Subgroups

A subgroup $H$ of $G$ is a subset $H \subseteq G$ that is itself a group under the operation of $G$.

Theorem 2.1 (Subgroup Criterion). A non-empty subset $H \subseteq G$ is a subgroup if and only if For all $a, b \in H$:

1. $a * b^{-1} \in H$ (closed under the operation and inverses).

Proof. If $H$ is a subgroup, closure gives $a * b^{-1} \in H$. Conversely, since $H$ is non-empty, Pick $a \in H$. Then $a * a^{-1} = e \in H$And $e * a^{-1} = a^{-1} \in H$. For any $a, b \in H$ We have $b^{-1} \in H$So $a * (b^{-1})^{-1} = a * b \in H$Proving closure under the group operation. $\blacksquare$

Corollary 2.2 (Finite Subgroup Criterion). A non-empty finite subset $H \subseteq G$ is a subgroup If and only if $H$ is closed under the group operation.

Example. For any group $G$, $\{e\}$ and $G$ itself are subgroups (the trivial subgroups).

Example. The set $SL_n(\mathbb{R})$ of $n \times n$ real matrices with determinant $1$ is a subgroup of $GL_n(\mathbb{R})$.

Example. The set $A_n$ of even permutations in $S_n$ is a subgroup called the alternating group. $|A_n| = n!/2$.

The center of $G$ is $Z(G) = \{z \in G : zg = gz \mathrm{\ for\ all\ } g \in G\}$.

Proposition 2.3. $Z(G)$ is a subgroup of $G$.

Proof. $e \in Z(G)$. If $z_1, z_2 \in Z(G)$ and $g \in G$Then $(z_1 z_2)g = z_1(z_2 g) = z_1(g z_2) = (z_1 g)z_2 = (g z_1)z_2 = g(z_1 z_2)$So $z_1 z_2 \in Z(G)$. Also, $z g = g z$ implies $g = z^{-1} g z$So $g z^{-1} = z^{-1} g$Giving $z^{-1} \in Z(G)$. $\blacksquare$

For $g \in G$The cyclic subgroup generated by $g$ is

$\langle g \rangle = \{g^n : n \in \mathbb{Z}\}$

Theorem 2.4. Every subgroup of a cyclic group is cyclic.

Proof. Let $G = \langle a \rangle$ and $H \leq G$. If $H = \{e\}$Then $H = \langle e \rangle$. Otherwise, let $m$ be the smallest positive integer with $a^m \in H$. We claim $H = \langle a^m \rangle$. For any $a^k \in H$Write $k = qm + r$ with $0 \leq r \lt m$. Then $a^r = a^{k - qm} = a^k (a^m)^{-q} \in H$. By minimality of $m$, $r = 0$So $a^k = (a^m)^q \in \langle a^m \rangle$. $\blacksquare$

Theorem 2.5. If $G = \langle a \rangle$ has order $n$Then $|\langle a^k \rangle| = n / \gcd(n, k)$.

Problem. Let $G = GL_2(\mathbb{R})$ and $H = \{A \in G : \det(A) = \pm 1\}$. Show that $H \leq G$.

Problem. Let $G = (\mathbb{Z}, +)$ and $H = \{5k - 3m : k, m \in \mathbb{Z}\}$. Show that $H = \mathbb{Z}$.

Theorem 2.6. If $\{H_i\}_{i \in I}$ is a family of subgroups of $G$Then $\bigcap_{i \in I} H_i$ is a subgroup of $G$.

Proof. Since $e \in H_i$ for all $i$We have $e \in \bigcap_{i \in I} H_i$So the intersection is Non-empty. If $a, b \in \bigcap_{i \in I} H_i$Then $a, b \in H_i$ for all $i$So $ab^{-1} \in H_i$ For all $i$ (since each $H_i$ is a subgroup). Thus $ab^{-1} \in \bigcap_{i \in I} H_i$. By the subgroup criterion, $\bigcap_{i \in I} H_i \leq G$. $\blacksquare$

Remark. The union of subgroups need not be a subgroup. For example, in $\mathbb{Z}$ $\langle 2 \rangle \cup \langle 3 \rangle$ is not a subgroup since $2 + 3 = 5 \notin \langle 2 \rangle \cup \langle 3 \rangle$.

Let $S \subseteq G$ be any subset (possibly empty). The subgroup generated by $S$ is

$\langle S \rangle = \bigcap\{H \leq G : S \subseteq H\}$

This is the smallest subgroup of $G$ containing $S$And it equals the set of all finite products Of elements of $S$ and their inverses:

$\langle S \rangle = \{s_1^{\epsilon_1} s_2^{\epsilon_2} \cdots s_k^{\epsilon_k} : s_i \in S,\ \epsilon_i \in \{1, -1\},\ k \geq 0\}$

When $S = \{g_1, \ldots, g_n\}$We write $\langle g_1, \ldots, g_n \rangle$.

Example. In $S_3$, $\langle (1\ 2), (2\ 3) \rangle = S_3$ since $(1\ 2)(2\ 3) = (1\ 2\ 3)$, $(1\ 2\ 3)(1\ 2) = (1\ 3)$And we obtain all six elements.

Example. In $D_4$, $\langle r^2, s \rangle = \{e, r^2, s, r^2s\} \cong V_4$ (the Klein four-group).

Problem. Show that $S_n = \langle (1\ 2), (1\ 2\ \ldots\ n) \rangle$.