Index of 2 is equivalent to normal

• Nov 26th 2007, 07:46 PM
epsilon
Index of 2 is equivalent to normal
Hello,

I have a homework problem which asks:
Let G be a group and let H be a subgroup of G.
(a) Show that if H is index 2 in G then H is a normal subgroup of G.
(b) Show that if H is a normal subgroup of G then H is index 2 in G.

I proved the first part like so:
Assume H has an index of 2
Thus it has 2 left cosets which are forced to be H and xH = G/H
and the 2 right cosets are H and Hx = G/H
So, Hx = xH for all x
Thus H is normal.

However, I can't figure out how to go the other way.

Any help would be much appreciated.
Thanks.
• Nov 26th 2007, 08:10 PM
ThePerfectHacker
Quote:

Originally Posted by epsilon
Hello,

I have a homework problem which asks:
Let G be a group and let H be a subgroup of G.
(a) Show that if H is index 2 in G then H is a normal subgroup of G.
(b) Show that if H is a normal subgroup of G then H is index 2 in G.

I proved the first part like so:
Assume H has an index of 2
Thus it has 2 left cosets which are forced to be H and xH = G/H
and the 2 right cosets are H and Hx = G/H
So, Hx = xH for all x
Thus H is normal.
.

(a)Consider H and H* (means H complement). We need to show xH = xH. If x in H then xH = H = hX. If x not in H then xH = H* = Hx.

(b)Garbage! Take H = { e } and G = Z_3. So?