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"
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.