2nd Law. Common sense is not common; lies are abundant and spreading, but the truth is hard to find -Do not kill/cancel the messenger-; nothing good comes easy, it requires sacrifice; and don’t ever underestimate the power of ignorance and stupidity, Apocalypse, Anawim, #justtothepoint.
Definition. Any two elements a and b of a group G are said conjugate if xax^{-1} = b for some x ∈ G.
Conjugacy is an equivalent relation on G, and the conjugacy class of “a”, denoted by cl(a) ={xax^{-1} | x ∈ G} is the equivalence class of “a” under conjugacy, and therefore we could partition any group G into disjoint conjugacy classes.
The Class Equation. Let G be a nontrivial finite group. Suppose that a_{1}, a_{2}, ···, a_{k} are the representatives of the conjugacy classes that have size > 1. Then, $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$
Sylow’s First Theorem. Let G be a finite group and p be a prime. If p^{k} | |G| for some integer k ∈ ℤ, then G has a subgroup of order p^{k}, i.e., ∃H ≤ G such that |H| = p^{k}.
Proof.
Let’s use induction on |G|.
If |G| = 1 ⇒ G is the trivial group, G = {e} ⇒ p^{k} ɫ |G| = 1 for any integer k, so the theorem is obviously true.
Let’s assume that the theorem holds for all groups of order less than |G|.
Let’s assume p^{k} | |G| for some integer k ∈ ℤ, then if there are some proper subgroup of G, say H < G, |H| = p^{k} (H is itself a group of order less than |G|), then, by our inductive hypothesis, H has a subgroup of order p^{k}, namely H itself, and the statement holds.
Let’s assume that there is no proper subgroup H of G such that p^{k} ɫ |H|. By the class equation, $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$ where this sum is done over representatives of conjugacy classes a_{k} such that a_{k} ∉ Z(G).
Since p^{k} | |G| = |G:C(a)|·|C(a)|, and p^{k} does not divide |C(a)| (By assumption, there is no such subgroup and C(a) is a subgroup of G) ⇒ p | |G:C(a)| ∀a ∉ Z(G) ⇒ [Cauchy's Theorem for Abelian Group. Let G be a finite Abelian group and let p be a prime that divides |G| ⇒ G has an element of order p] Z(G) is a finite Abelian group, p | |Z(G)|🚀 ⇒ ∃z ∈ Z(G) of order p.
🚀It follows from the class equation, $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$ and p | |G:C(a)| ∀a ∉ Z(G), p^{k} | |G| ⇒ p | |G| - $\sum_{i=1}^k |G:C(a_k)|$ = |Z(G)|
Let Z = ⟨z⟩ be the cyclic group generated by z ⇒ [Because any subgroup of Z(G) is normal. Remember that a group is normal if it is closed under conjugation] Z = ⟨z⟩ is a normal subgroup of G and we may form the factor or quotient group H = G/Z = G/⟨z⟩ ⇒ Since the order of H is given by |H| = |G|/|Z| = |G|/p ⇒ |H| < |G| ⇒ [The induction hypothesis applies, |H| < |G|, p^{k} | |G| and p^{k-1}| |H| = |G|/|Z| = |G|/p] there exists a subgroup of H = G/Z = G/⟨z⟩ of order p^{k-1} ⇒ [Correspondence Theorem, Let G be a group, N ⊲ G be a normal subgroup, then every subgroup of the quotient group G/N is of the form S/N = {sN: s ∈ S}, where N ≤ S ≤ G] this subgroup has the form S/⟨z⟩ where S is a subgroup of G ⇒ |S/⟨z⟩| = p^{k-1} and |⟨z⟩| = p ⇒ |S| = p^{k}∎
Definition. Let G be a finite group, and let p be a prime dividing the order of G. Then, a Sylow p-subgroup of G is a maximal p-subgroup of G, that is, p^{k} divides |G| and p^{k+1} does not divide |G|. Let Syl_{p}(G) denote the set of Sylow p-subgroups of G.
Let G = ℤ_{12}, ℤ_{12} = 2^{2}·3 then ⟨4⟩ ≋ ℤ_{3} is the unique Sylow 3-subgroup of ℤ_{12} and ⟨3⟩ ≋ ℤ_{3} is the unique Sylow 2-subgroup of ℤ_{12}.
In our previous example, any subgroup of order 8 is a Sylow 2-subgroup of G, any subgroup of order 125 is a Sylow 5-subgroup of G and any subgroup of order 7 is a Sylow 7-subgroup of G; in other words, p^{k} divides |G|, but p^{k+1} does not divide |G|, i.e., the largest power of p consistent with Lagrange's theorem.
Let G = S_{3}, with |S_{3}| = 6 = 2·3. Then, S_{3} has a unique Sylow 3-subgroup, namely A_{3}, but three distinct Sylow 2-subgroups: ⟨12⟩, ⟨13⟩, and ⟨23⟩.
Let G = A_{4}, with |A_{4}| = 12 = 2^{2}·3. Then, A_{4} has a unique Sylow 4-subgroup, namely the Klein 4-group V_{4} = {1, (12)(34), (13)(24), (14)(23)}, and four Sylow (cyclic) 3-subgroups of order 3.
Cauchy’s Theorem. Let G be a finite group and let p be a prime that divides the order of G. Then, G has an element of order p.
Proof.
Let G be a finite group and let p be a prime that divides the order of G, p | |G| ⇒ [Sylow’s First Theorem. Let G be a finite group and p be a prime. If p^{k} | |G| for some integer k ∈ ℤ, then G has a subgroup of order p^{k}, i.e., ∃H ≤ G such that |H| = p^{k}.] G has a subgroup of order p = p^{k} for k = 1 ⇒ [Since any subgroup of order p prime is cyclic] G has an element of order p.
p needs to be prime. Klein 4-group does not contain an element of order 4 (4 | 4), but it contains an element of order 2.