Freegroup's normal subgroup

hey,

i'm trying to study for my up coming exams with these past papers but i've gotten stuck with part c,

i've tried a bunch of different spanning tree's some gave better generators and some gave worse but the ones in the picture were the first ones i found,

i've been trying to use the generators to show that the its not possible to get the conjugate of one of the provided elements of H by multiplying the generators but it hasn't gotten me anywhere,

http://img99.imageshack.us/img99/6621/mathq.jpg

i also have a feeling the answer to part d) is the whole group F but im not sure why it's just a guess.

could anyone give me a hand working out these two parts?

thanks in advance

Re: Freegroup's normal subgroup

i worked out part c,

still trying find a solid process of showing that F is the smallest normal subgroup containing H

Re: Freegroup's normal subgroup

let K be the smallest normal subgroup of F that contains H.

H partitions F into 5 cosets, so H must partition K into at most 5 cosets (two elements x,y of F are in the same coset of H even when both are in K). thus [K:H] is finite. similarly, [F:K] must also be finite (each coset of H is entirely contained in a coset of K).

hence [F:H] = [F:K][K:H]. since [F:H] = 5 is prime, either [F:K] = 1, or [K:H] = 1. but [K:H] = 1 is not possible, since that implies H = K, but H is not normal.

thus [F:K] = 1, which means F = K.