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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.

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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., Φa defined by Φa(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)) = Φg where Φg(x) = gxg-1 ∀x ∈ G.

∀x ∈ G, Φg(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 ⇒ Φg(x) = gxg-1 = hxh-1 = Φh(x) ⇒ ∀x ∈ G, Φg(x) = Φh(x) ⇒ Φg = Φh.

∀x ∈ G, Φg(x) = Φh(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).

Example. Inn(D6) ≋ D3.

  1. The center of the dihedral group Dn, Z(Dn) = {e, αn/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 ≋ ℤ2p or Dp. 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 Dn 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, xm = e ⇒ (xH)m = xmH = eH = H in G/H ⇒[|xH| = p] p divides m ⇒ xm/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 ℤp2 or ℤp⊕ℤp. More concisely, let G be a group, |G| = p2, p prime, G ≋ ℤp2 or ℤp⊕ℤp.

Proof:

|G| = p2, p prime. If G has an element of order p2, then G ≋ ℤp2 that’s because if G is a finite cyclic group of order n, then it is isomorphic to (ℤn, +) 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⟩, … ap-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(bajb-1) ⇒ b-1 = ai(bajb-1) ⇒ [Cancelling terms] e = aibaj ⇒ 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⟩ ≋ ℤp ⊕ ℤp 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⟩ ≋ ℤp⊕ ℤp

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

Theorem. If m = n1n2 ··· nk where gcd(ni, nj) = 1 for i ≠ j, then U(m) = Um/n1(m) x Um/n2(m) x ··· Um/nk(m) ≋ U(n1)⊕U(n2)⊕ ··· ⊕ U(nk).

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), Us(st)≈U(t) and Ut(st)≈U(s). Besides, m = n1n2 ··· nk where gcd(ni, nj) = 1 for i ≠ j, then U(m) = Um/n1(m) x Um/n2(m) x ··· Um/nk(m) ≋ U(n1)⊕U(n2)⊕ ··· ⊕ U(nk).

Examples:

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
  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.
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