Question:

I have a question about cosets. I'm trying to do a problem from sec 10.

"Show that if a subgroup H of finite group G has an index of 2, then

every left coset of H is a also a right coset of H." It's problem #39.

I haven't started the proof but I'm puzzled by something I did.

I looked at table 8.8 on page 79, let H = {P0, U1,U2} and found the

following left and right cosets:

Left:

P0H = {P0, U1, U2}

P1H = {P1, U3, U1}

P2H = {P2, U2, U3}

U1H = {U1, P0, P1}

U2H = {U2, P2, P0}

U3H = {U3, P1, P2}

Right:

HP0 = (P0, U1, U2}

HP1 = {P1, U2, U3}

HP2 = {P2, U3, U1}

HU1 = {U1, P0, P2}

HU2 = {U2, P1, P0}

HU3 = {U3, P2, P1}

My question is: How many left (right) cosets does H have?

Each one is different from another, although each one can pair with

another to form G. I suppose the answer is two because the order of G

divide by the order of H is two. So the index of H in G is two. But

then not every left coset is a right coset. Only P0H and U3H are. I

must have done something silly. Please let me know what the problem is.

(To do the proof, am I to show that if x in a1h, then x in hb2, etc?)

Answer: