Thread: Subgroup of index 2 is normal

Prove that every subgroup of index 2 is normal.

The problem with me here is that I'm not quite sure what subgroup of index 2 means. Does that have anything to do with a family? So like a family of subgroup with an index that contains two elements?

Thanks.

The index of a subgroup of a group is the number of distinct left or right cosets of that subgroup. This is denoted

If and are finite, Lagrange’s theorem states that Hence, in the finite case, the index of in is the order of divided by the order of . Be careful though: and are not always finite. In the infinite case, the index is taken to be the number of cosets of in (this can be finite even if and are infinite).

Suppose is a subgroup of of index 2. Let .

If then and , so

If then is the disjoint union of the left cosets and and the disjoint union of the right cosets and Hence, again

Since for all is normal in The beauty of this result is that it doesn’t assume that or is finite, only that the index of is finite.

index is the number of the left coset of group G,why gH=Hg for every element of G?

Index is 2. So there are just two right cosets - H itself and Hx where x doesn't belong to H. every element g in G belongs one of the two cosets (as they are mutually exclusive and exhaistive over G). So Hx = G - H

As there is one-to-one mapping between left and right cosets, similarly there are two left cosets H and yH. So yH = G - H = Hx

So for every x,y not in H, Hx = yH.

If x is in H, Hx = xH = H. This is very trivial prove using the fact H is a sub-group hence closed.

@11235813 - Infact you should try this - Prove that if every right coset of H is a left coset as well, then H is normal. Then prove for index 2 is just a corollary to this result.

Originally Posted by aman_cc

Index is 2. So there are just two right cosets - H itself and Hx where x doesn't belong to H. every element g in G belongs one of the two cosets (as they are mutually exclusive and exhaistive over G). So Hx = G - H

As there is one-to-one mapping between left and right cosets, similarly there are two left cosets H and yH. So yH = G - H = Hx

So for every x,y not in H, Hx = yH.

If x is in H, Hx = xH = H. This is very trivial prove using the fact H is a sub-group hence closed.

Thank You!