Group Theory Properties
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Let G be a group. Let p be a prime. #math #mathematics #algebra #grouptheory #science
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Let K be a characteristic subgroup of N. And let N be a normal subgroup of G. Then, K is a normal subgroup of G.
Let G be field, and ρ be an irreducible representation over the field of complex numbers. Then |G| is devided by the degree of ρ.
【Burnside's theorem】Let P be a Sylow p-subgroup of G. Suppose that P is contained in the center of the normalizer of P. Then, G has a normal p-complement.
Let Sylp(G) be the set of Sylow p-subgroups of G. And let S be a Sylow p-subgroup. Then, the following are equivalent. (1) S is normal of G. (2) Sylp(G)={S}
Let N be an arbitrary normal subgroup such that G/N is abelian. Then, the commutator subgroup G' is contained in N. In other words, G' is the unique normal subgroup of G with the minimal property that the corresponding factor group is abelian.
【Maschke】Let k be a field, and (ρ, V) be a representation of G over k. And suppose the character of k doesn't divide the order of G. Then, (ρ, V) is completely reducible. In particular, if the character of k is 0, (ρ, V) is completely reducible.
Let M be the Frattini subgroup of P, where P is a p-group. And let N be normal of P. P/N is elementary abelian if and only if M is contained in N.
Let Sylp(G) be the set of Sylow p-subgroups of G. And let S be a Sylow p-subgroup. Then, the following are equivalent. (1) S is normal of G. (2) Sylp(G)={S}
【Dedekind's lemma】Let H, K, L be subgroups of G. And assume H is contained in L. Then, H(K∩L)=HK∩L
Let G be field, and ρ be an irreducible representation over the field of complex numbers. Then |G| is devided by the degree of ρ.
【Schur-Zassenhaus theorem】Let G be finite, and N be a normal subgroup of G. And assume (|G|, |G:N|)=1. Then, N is complemented in G.
【Burnside】Let G be finite. And let Χ be an irreducible character of G with degΧ is not equal to 1. Then, there exists x in G such that Χ(x)=0.
【Maschke】Let k be a field, and (ρ, V) be a representation of G over k. And suppose the character of k doesn't divide the order of G. Then, (ρ, V) is completely reducible. In particular, if the character of k is 0, (ρ, V) is completely reducible.
Let G be finite, and Χ be an irreducible character of G. And suppose x in K, where K is a conjugacy class of G. Then, |K|Χ(x)/Χ(1) is an algebraic integer.
For any subgroup H of G, let core(H) be the intersection of H^x (x runs over G). Then, core(H) is the unique largest subgroup of G contained in H.
The Frattini subgroup of a p-group is the unique normal subgroup with the minimal property that the corresponding factor group is elementary abelian.
Let U be the intersection of all Sylow p-subgroups of G. Then, U is the unique largest normal p-subgroup.
【Burnside】Let G be finite. And let Χ be an irreducible character of G with degΧ is not equal to 1. Then, there exists x in G such that Χ(x)=0.
【Maschke】Let k be a field, and (ρ, V) be a representation of G over k. And suppose the character of k doesn't divide the order of G. Then, (ρ, V) is completely reducible. In particular, if the character of k is 0, (ρ, V) is completely reducible.
The commutator subgroup G' is trivial if and only if G is abelian.
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