Recent content by tiger4

Does this work? Since H is abelian then for all a,b in H we have that (ab)^m = a^m b^m. Thus phi(ab)=(ab)^m = a^m b^m = phi(a)phi(b) so phi is a group homomorphism. Since H is finite it suffices to show phi is surjective. Let x be in H , we need to find a in H such that a^m = x. By...

I think I proved the contrapositive, Let a be a non-zero nilpotent element in R such that a^m = 0. If psi(a) is 0, then a is in the kernel and we're done. Otherwise, psi(a) is non-zero, and psi(a)^m = psi(a^m) = psi(0) = 0, which means that psi(a) is a non-zero nilpotent element of the image.

let x,y be in aKa^-1. thus x = aka^-1, for some k in K, and y = ak'a^-1 for some k' in K. then xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1, and since K is a subgroup kk'^-1 is in K whenever k,k' are, so xy^-1 is in aKa^-1. thus aKa^-1 is certainly a subgroup...

Homework Statement Let G be an abelian group of order n. Define phi: G --> G by phi(a) = a^m, where a is in G. Prove that if gcd(m,n) = 1 then phi is an isomorphism Homework Equations phi(a) = a^m, where a is in G gcd(m,n) = 1 The Attempt at a Solution I know since G is an...

Homework Statement Let G be and group and let N and H be normal subgroups of G with $N \subset H \subset G. Prove that H/N is a subgroup of G/N, and that it is a normal subgroup. Note that aKa^{-1} = {aKa^{-1} | k in K}.Homework Equations The Attempt at a Solution I understand that since H is...

Homework Statement Let H and K be subgroups of the group G. Let a,b \in G and define a relation on G by a ~ b if and only if a = hbk for some h \in H and k \in K. Prove that this is an equivalence relation.Homework Equations a = hbkThe Attempt at a Solution The goal is to prove the reflexive...