Groups

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:

$D_n = \{e, r, r^2, \ldots, r^{n-1}, s, rs, r^2s, \ldots, r^{n-1}s\}$

The group satisfies the presentation:

$D_n = \langle r, s \mid r^n = s^2 = e,\ srs = r^{-1} \rangle$

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: $(a_1\ a_2\ \ldots\ a_k) = (a_1\ a_k)(a_1\ a_{k-1})\cdots(a_1\ a_2)$

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 $12$.

:::caution Common Pitfall Not every group of order $n$ has a subgroup of order $d$ for each divisor $d$ of $n$. The converse of Lagrange’s theorem is false. For example, $A_4$ has order $12$ but no subgroup of order $6$. However, every subgroup of a cyclic group of order $n$ has order dividing $n$And for each divisor There is exactly one such subgroup.

:::