Unit 2: Groups, Subgroups, Cyclic Structures & Element Orders

1. The Notion of a Subgroup

Let $(G, *)$ be a group.

Definition 2.1 (Subgroup): A subset $H \subseteq G$ is called a subgroup of $G$, denoted by $H \le G$, if $H$ is itself a group under the operation of $G$. If $H \le G$ and $H \ne G$, we write $H < G$ ($H$ is a proper subgroup). The trivial subgroups of any group $G$ are $\{e\}$ and $G$ itself.

2. Subgroup Testing Criteria

Rather than verifying all four group axioms, we can determine subgroup status using specialized, concise tests:

Theorem 2.1 (Two-Step Subgroup Test): A non-empty subset $H \subseteq G$ is a subgroup of $G$ if and only if:

1. Closure under operation: $\forall a, b \in H \implies a * b \in H$.
2. Closure under inverses: $\forall a \in H \implies a^{-1} \in H$.

Theorem 2.2 (One-Step Subgroup Test): A non-empty subset $H \subseteq G$ is a subgroup of $G$ if and only if:

Complete Proof of the One-Step Test:

$(\implies)$ If $H \le G$, then for any $b \in H$, $b^{-1} \in H$ by inverse axiom. By closure, $a * b^{-1} \in H$.

$(\impliedby)$ Suppose $H \ne \emptyset$ and $a * b^{-1} \in H$ for all $a, b \in H$.

1. Identity: Since $H \ne \emptyset$, choose any element $x \in H$. Setting $a = x$ and $b = x$ gives:

Thus the identity of $G$ belongs to $H$.

2. Inverses: For any $x \in H$, setting $a = e$ and $b = x$ gives:

Thus every element in $H$ has its inverse in $H$.

3. Closure: For any $x, y \in H$, since $y \in H \implies y^{-1} \in H$. Setting $a = x$ and $b = y^{-1}$ gives:

4. Associativity: Inherited automatically from $G$ since $H \subseteq G$.

Therefore, $H \le G$. $\blacksquare$

Theorem 2.3 (Finite Subgroup Test): Let $H$ be a non-empty finite subset of a group $G$. Then $H \le G$ if and only if $H$ is closed under multiplication:

(Inverses are guaranteed automatically by pigeonhole finiteness: the sequence $a, a^2, a^3, \dots$ must eventually repeat, generating $a^{-1} = a^{k-1}$!)

3. Centralizers, Normalizers and the Center

Let $G$ be a group.

• The Center of $G$, $Z(G)$: The set of elements that commute with every element of $G$:

• The Centralizer of an Element $a \in G$, $C_G(a)$:

• The Normalizer of a Subset $S \subseteq G$, $N_G(S)$:

Theorem: $Z(G)$, $C_G(a)$, and $N_G(S)$ are always subgroups of $G$.

1. Order of an Element

Let $G$ be a group and $a \in G$.

Definition 2.2 (Order of an Element): The order of an element $a \in G$, denoted by $|a|$ or $\text{ord}(a)$, is the smallest positive integer $n \in \mathbb{Z}^+$ such that:

If no such positive integer exists, $a$ is said to have infinite order ($|a| = \infty$).

Theorem 2.4 (Properties of Element Order):

Let $a \in G$ with $|a| = n$.

1. $a^k = e$ if and only if $n$ divides $k$ ($n \mid k$).
2. $a^i = a^j$ if and only if $i \equiv j \pmod n$.
3. For any integer $k \in \mathbb{Z}$, the order of $a^k$ is given by:

2. Cyclic Groups

Definition 2.3 (Cyclic Group): A group $G$ is called cyclic if there exists an element $a \in G$ such that every element of $G$ is an integral power of $a$:

The element $a$ is called a generator of $G$.

Theorem 2.5 (Abelian Nature of Cyclic Groups):

Every cyclic group is Abelian. Proof: Let $x, y \in \langle a \rangle$. Then $x = a^r$ and $y = a^s$ for some $r, s \in \mathbb{Z}$.

3. The Fundamental Theorem of Cyclic Groups

Theorem 2.6 (Fundamental Theorem of Cyclic Groups): Let $G = \langle a \rangle$ be a cyclic group of order $n$.

1. Subgroups are cyclic: Every subgroup of $G$ is cyclic.
2. Order of Subgroups: The order of any subgroup $H \le G$ is a divisor of $n$.
3. Existence and Uniqueness: For each positive divisor $d$ of $n$, there exists one and only one subgroup of $G$ of order $d$, given explicitly by:

Complete Proof of Part 1 (Subgroups are Cyclic):

Let $H \le G$. If $H = \{e\}$, then $H = \langle e \rangle$ is cyclic. Suppose $H \ne \{e\}$. Since $G = \langle a \rangle$, every element in $H$ has the form $a^k$ for some $k \in \mathbb{Z}$. Since $H$ is a subgroup, if $a^k \in H$ with $k < 0$, then $(a^k)^{-1} = a^{-k} \in H$ where $-k > 0$. Thus $H$ must contain positive powers of $a$. Let $m$ be the smallest positive integer such that $a^m \in H$ (well-ordering principle of $\mathbb{Z}^+$). We claim that $H = \langle a^m \rangle$.

• Clearly $\langle a^m \rangle \subseteq H$ since $a^m \in H$ and $H$ is closed under powers.
• Conversely, let $h \in H$. Then $h = a^s$ for some $s \in \mathbb{Z}$.

By the Division Algorithm, divide $s$ by $m$:

Then:

Since $a^s \in H$ and $a^m \in H \implies (a^m)^{-q} \in H$, their product $a^r \in H$. Since $0 \le r < m$ and $m$ was defined as the minimal positive integer with $a^m \in H$, we must have $r = 0$. Therefore $s = q m$, which means:

Hence $H \subseteq \langle a^m \rangle$. Thus $H = \langle a^m \rangle$, proving that $H$ is cyclic! $\blacksquare$

1. Counting Generators of a Finite Cyclic Group

Let $G = \langle a \rangle$ be a cyclic group of order $n$. By Theorem 2.4, an element $a^k$ has order:

For $a^k$ to be a generator of $G$, we require $|a^k| = n$, which is true if and only if:

Theorem 2.7 (Number of Generators): A cyclic group of order $n$ has exactly $\phi(n)$ generators, where $\phi$ is Euler's totient function. Specifically, the generators of $\langle a \rangle$ are:

Gauss's Identity:

Every element of $\langle a \rangle$ generates a cyclic subgroup of some divisor order $d \mid n$. Since the number of elements of exact order $d$ is $\phi(d)$, partitioning the group by element orders proves Gauss's classical identity:

2. Subgroup Lattices

The collection of all subgroups of a group $G$, ordered by set inclusion $\subseteq$, forms a complete algebraic lattice called the Subgroup Lattice $\mathcal{L}(G)$.

• In a cyclic group $\mathbb{Z}_n$, by the Fundamental Theorem, the lattice of subgroups is isomorphic to the divisor lattice of $n$, ordered by divisibility:

• In prime-power cyclic groups $\mathbb{Z}_{p^k}$, the subgroup lattice is a strictly linear chain:

1. External Direct Products

Let $G_1, G_2, \dots, G_n$ be groups.

Definition 2.4 (External Direct Product): The external direct product $G_1 \times G_2 \times \dots \times G_n$ is the Cartesian product equipped with component-wise group operations:

• Identity element: $(e_1, e_2, \dots, e_n)$.
• Inverse element: $(g_1, \dots, g_n)^{-1} = (g_1^{-1}, \dots, g_n^{-1})$.

2. Order of Elements in Direct Products

Theorem 2.8 (Order Formula for Direct Products): The order of an element $(g_1, g_2, \dots, g_n) \in G_1 \times G_2 \times \dots \times G_n$ is the least common multiple of the orders of its components:

Proof:

Let $m = |(g_1, \dots, g_n)|$ and $L = \text{lcm}(|g_1|, \dots, |g_n|)$.

Since each $|g_i|$ divides $L$, $g_i^L = e_i$ for all $i$. Thus $(g_1, \dots, g_n)^L = (e_1, \dots, e_n)$. By Theorem 2.4, this implies $m \mid L$. Conversely, $(g_1, \dots, g_n)^m = (e_1, \dots, e_n) \implies g_i^m = e_i$ for all $i$. Hence each $|g_i| \mid m$. By definition of least common multiple, $L \mid m$. Since $m \mid L$ and $L \mid m$ with $m, L > 0$, we have $m = L$. $\blacksquare$

3. Criterion for Cyclicity of Direct Products

Theorem 2.9 (Cyclicity of $\mathbb{Z}_m \times \mathbb{Z}_n$): The direct product $\mathbb{Z}_m \times \mathbb{Z}_n$ is cyclic if and only if $m$ and $n$ are relatively prime:

Proof:

$(\impliedby)$ If $\gcd(m, n) = 1$, consider the element $(1, 1) \in \mathbb{Z}_m \times \mathbb{Z}_n$.

Since the order of $(1, 1)$ equals the order of the group $|\mathbb{Z}_m \times \mathbb{Z}_n| = mn$, $(1, 1)$ is a generator, so the group is cyclic.

$(\implies)$ Suppose $\gcd(m, n) = d > 1$. Then $\text{lcm}(m, n) = \frac{mn}{d} < mn$. For any element $(x, y) \in \mathbb{Z}_m \times \mathbb{Z}_n$, $|(x, y)| = \text{lcm}(|x|, |y|)$. Since $|x| \mid m$ and $|y| \mid n$, $\text{lcm}(|x|, |y|) \mid \text{lcm}(m, n) = \frac{mn}{d}$. Hence no element can have order $mn$. Thus the group cannot be cyclic. $\blacksquare$