Groups | Mathematics - Wyatt's Notes

A group is a set $G$ together with a binary operation $* : G \times G \to G$ satisfying:

1. Associativity: $(a * b) * c = a * (b * c)$ for all $a, b, c \in G$.
2. Identity: There exists $e \in G$ such that $e * a = a * e = a$ for all $a \in G$.
3. Inverse: For each $a \in G$ There exists $a^{-1} \in G$ such that $a * a^{-1} = a^{-1} * a = e$.

If $a * b = b * a$ for all $a, b \in G$ The group is abelian (or commutative).

Example 1. $(\mathbb{Z}, +)$ is an abelian group with identity $0$ and inverse $-n$.

Example 2. $(\mathbb{R}^*, \cdot)$ (non-zero reals under multiplication) is an abelian group with identity $1$ and inverse $1/a$.

Example 3. The symmetric group $S_n$ of all permutations of $\{1, 2, \ldots, n\}$ under composition is a non-abelian group for $n \geq 3$. It has order $n!$.

Example 4. $(\mathbb{Z}/n\mathbb{Z}, +)$ is an abelian group of order $n$ Called the cyclic group of order $n$.

Example 5. The set $GL_n(\mathbb{R})$ of all $n \times n$ invertible real matrices, under matrix multiplication, forms the general linear group.

The dihedral group $D_n$ is the group of symmetries of a regular $n$-gon (for $n \geq 3$). It consists of $n$ rotations and $n$ reflections, giving $|D_n| = 2n$ elements.

Let $r$ denote rotation by $2\pi/n$ and $s$ denote any fixed reflection. Then:

The group satisfies the presentation:

For $n \geq 3$, $D_n$ is non-abelian (e.g., $rs \neq sr$ since $sr = r^{-1}s \neq rs$ when $n > 2$).

Example 1.6. $D_3$ (symmetries of an equilateral triangle) has $6$ elements: $\{e, r, r^2, s, rs, r^2s\}$. The map sending each symmetry to its permutation of the three vertices Gives an isomorphism $D_3 \cong S_3$. The rotations $\{e, r, r^2\}$ correspond to $A_3$ And the three Reflections correspond to the three transpositions.

Example 1.7. $D_4$ (symmetries of a square) has $8$ elements. Its center is $Z(D_4) = \{e, r^2\}$ Where $r^2$ is the $180°$ rotation. The subgroups of $D_4$ include the rotation subgroup $\langle r \rangle \cong \mathbb{Z}/4\mathbb{Z}$ of order $4$ Four subgroups of order $2$ generated By reflections, and $\langle r^2, s \rangle \cong V_4$ (the Klein four-group).

Every element of $S_n$ can be written uniquely (up to reordering of disjoint cycles) as a product Of disjoint cycles. The cycle type of a permutation is the multiset of its cycle lengths (including $1$-cycles).

Example. In $S_5$ The permutation $\sigma = (1\ 2\ 3)(4\ 5)$ has cycle type $(3, 2)$.

A transposition is a $2$-cycle $(i\ j)$. Every permutation factors into transpositions:

The number of transpositions in a factorization is not unique, but its parity is.

Proposition 1.6. The sign of a permutation $\sigma$ Defined by $\mathrm{sgn}(\sigma) = (-1)^{N(\sigma)}$ Where $N(\sigma)$ is the number of inversions of $\sigma$ Is a well-defined group homomorphism $\mathrm{sgn} : S_n \to \{1, -1\}$. Its kernel is the alternating group $A_n$ of even permutations.

Proof. We first show $\mathrm{sgn}$ is a homomorphism. For any transposition $\tau$ and any Permutation $\sigma$ Multiplying by $\tau$ changes the number of inversions by an odd number, So $\mathrm{sgn}(\tau\sigma) = -\mathrm{sgn}(\sigma)$. For any $\sigma, \rho \in S_n$ Write $\rho$ As a product of $k$ transpositions. Then $\mathrm{sgn}(\sigma\rho) = (-1)^k\mathrm{sgn}(\sigma) = \mathrm{sgn}(\sigma)\mathrm{sgn}(\rho)$. The kernel is precisely the set of even permutations, and $|A_n| = n!/2$. $\blacksquare$

Proposition 1.7. The number of $k$-cycles in $S_n$ is $\frac{n!}{k(n-k)!}$.

Proof. Choose $k$ elements from $n$: $\binom{n}{k}$ ways. Arrange them in a cycle: $(k-1)!$ distinct $k$-cycles (since a $k$-cycle has $k$ equivalent representations by cyclic rotation). Total: $\binom{n}{k}(k-1)! = \frac{n!}{k(n-k)!}$. $\blacksquare$

Proposition 1.8. The identity element of a group is unique.

Proof. Suppose $e$ and $e"$ are both identity elements. Then $e = e * e' = e'$. $\blacksquare$

Proposition 1.9. The inverse of each element is unique.

Proof. If $b$ and $c$ are both inverses of $a$ Then $b = b * e = b * (a * c) = (b * a) * c = e * c = c$. $\blacksquare$

Proposition 1.10 (Cancellation Law). If $a * b = a * c$ Then $b = c$. Similarly, if $b * a = c * a$ Then $b = c$.

Proof. Multiply on the left by $a^{-1}$: $a^{-1} * (a * b) = a^{-1} * (a * c)$ So $(a^{-1} * a) * b = (a^{-1} * a) * c$ Giving $e * b = e * c$ I.e., $b = c$. $\blacksquare$

Proposition 1.11. $(a^{-1})^{-1} = a$ and $(a * b)^{-1} = b^{-1} * a^{-1}$.

The order of an element $g \in G$ Denoted $|g|$ Is the smallest positive integer $n$ such that $g^n = e$. If no such $n$ exists, $|g| = \infty$.

Proposition 1.12. $g^k = e$ if and only if $|g|$ divides $k$.

Proof. Write $k = q|g| + r$ with $0 \leq r \lt |g|$. Then $e = g^k = g^{q|g| + r} = (g^{|g|})^q * g^r = e^q * g^r = g^r$. Since $r \lt |g|$ and $|g|$ is the smallest positive exponent giving $e$ We must have $r = 0$ So $|g|$ divides $k$. Conversely, if $|g|$ divides $k$ Say $k = m|g|$ Then $g^k = (g^{|g|})^m = e^m = e$. $\blacksquare$

The subgroup lattice of $G$ is the set of all subgroups of $G$ Partially ordered by inclusion, Visualized as a Hasse diagram (edges connect each subgroup to its immediate supergroups).

Example. The subgroup lattice of $S_3$ (order $6$):

• $S_3$ (order $6$)
• $A_3 = \langle (1\ 2\ 3) \rangle$ (order $3$)
• $\{e\}$ (order $1$)
• $\langle (1\ 2) \rangle$ (order $2$)
• $\{e\}$
• $\langle (1\ 3) \rangle$ (order $2$)
• $\{e\}$
• $\langle (2\ 3) \rangle$ (order $2$)
• $\{e\}$

The only proper non-trivial normal subgroup is $A_3$ (it has index $2$).

Example. The subgroup lattice of $\mathbb{Z}/12\mathbb{Z}$:

• $\mathbb{Z}/12\mathbb{Z}$ (order $12$)
• $\langle 2 \rangle \cong \mathbb{Z}/6\mathbb{Z}$ (order $6$)
• $\langle 4 \rangle \cong \mathbb{Z}/3\mathbb{Z}$ (order $3$)
• $\{0\}$
• $\langle 6 \rangle \cong \mathbb{Z}/2\mathbb{Z}$ (order $2$)
• $\{0\}$
• $\langle 3 \rangle \cong \mathbb{Z}/4\mathbb{Z}$ (order $4$)
• $\langle 6 \rangle \cong \mathbb{Z}/2\mathbb{Z}$ (order $2$)
• $\{0\}$
• $\langle 4 \rangle \cong \mathbb{Z}/3\mathbb{Z}$ (order $3$)
• $\{0\}$
• $\langle 6 \rangle \cong \mathbb{Z}/2\mathbb{Z}$ (order $2$)
• $\{0\}$

By Theorem 2.4, every subgroup of $\mathbb{Z}/12\mathbb{Z}$ is cyclic, and there is exactly one subgroup Of order $d$ for each divisor $d$ of $12$.

A symmetry of an object is a transformation that preserves the object’s structure. Groups capture symmetry because the composition of two symmetries is again a symmetry (closure), doing nothing is a symmetry (identity), and every symmetry can be undone (inverse).

Example: The equilateral triangle. The equilateral triangle has six symmetries: three rotations ($0°$, $120°$, $240°$) and three reflections (one through each altitude). These symmetries form the group $D_3 \cong S_3$. The group multiplication table (Cayley table) is:

Reading the table: The entry in row $a$, column $b$ is $a \cdot b$. For example, $r \cdot s = rs$ and $s \cdot r = r^2s$. Since $rs \neq r^2s$, the group is non-abelian.

Example: The square. The square has eight symmetries forming $D_4$: four rotations ($0°$, $90°$, $180°$, $270°$) and four reflections. The center of $D_4$ is $Z(D_4) = \{e, r^2\}$, where $r^2$ is the $180°$ rotation. The quotient $D_4/Z(D_4) \cong V_4$ (the Klein four-group), reflecting the fact that modding out by the central $180°$ rotation collapses the four rotations to two elements.

Why abelian groups are simpler. In an abelian group, the Cayley table is symmetric across the main diagonal (since $a \cdot b = b \cdot a$). Every subgroup of an abelian group is normal, so the quotient construction always works. The structure theory of finite abelian groups (every finite abelian group is a direct product of cyclic groups) is much simpler than the classification of all finite groups.

The cyclic group $\mathbb{Z}/4\mathbb{Z} = \{0, 1, 2, 3\}$ under addition modulo 4 has the following Cayley table:

This table is symmetric across the main diagonal (the group is abelian). Each row and column is a permutation of $\{0, 1, 2, 3\}$ (the Latin square property, which holds for all groups). The element $1$ generates the entire group: $\langle 1 \rangle = \{0, 1, 2, 3\} = \mathbb{Z}/4\mathbb{Z}$.

• The order of a group is the number of elements, not the order of any particular element. $|S_3| = 6$ but the elements have orders $1, 2, 2, 2, 3, 3$.
• A group of order $n$ need not be cyclic. The Klein four-group $V_4 = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ has order 4 but is not cyclic (every non-identity element has order 2).
• The Cayley table determines the group up to isomorphism. Two groups with the same Cayley table (up to relabeling of elements) are isomorphic.
• Not every subset of a group is a subgroup. The subset must contain the identity, be closed under the operation, and contain inverses. For example, $\{1, 2\} \subset \mathbb{Z}/4\mathbb{Z}$ is not a subgroup because $1 + 1 = 2 \notin \{1, 2\}$ is fine, but $2 + 2 = 0 \notin \{1, 2\}$.

• Subgroups: Subgroups are subsets of a group that inherit the group structure.

• Lagrange’s Theorem: Lagrange’s theorem constrains the order of subgroups based on the order of the group.

• Homomorphisms and Isomorphism Theorems: Homomorphisms preserve group structure and lead to the isomorphism theorems.

• Quantum Mechanics

• Graph Theory

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