# Thread: [SOLVED] Normal Subgroups: Prove g^2 is in H for all g in G.

1. ## [SOLVED] Normal Subgroups: Prove g^2 is in H for all g in G.

Can someone help me with this proof?

Let H be a subgroup of G with index 2.
Prove that g² ∈ H for all g ∈ G.

Thanks!

2. Originally Posted by yvonnehr
Can someone help me with this proof?

Let H be a subgroup of G with index 2.
Prove that g² ∈ H for all g ∈ G.

Thanks!
A subgroup of index 2 is normal, and the factor group $\displaystyle G/H$ is a group of order 2. Hence for any $\displaystyle g\in G,$ $\displaystyle g^2H=(gH)^2=H$ $\displaystyle \implies$ $\displaystyle g^2\in H.$

In general, if $\displaystyle |G:H|=2,$ then $\displaystyle \forall\,x,y\in G,$ $\displaystyle xy\in H\ \iff\ x,y\in H\ \mbox{or}\ x,y\notin H.$ The result you are proving is a special case with $\displaystyle x=y.$

3. ## Very nice!

Thanks for the help. That was very clear!

4. ## (gH)^2=H ?

Originally Posted by TheAbstractionist
A subgroup of index 2 is normal, and the factor group $\displaystyle G/H$ is a group of order 2. Hence for any $\displaystyle g\in G,$ $\displaystyle g^2H=(gH)^2=H$ $\displaystyle \implies$ $\displaystyle g^2\in H.$

In general, if $\displaystyle |G:H|=2,$ then $\displaystyle \forall\,x,y\in G,$ $\displaystyle xy\in H\ \iff\ x,y\in H\ \mbox{or}\ x,y\notin H.$ The result you are proving is a special case with $\displaystyle x=y.$
One more question, how did you get from

(gH)^2=H ?

Thanks.

5. In a factor group $\displaystyle G/H$ the group operation is defined to be $\displaystyle (aH)(bH)=abH$. As long as H is normal, this is well defined and is a group.

You are given that $\displaystyle 2=[G:H]=|G/H|$

By Lagrange's theorem since $\displaystyle |G/H|=2$ any element of G/H must have order dividing 2, so in this case $\displaystyle g^2H=(gH)(gH)=(gH)^2=1H=H$ since $\displaystyle gH \in G/H$ $\displaystyle \forall g\in G$.

This means $\displaystyle g^2H=H \Rightarrow g^2h_1=h_2 \Rightarrow g^2=h_2h_1^{-1}\in H$ as desired.