Let G be a finite group and G rightarrow H be a subjective homomorphism of Prove

Question

Let G be a finite group and G rightarrow H be a subjective homomorphism of Prove that |H| divides |G| (Hint: Use Fundamental Homomorphism Theorem). Let now G be an abelian group. Claim. H is abelian. Prove or disprove this claim.

Explanation / Answer

(a) let K=ker(), i.e. K={g| (g)=1 in H}. Thefundamental homomorphism theorem states that K is a normal subgroupof G (this is rather straightforward - in fact any normal subgroupoccurs as a kernel of some group homomorphism into some othergroup) and that H is isomorphic to the quotient group G/K. Hence|H|=|G|/|K| => |K|=|G|/|H| => |H| | |G|. (b) Since G is finite and the homomorphism into H is surjective, asin part (a) H is isomorphic to the quotient group G/K. Now G isabelian, so gh=hg for all g,h in G. Hence gK={gk|k in K}={kg|k inK}=Kg. Thus, (gK)(hK)=ghK= Kgh = Khg=(Kh)(Kg), so indeed G/H isabelian.

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