Abstract Algebra: Solving with Cosets

[ZZ Specs]

Homework Statement

Suppose H is a subgroup of G. For g in G, define f_g : G/H > G/H by f_g (aH) = gaH for a in G, where G/H is the set of left cosets of H in G.

I know that f_g is a well-defined permutation. However, we have not established (yet) that G/H is a group.

2 parts to the question:

1) for a given aH in G/H, find the set {g in G : f_g(aH) = aH }

2) find the set {g in G : f_g = the identity permutation in G/H}

The Attempt at a Solution

I have done part (1), finding the solution set {g in G : g = aha^-1 for some h in H}.

However, I struggle with part (2), as we have no information on a or H so I'm not sure what counts as a solution. I feel that normality may be involved but I cannot find out how to use it.

I know we want g such that f_g(aH) = gaH = aH for all cosets aH ; this is the identity permutation. By equality of cosets, we can say that a^-1ga = h for some a in G and h in H, or that g = aha^-1 for some a in H and g in G, but I'm not sure if this consitutes a solution.

Any help is very much appreciated. Thank you.

 

[Dick]

Homework Statement

Suppose H is a subgroup of G. For g in G, define f_g : G/H > G/H by f_g (aH) = gaH for a in G, where G/H is the set of left cosets of H in G.

I know that f_g is a well-defined permutation. However, we have not established (yet) that G/H is a group.

2 parts to the question:

1) for a given aH in G/H, find the set {g in G : f_g(aH) = aH }

2) find the set {g in G : f_g = the identity permutation in G/H}

The Attempt at a Solution

I have done part (1), finding the solution set {g in G : g = aha^-1 for some h in H}.

However, I struggle with part (2), as we have no information on a or H so I'm not sure what counts as a solution. I feel that normality may be involved but I cannot find out how to use it.

I know we want g such that f_g(aH) = gaH = aH for all cosets aH ; this is the identity permutation. By equality of cosets, we can say that a^-1ga = h for some a in G and h in H, or that g = aha^-1 for some a in H and g in G, but I'm not sure if this consitutes a solution.

Any help is very much appreciated. Thank you.

The only difference between 1) and 2) is that for 1) it has to be true for single value of a. For 2) it has to be true for all values of a in G.