10.2 Cosets, Lagrange's Theorem, Normal Subgroups, and Homomorphisms

10.2 Cosets, Lagrange's Theorem, Normal Subgroups, and Homomorphisms

Coset partitions, normal subgroups, and homomorphisms form the foundation of structural group theory. GRE Mathematics Subject Test problems frequently test the consequences of Lagrange's theorem, index calculations, quotient group structures, group center theorems, and the First Isomorphism Theorem.

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Cosets and Equivalence Partitions

Let $G$ be a group and $H \le G$ a subgroup. For any element $g \in G$:

• The left coset of $H$ containing $g$ is $gH = {gh : h \in H}$.
• The right coset of $H$ containing $g$ is $Hg = {hg : h \in H}$.

Fundamental Properties of Cosets

1. Uniform Cardinality: The map $h \mapsto gh$ is a bijection from $H$ to $gH$, so $|gH| = |H|$ for every $g \in G$.
2. Partition of $G$: Define an equivalence relation on $G$ by $a \sim b \iff a^{-1}b \in H$ (or equivalently $b \in aH$). The equivalence classes under this relation are precisely the left cosets of $H$. Therefore, the left cosets partition $G$: any two cosets $aH$ and $bH$ are either identical ($aH = bH$) or completely disjoint ($aH \cap bH = \emptyset$).
3. Coset Equality Criterion: $aH = bH \iff b^{-1}a \in H \iff a \in bH$.
4. Asymmetry: In a non-abelian group, a left coset $gH$ is generally not equal to the right coset $Hg$.

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Lagrange's Theorem and Its Direct Corollaries

Lagrange's Theorem: If $G$ is a finite group and $H \le G$, then the order of $H$ divides the order of $G$. The number of distinct left (or right) cosets of $H$ in $G$ is called the index of $H$ in $G$, denoted $[G : H]$: $|G| = [G : H] \cdot |H|$

Corollaries Frequently Tested on the GRE

1. Element Order Divisibility: For every element $g \in G$, the order $|g| = |\langle g \rangle|$ divides $|G|$.
2. Exponent Identity: For every $g \in G$, $g^{|G|} = e$.
   • In the multiplicative group $(\mathbb{Z}_p^*, \cdot)$ of order $p - 1$, this yields Fermat's Little Theorem: $a^{p-1} \equiv 1 \pmod p$ for $\gcd(a, p) = 1$.
   • In the group of units $(\mathbb{Z}_n^*, \cdot)$ of order $\phi(n)$, this yields Euler's Totient Theorem: $a^{\phi(n)} \equiv 1 \pmod n$ for $\gcd(a, n) = 1$.
3. Groups of Prime Order: If $|G| = p$ (where $p$ is prime), then $G$ has no non-trivial proper subgroups. Every non-identity element $g \neq e$ generates $G$, meaning $G$ is cyclic and isomorphic to $\mathbb{Z}_p$.
4. Converse of Lagrange's Theorem is False: If $d$ divides $|G|$, $G$ does not necessarily have a subgroup of order $d$. The standard counterexample is the alternating group $A_4$: $|A_4| = 12$, and 6 divides 12, but $A_4$ contains no subgroup of order 6.

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Normal Subgroups and Quotient Groups

A subgroup $N \le G$ is a normal subgroup (written $N \trianglelefteq G$) if it is invariant under conjugation by all elements of $G$: $gNg^{-1} = N \quad \text{for all } g \in G$

Equivalent Conditions for Normality

• For every $g \in G$, $gNg^{-1} \subseteq N$.
• Left cosets coincide with right cosets: $gN = Ng$ for all $g \in G$.
• $N$ is the kernel of a group homomorphism with domain $G$.

Subgroups of Index 2 Are Always Normal

If $[G : H] = 2$, then $H \trianglelefteq G$. Proof: The left cosets partition $G$ into $H$ and $G \setminus H$. The right cosets also partition $G$ into $H$ and $G \setminus H$. For any $g \notin H$, $gH = G \setminus H = Hg$. For $g \in H$, $gH = H = Hg$. Thus $gH = Hg$ for all $g \in G$. Example: $A_n \trianglelefteq S_n$ because $[S_n : A_n] = n! / (n!/2) = 2$.

The Quotient Group $G/N$

When $N \trianglelefteq G$, the set of cosets $G/N = {gN : g \in G}$ forms a group under the operation: $(aN)(bN) = (ab)N$

• Well-Definedness: Normality of $N$ is required to ensure that $(an_1)(bn_2) \in abN$ regardless of the chosen coset representatives.
• Group Data: The identity element is $eN = N$, and the inverse is $(aN)^{-1} = a^{-1}N$.
• Order: $|G/N| = [G : N] = |G|/|N|$.

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Center of a Group and the $G/Z(G)$ Theorem

The center $Z(G) = {z \in G : zg = gz \text{ for all } g \in G}$ is always a normal abelian subgroup of $G$.

The $G/Z(G)$ Cyclic Theorem

If the quotient group $G/Z(G)$ is cyclic, then $G$ is abelian (and hence $G = Z(G)$). Proof: Let $G/Z(G) = \langle g Z(G) \rangle$. Every $x, y \in G$ can be expressed as $x = g^i z_1$ and $y = g^j z_2$ for $z_1, z_2 \in Z(G)$. Then $xy = g^i z_1 g^j z_2 = g^{i+j} z_1 z_2 = g^{j+i} z_2 z_1 = g^j z_2 g^i z_1 = yx$.

Prime-Power Center Property

If $G$ is a finite $p$-group (order $p^n$ with $p$ prime and $n \ge 1$), its center is non-trivial: $|Z(G)| > 1$.

• Groups of order $p^2$: Since $|Z(G)| \in {p, p^2}$, if $|Z(G)| = p$, then $|G/Z(G)| = p$, making $G/Z(G)$ cyclic. By the theorem above, this forces $G$ to be abelian, contradicting $|Z(G)| = p$. Thus $|Z(G)| = p^2$, proving that every group of order $p^2$ is abelian (isomorphic to either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$).

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Group Homomorphisms and the First Isomorphism Theorem

A map $\phi: G \to H$ between groups is a homomorphism if for all $a, b \in G$: $\phi(ab) = \phi(a)\phi(b)$

• Basic Identities: $\phi(e_G) = e_H$, $\phi(a^{-1}) = (\phi(a))^{-1}$, and $|\phi(a)|$ divides $|a|$.
• Kernel: $\ker(\phi) = {g \in G : \phi(g) = e_H}$. The kernel is always a normal subgroup of $G$ ($\ker(\phi) \trianglelefteq G$).
• Injectivity Criterion: $\phi$ is injective (a monomorphism) if and only if $\ker(\phi) = {e_G}$.
• Image: $\operatorname{im}(\phi) = \phi(G) = {\phi(g) : g \in G}$ is a subgroup of $H$.

The First Isomorphism Theorem for Groups

If $\phi: G \to H$ is a group homomorphism, then: $G / \ker(\phi) \cong \operatorname{im}(\phi)$ The isomorphism is given by $g\ker(\phi) \mapsto \phi(g)$.

Cayley's Theorem

Every group $G$ is isomorphic to a subgroup of the symmetric group $\operatorname{Sym}(G)$ on the set $G$. The embedding maps $g \in G$ to the left-multiplication permutation $\lambda_g(x) = gx$.

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Subgroups, Normality, and Quotients Summary

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Step-by-Step Worked Problem

Problem: Determine all group homomorphisms $\phi: \mathbb{Z}{24} \to \mathbb{Z}{18}$. For the non-trivial homomorphism satisfying $\phi(1) = 6$, compute its kernel, image, and verify the First Isomorphism Theorem.

Solution:

• Step 1: Determine all homomorphisms. Since $\mathbb{Z}{24}$ is cyclic generated by $1$, any homomorphism $\phi$ is completely determined by $\phi(1) = k \in \mathbb{Z}{18}$. The condition that $\phi$ is a well-defined group homomorphism is that the order of $\phi(1)$ in $\mathbb{Z}{18}$ must divide $|1| = 24$, which is equivalent to: $24k \equiv 0 \pmod{18}$ Divide by $\gcd(24, 18) = 6$: $4k \equiv 0 \pmod 3 \iff k \equiv 0 \pmod 3$ In $\mathbb{Z}{18}$, the multiples of 3 are ${0, 3, 6, 9, 12, 15}$. Thus, there are exactly $\gcd(24, 18) = 6$ distinct homomorphisms.

• Step 2: Analyze $\phi(1) = 6$.

• Image: The image is the cyclic subgroup generated by $6$ in $\mathbb{Z}_{18}$: $\operatorname{im}(\phi) = \langle 6 \rangle = \{0, 6, 12\} \le \mathbb{Z}_{18} \quad \implies |\operatorname{im}(\phi)| = 3$

• Kernel: $\ker(\phi) = {x \in \mathbb{Z}{24} : \phi(x) = 6x \equiv 0 \pmod{18}}$. $18 \mid 6x \iff 3 \mid x$ Thus $\ker(\phi) = 3\mathbb{Z}{24} = {0, 3, 6, 9, 12, 15, 18, 21}$, so $|\ker(\phi)| = 8$.

• Verify First Isomorphism Theorem: $|\mathbb{Z}_{24} / \ker(\phi)| = \frac{24}{8} = 3 = |\operatorname{im}(\phi)|$ Both $\mathbb{Z}{24} / \ker(\phi)$ and $\operatorname{im}(\phi)$ are cyclic groups of order 3, confirming $\mathbb{Z}{24} / \ker(\phi) \cong \mathbb{Z}_3 \cong \operatorname{im}(\phi)$.

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GRE Exam Traps & Pitfalls

Trap 1: Transitivity of Normality Fails If $K \trianglelefteq H$ and $H \trianglelefteq G$, it is not generally true that $K \trianglelefteq G$. In the dihedral group $D_4$, a reflection subgroup of order 2 is normal in $V_4 \le D_4$, but is not normal in $D_4$.

Trap 2: Multiplying Cosets of Non-Normal Subgroups If $H$ is not normal in $G$, the product set $(aH)(bH)$ is typically a union of multiple cosets rather than a single coset. The coset space $G/H$ does not form a group.

Trap 3: Assuming Element Orders Are Preserved by Homomorphisms A homomorphism only guarantees that $|\phi(a)|$ divides $|a|$. For example, the trivial homomorphism maps all elements to $e$, so $|\phi(a)| = 1$ for all $a$.

Trap 4: Misapplying the $G/Z(G)$ Cyclic Theorem If $G/Z(G)$ is non-cyclic (such as $D_4 / Z(D_4) \cong V_4$), $G$ can be non-abelian. The theorem only establishes that a cyclic quotient forces $G$ to be abelian.