Cauchy's Theorem for Abelian Groups

Recall

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Definition. A subgroup H of a group G, H ≤ G, is called a normal subgroup of G if aH = Ha, ∀a ∈ G. In other words, a normal subgroup of a group G is one in which the right and left cosets are precisely the same. The usual notation for this relation is H ◁ G. It means ∀a ∈ G, h ∈ H, ∃h’, h’’∈ H: ah = h’a and ha = ah’’.

Normal Subgroup Test or alternative definition. A subgroup H of a group G is normal in G if and only if is invariant under conjugation by members of the group G. H ◁ G iff gHg^-1 ⊆ H, ∀g ∈ G, i.e., ∀g ∈ G, h ∈ H, ghg^-1 ∈ H.

Recall. An automorphism is an isomorphism from a group to itself.

Let G be a group, and let a be a fixed or given element of G, a ∈ G.An inner automorphism of G (induced or given by a) is defined by the conjugation action of the fixed element a, called the conjugating element i.e., Φ_a defined by Φ_a(x) = a·x·a^-1 ∀x ∈ G.

Theorem. For any group G, G/Z(G) is isomorphic to Inn(G), i.e., G/Z(G) ≋ Inn(G).

Proof:

Let’s consider the mapping from T: G/Z(G) → Inn(G), T(gZ(G)) = Φ_g where Φ_g(x) = gxg^-1 ∀x ∈ G.

• Is it well-defined? Or, alternatively, does the image depend only on the coset itself and not on the element representing the coset? Let’s suppose that gZ(G) = hZ(G), Φ_g = Φ_h?

∀x ∈ G, Φ_g(x) = gxg^-1.

gZ(G) = hZ(G) ⇒ h^-1gZ(G) = Z(G) ⇒[aH = H ↭ a ∈ H.] h^-1g ∈ Z(G) ⇒ h^-1gx = xh^-1g ⇒ gx = hxh^-1g ⇒ gxg^-1 = hxh^-1 ⇒ Φ_g(x) = gxg^-1 = hxh^-1 = Φ_h(x) ⇒ ∀x ∈ G, Φ_g(x) = Φ_h(x) ⇒ Φ_g = Φ_h.

• is T one-to-one? Suppose Φ_g = Φ_h.

∀x ∈ G, Φ_g(x) = Φ_h(x) ⇒ gxg^-1 = hxh^-1 ⇒ ∀x ∈ G, h^-1gxg^-1 = xh^-1 ⇒ ∀x ∈ G, h^-1gx = xh^-1g ⇒ h^-1g ∈ Z(G) ⇒[aH = H ↭ a∈H.] h^-1gZ(G) = Z(G) ⇒ gZ(G) = hZ(G).

• T is onto by construction and operation preserving, ∀x ∈G: Φ_gh(x)= (gh)x(gh)^-1 =[Sock- Shoe Property] (gh)x(h^-1g^-1) = g(hxh^-1)g^-1 = gΦ_h(x)g^-1 = Φ_g(Φ_h(x)) ⇒ Φ_gh = Φ_gΦ_h∎

Example. Inn(D₆) ≋ D₃.

1. The center of the dihedral group D_n, Z(D_n) = {e, α^n/2}, if n is even. Therefore, |Z(D₆)| = 2.
2. |D₆/Z(D₆)| = 12/2 = 6 = 2·3, 3 prime.
3. Classification of Groups of Order 2p. Let G be a group, |G| = 2p, p is prime, p ≥ 2 ⇒ G ≋ ℤ_2p or D_p. Therefore D₆/Z(D₆) ≋ Inn(D₆) is isomorphic to ℤ₆ or D₃.
4. If Inn(D₆) were to be cyclic, then D₆/Z(D₆) would be cyclic ⇒ [Theorem. If G/Z(G) is cyclic, then G is Abelian] However, D₆ = ⟨a, b | a⁶ = b² = e, ba = a^-1b⟩ = {e, a, a², a³, a⁴, a⁵, b, ab, a²b, , a³b, a⁴b, a⁵b} is the dihedral group of order 2n, but we know the dihedral groups D_n are non-Abelian for n ≥ 3. Therefore, Inn(D₆) ≋ D₃∎

We know the importance of Lagrange’s Theorem. Let G be a finite group and let H be a subgroup of G. Then, the order of H is a divisor of the order of G.

The converse of Lagrange’s Theorem is not true, e.g., A₄, |A₄| = 12, but even though 6 is a divisor of 12, A₄ does not have a subgroup of order 6. Cauchy’s Theorem is a partial converse of Lagrange’s Theorem.

Cauchy’s Theorem for Abelian Groups. Let G be a finite Abelian group, |G| = n, and let p be a prime number dividing the order of G, then G contains an element of order p.

Proof.

If G has order 2 ⇒ G ≋ ℤ/2ℤ. If p is a prime such that p divides 2, then p = 2, and ℤ/2ℤ has an element of order 2, namely 1 ⇒ [Group isomorphisms preserve the order of an element] G has an element of order 2, too.

Let’s use induction on |G|. Let’s assume that the statement is true for all Abelian groups with fewer elements than G.

Let a ∈ G, a ≠ e, and let H be the cyclic group it generates, H = ⟨a⟩ ⇒ [Normal. Every subgroup of an Abelian group is normal] H = ⟨x⟩ ◁ G. There are two options:

1. If p divides |H|, then $a^{\frac{|H|}{p}}$ is an element of order p, just consider $(a^{\frac{|H|}{p}})^{p}= e$
2. If p does not divide |H|, and by assumption p divides |G| ⇒ p divides the order of G/H (|G/H| = |G|/|H|), and by the induction hypothesis (|G/H| < |G|), there is a closet of order p, i.e., there is a xH, for some x ∈ G. Let be m the order of x in G, x^m = e ⇒ (xH)^m = x^mH = eH = H in G/H ⇒[|xH| = p] p divides m ⇒ x^m/p is now an element of order p in G ∎

Corollary. Let G be a finite group and p be a prime. If p | |G|, then G has a subgroup of order p.

Proof.

We already know that there exist a “g” ∈ G, |g| = p ⇒ Consider H = ⟨g⟩, |H| = p∎

Example. Let G = A₄. |A₄| = 4!/2 = 12 = 2²·3 ⇒ [Cauchy’s Theorem] A₄ has elements (and hence cyclic subgroups) of order 2 and 3, and these are the 2-2 cycles and the 3-cycles in A₄, but there is no guarantee for elements of order 4, 6, and 12. In fact, no such elements exists in A₄.

Classification of groups of order p². Let G be a group of order p², where p is a prime number, then G is isomorphic to ℤ_p² or ℤ_p⊕ℤ_p. More concisely, let G be a group, |G| = p², p prime, G ≋ ℤ_p² or ℤ_p⊕ℤ_p.

Proof:

|G| = p², p prime. If G has an element of order p², then G ≋ ℤ_p² that’s because if G is a finite cyclic group of order n, then it is isomorphic to (ℤ_n, +) or, in other words, we have previously demonstrated that there is only one cyclic group up to isomorphism of any given order.

From now on, let’s suppose that G has not an element of order p² (G is not cyclic) ⇒[By Lagrange’s Theorem Corollary, the order of an element divides the order of the finite group] ∀a ∈ G ⇒ |a| = 1, i.e., it is the trivial element (a = e) or has order p, |a| = p.

Claim: The subgroup generated by any arbitrary element a ∈ G, |a| = p, ⟨a⟩ is normal in G, ⟨a⟩ ◁ G

For the sake of contradiction, let’s suppose that this is not the case. Then, ∃b ∈ G: bab^-1 ∉ ⟨a⟩ ⇒ [By previous assumption, G is not cyclic] ⟨bab^-1⟩ and ⟨a⟩ are distinct subgroups of order p ⇒ ⟨bab^-1⟩ ∩ ⟨a⟩ is a subgroup of both of them, and they are distinct subgroups, then ⟨bab^-1⟩ ∩ ⟨a⟩ = {e} ⇒ Since |G| = p², |⟨a⟩| = p, and all cosets have the same order, ⟨bab^-1⟩, a⟨bab^-1⟩, a²⟨bab^-1⟩, … a^p-1⟨bab^-1⟩ are p distinct left cosets of order p, so every element of G is in one of them (as distinct coset partition G into equivalent classes).

In particularly, b^-1 must be in one of these cosets, say b^-1 ∈ aⁱ⟨bab^-1⟩ ⇒∃j: b^-1 = aⁱ(bab^-1)^j = [Observe that bab^-1bab^-1 = ba(b^-1b)ab^-1 = ba²b^-1] aⁱ(ba^jb^-1) ⇒ b^-1 = aⁱ(ba^jb^-1) ⇒ [Cancelling terms] e = aⁱba^j ⇒ b = a^-i-j ∈ ⟨a⟩ ⊥

Therefore, every subgroup of the form ⟨a⟩ is normal in G and of order p, let a be a non-identity element in G, and b any element of G not in ⟨a⟩, then G = ⟨a⟩ x ⟨b⟩ ≋ ℤ_p ⊕ ℤ_p because of a previous theorem’s requirements are satisfied, namely (i) ⟨a⟩ ◁ G, ⟨b⟩ ◁ G, (ii) G = ⟨a⟩ x ⟨b⟩ -⟨a⟩ and ⟨b⟩ = b⟨a⟩ are a partition of G, |G/⟨a⟩| = p²/p = 2, and |G| = 4 = |⟨a⟩ x ⟨b⟩|/|⟨a⟩ ∩ ⟨b⟩| = 2 x 2 / 1 -, (iii) ⟨a⟩ ∩ ⟨b⟩ = {e} ⇒ G = ⟨a⟩ x ⟨b⟩ ≋ ⟨a⟩ ⊕ ⟨b⟩ ≋ ℤ_p⊕ ℤ_p∎

Corollary. If G is a group of order p², where p is a prime, then G is Abelian.

Theorem. If m = n₁n₂ ··· n_k where gcd(n_i, n_j) = 1 for i ≠ j, then U(m) = U_m/n1(m) x U_m/n2(m) x ··· U_m/nk(m) ≋ U(n₁)⊕U(n₂)⊕ ··· ⊕ U(n_k).

Proof.

We used the previous theorem and we have already demonstrated that if s and t are relatively prime, U(st)≈U(s)⊕U(t), U_s(st)≈U(t) and U_t(st)≈U(s). Besides, m = n₁n₂ ··· n_k where gcd(n_i, n_j) = 1 for i ≠ j, then U(m) = U_m/n1(m) x U_m/n2(m) x ··· U_m/nk(m) ≋ U(n₁)⊕U(n₂)⊕ ··· ⊕ U(n_k).

Examples:

• U(165) = U(5· 33) = U₅(165) x U₃₃(165) ≋ U(33)⊕U(5).
• U(165) = U(3· 55) = U₃(165) x U₅₅(165) ≋ U(55)⊕U(3).
• U(165) = U(15· 11) = U₁₅(165) x U₁₁(165) ≋ U(11)⊕U(15).
• U(165) = U(3· 5· 11) = U₅₅(165) x U₃₃(165)x U₁₅(165) ≋ U(3)⊕U(5)⊕U(11) ≋ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₁₀.
• Analogously, U(105) = U(15 · 7) = U₁₅(105) x U₇(105) ≋ U(7)⊕U(15) or, alternatively, U(105) = U(5 · 21) = U₅(105) x U₂₁(105) ≋ U(21)⊕U(5). Besides, U(105) = U(3 · 5 · 7) = U₃₅(105) x U₂₁(105) x U₁₅(105) ≋ U(3)⊕U(5)⊕U(7) ≋ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.