# Math Help - Subgroup of index 2 is normal

1. ## 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.

2. The index of a subgroup $H$ of a group $G$ is the number of distinct left or right cosets of that subgroup. This is denoted $|G:H|.$

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

Suppose $H$ is a subgroup of $G$ of index 2. Let $g\in G$.

If $g\in H$ then $gH=H$ and $Hg=H$, so $gH=Hg.$

If $g\notin H,$ then $G$ is the disjoint union of the left cosets $H=eH$ and $gH,$ and the disjoint union of the right cosets $H=He$ and $Hg.$ Hence, again $gH=Hg.$

Since $gH=Hg$ for all $g\in G,$ $H$ is normal in $G.$ The beauty of this result is that it doesn’t assume that $G$ or $H$ is finite, only that the index of $H$ is finite.

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

4. 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.

5. @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!