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

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

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:

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

A subgroup is a subset of a group that is itself a group under the same operation. The subgroup criterion, $ab^{-1} \in H$ for all $a, b \in H$, packages closure, identity, and inverses into one condition. The cyclic subgroup generated by an element $g$ consists of all powers of $g$: it is the smallest subgroup containing $g$, like the orbit of a single rotation. Lagrange’s theorem says the size of any subgroup divides the size of the whole group, which severely limits the possible subgroup structures. Normal subgroups are the ones that are compatible with coset multiplication, enabling quotient groups, the algebraic analogue of factoring out by a symmetry.

Mistake 1: Forgetting to check closure under inverses when verifying subgroups The subgroup criterion requires $ab^{-1} \in H$ for all $a, b \in H$. Students sometimes only check closure under the operation ($ab \in H$) and forget to verify that inverses exist in $H$. For example, $\{1, 2, 3\}$ under multiplication mod $4$ is closed but not a subgroup because $2$ has no inverse. Always check the full criterion.

Mistake 2: Assuming the identity element is obvious When proving a subset is a subgroup, explicitly verify that the identity element of $G$ belongs to $H$. This is part of the non-empty requirement in the subgroup criterion. For example, $\{x \in \mathbb{R} : x > 0\}$ under addition does not contain $0$ and is not a subgroup of $(\mathbb{R}, +)$.

Mistake 3: Confusing quotient groups with cosets A coset $gH$ is a set of elements, not a group (unless $H$ is normal). The quotient group $G/H$ consists of cosets as elements, but its group operation is only well-defined when $H \trianglelefteq G$. Never assume a set of cosets forms a group without first verifying normality of $H$.

• Groups: Every subgroup is itself a group under the inherited operation.

• Lagrange’s Theorem: Lagrange’s theorem shows the order of a subgroup divides the order of the group.

• Normal Subgroups and Quotient Groups: Normal subgroups are those for which the quotient group construction is well-defined.

• Classical Mechanics

• Electromagnetism

• Additional Results
• Classification of Groups of Small Order
• Common Pitfalls
• Euclidean Domains, PIDs, and UFDs