Theorem 54: Cauchy s Theorem • Statement ○ If G is a finite group, and p is a prime divisor of |G|, then ∃H≤G of order p • Proof ○ Write |G|=mp. We argue by strong induction on m ○ When m=1, this is trivial, since any non-identity element of G has order p ○ Suppose m 1, and ∀n∈{1,…,m−1} if |G^′ |=np, then ∃H′≤G′ of order p ○ If G is abelian § Let x∈G∖{1} § If ⟨x⟩=G □ By the Fundamental Theorem of Cyclic Groups, □ G=⟨x⟩ contains a (unique) subgroup of order p § If ⟨x⟩≠G □ Set H≔⟨x⟩⊴G □ By the Lagrange s Theorem, |G|=|H|[G:H]=|H|⋅|G/H| □ Since ├ p┤ ||G|┤,either ├ p┤ ||H|┤ or ├ p┤ ||G/H|┤ □ If ├ p┤ ||H|┤ ® Since H is cyclic, H contains a (unique) subgroup of order p ® It follows that G contains a subgroup of order p □ If ├ p┤ ||G/H|┤ ® |G/H| |G|, so, by induction, ∃gH∈G\H s.t. |gH|=p ® So we only need to prove ├ |gH|┤ ||g|┤ ◊ If K ⟶┴f K′ is a group homomorphism, ├ |f(k)|┤ ||k|┤,∀k∈K ◊ Now, take K=G, K^′=G\H, f the usual surjection g↦gH ® Therefore ├ p┤ ||g|┤ ® Since ⟨g⟩ is cyclic, ⟨g⟩ contains a (unique) subgroup of order p ® It follows that G contains a subgroup of order p ○ If G is not abelian § By the Lagrange s Theorem, |G|=|C_G (g_i )|⋅[G:C_G (g_i )], ∀i∈{1,…,r} § Since ├ p┤ ||G|┤,either ├ p┤ ||C_G (g_i )|┤ or ├ p┤ |[G:C_G (g_i )]┤ § If ├ p┤ ||C_G (g_i )|┤ for some i □ Since G is not abelian, C_G (g_i )≨G for all i □ Apply the induction hypothesis, C_G (g_i ) contains a subgroup of order p □ It follows that G contains a subgroup of order p § If ├ p┤ |[G:C_G (g_i )]┤,∀i □ By the Class Equation, |G|=|Z(G)|+∑_(i=1)^r▒[G:C_G (g_i )] where g_1,…,g_r∈G □ are the representatives of the conjugate classes not contained in Z(G) □ It follows that ├ p┤ |(|G|−∑_(i=1)^r▒[G:C_G (g_i )] )┤=|Z(G)| □ G is not abelian, so Z(G)≨G □ Apply the induction hypothesis, Z(G) contains a subgroup of order p □ It follows that G contains a subgroup of order p Lemma 55: Recognizing Direct Products • Statement ○ Let G be a group with normal subgroups N_1,N_2 ○ The map α:N_1×N_2→G given by (n_1,n_2 )↦n_1 n_2 is an isomorphism ○ if and only if N_1 N_2=G and N_1∩N_2={1} • Proof (⟹) ○ Since α is surjective, N_1 N_2=G ○ Suppose n∈N_1∩N_2 ○ Then α(n,1)=n=α(1,n) ○ Since α is injective, (1,n)=(n,1)⇒n=1 ○ So N_1∩N_2={1} • Proof (⟸) ○ α is surjective § This is true since N_1 N_2=G ○ α is a homomorphism § α((n_1,n_2 ),(n_1′,n_2′ ))=α((n_1 n_1^′,n_2 n_2^′ ))=n_1 n_1^′ n_2 n_2^′ § α(n_1,n_2 )α(n_1′,n_2′ )=n_1 n_2 n_1^′ n_2^′ § We want show that α((n_1,n_2 ),(n_1′,n_2′ )) (α(n_1,n_2 )α(n_1′,n_2′ ))^(−1)=1 § (n_1 n_1^′ n_2 n_2^′ ) (n_1 n_2 n_1^′ n_2^′ )^(−1)=n_1 n_1^′ n_2 n_2^′ (n_2^′ )^(−1) (n_1^′ )^(−1) n_2^(−1) n_1^(−1) § =n_1 ⏟(n_1^′ n_2 (n_1^′ )^(−1) )┬(∈N_2 ) n_2^(−1) n_1^(−1)=n_1 ⏟(n_1^′ n_2 (n_1^′ )^(−1) n_2^(−1) )┬(∈N_2 ) n_1^(−1)∈N_2 § =n_1 n_1^′ ⏟(n_2 (n_1^′ )^(−1) n_2^(−1) )┬(∈N_1 ) n_1^(−1)=n_1 ⏟(n_1^′ n_2 (n_1^′ )^(−1) n_2^(−1) )┬(∈N_1 ) n_1^(−1)∈N_1 § Thus (n_1 n_1^′ n_2 n_2^′ ) (n_1 n_2 n_1^′ n_2^′ )^(−1)∈N_1∩N_2={1} § Therefore α((n_1,n_2 ),(n_1′,n_2′ ))=α((n_1,n_2 ),(n_1′,n_2′ )) ○ α is injective § If (n_1,n_2 )=1 § ⇒n_1 n_2=1 § ⇒n_1=n_2^(−1) § ⇒n_1∈N_2,n_2∈N_1 § ⇒n_1=n_2=1 § ⇒(n_1,n_2 )=(1,1) § ⇒α is injective
