \in \overline A\) with \(a = a_1^{\epsilon_1}a_2^{\epsilon_2}…a_n^{\epsilon_n}\) and \(b = b_1^{\delta_1}b_2^{\delta_2}…b_m^{\delta_m}\). Then

$$ ab^{-1} = a_1^{\epsilon_1}a_2^{\epsilon_2}…a_n^{\epsilon_n} \cdot b_m^{-\delta_m}b_{m-1}^{-\delta_{m-1}}…b_1^{-\delta_1}. $$

That is, \(ab^{-1}\) is a product of elements of \(A\) raised to powers \(\pm 1\). Therefore, \(ab^{-1} \in \overline A\). By the Subgroup Criterion, \(\overline A\) is a subgroup of \(G\).

Every \(a \in A\) can be written as \(a^1\), so \(A \subseteq \overline A\). Therefore, \(\overline A\) is a subgroup of \(G\) that contains \(A\). Hence, \(\langle A \rangle \subseteq \overline A\).

\(\langle A \rangle\) is a subgroup containing \(A\), and is closed under the group operation and the process of taking inverses. Therefore, \(\langle A \rangle\) contains each element of the form \(a_1^{\epsilon_1}a_2^{\epsilon_2}…a_n^{\epsilon_n}\). Hence, \(\overline A \subseteq \langle A \rangle\).

Thus, \(\overline A = \langle A \rangle\).

$$\tag*{$\blacksquare$}$$