- #1

ssayani87

- 10

- 0

## Homework Statement

I think I've got this one about figured out, I just wanted someone to check it over. (For this problem, (a-1) is a inverse, (b-1) b inverse, etc.)

"Let G be a group, H a subgroup of G.

Then, H is normal in G iff every left coset of H is equal to some right coset of H"

## Homework Equations

H normal in G implies aH(a-1) = H for all a in G

## The Attempt at a Solution

Let a,b be elts of G, and H a normal sbgp of G.

(=>):

By normal condition, aH(a-1) = H for all a in G

-> aH = Ha (right multiply by a)

H normal in G implies the left coset of H by a coincides with the right coset of H by a.

Implies that if H is normal, every left coset of H is equal to some right coset of H.

(<=):

Let the left coset of H by a be equal to some right coset of H by b.

So, aH = Hb

-> aH(b-1) = H (right multiply by (b-1) )

H = aH(b-1) implies (by taking inverses) that H = bH(a-1)

Finally, H = HH = (aH(b-1))(bH(a-1)) = aH((b-1)b)H(a-1) = aH(e)H(a-1) = aHH(a-1) = aH(a-1)

Then we have H = aH(a-1)

Similar argument shows that H = bH(b-1)

So, H is normal.

In conclusion, if every left coset of H is equal to some right coset of H, then H is normal in G.

I think that handles both directions (<=) & (=>).

To me, this seems like a pretty pertinent question regarding normal sbgps so I'd like to make sure I hit it on the head.

Thanks!