# Prove normal subgroup

• Apr 27th 2010, 11:55 AM
rainyice
Prove normal subgroup
Let H be a subgroup of G with index 2.

a. Prove that H is a normal subgroup of G.

b. Prove that g^(2) E H for all g E G.
• Apr 27th 2010, 12:01 PM
Drexel28
Quote:

Originally Posted by rainyice
Let H be a subgroup of G with index 2.

a. Prove that H is a normal subgroup of G.

There is a fairly well-known theorem which says that "if all the right cosets of a subgroup are left cosets of the subgroup then the subgroup is normal". So, notice that if $L_1,L_2$ are the left cosets and $R_1,R_2$ the right then $L_1\amalg L_2=R_1\amalg R_2$ (where that funny symbol just means union but the sets are pairwise disjoint). What now?

Quote:

b. Prove that g^(2) E H for all g E G.
A more general theorem says that if $N\unlhd G$ then $g^{\left[G:N\right]}\in N$ for all $g\in G$. To see this think about $G/N$. What's it's order? what is $\left(gN\right)^{[G:N]}=?$
• Apr 27th 2010, 01:22 PM
glasssocks
Quote:

Originally Posted by rainyice
Let H be a subgroup of G with index 2.

a. Prove that H is a normal subgroup of G.

b. Prove that g^(2) E H for all g E G.

For the first part:

for g E G

if g E H , then gH = H = Hg. Therefore H is a normal subgroup.

if g is not an element of H, G = eH U gH because there are only 2 cosets.
also G = He U Hg.

So G = H U Hg and G = H U gH (H = eH = He)

Therefore, gH = Hg. H is a normal subgroup.

I don't know how to prove part b. Sorry
• Apr 27th 2010, 02:35 PM
rainyice
Quote:

Originally Posted by Drexel28
There is a fairly well-known theorem which says that "if all the right cosets of a subgroup are left cosets of the subgroup then the subgroup is normal". So, notice that if $L_1,L_2$ are the left cosets and $R_1,R_2$ the right then $L_1\amalg L_2=R_1\amalg R_2$ (where that funny symbol just means union but the sets are pairwise disjoint). What now?

A more general theorem says that if $N\unlhd G$ then $g^{\left[G:N\right]}\in N$ for all $g\in G$. To see this think about $G/N$. What's it's order? what is $\left(gN\right)^{[G:N]}=?$

about the Quotient subgroup, I haven't learn that yet
• Apr 27th 2010, 02:54 PM
Drexel28
Quote:

Originally Posted by rainyice
about the Quotient subgroup, I haven't learn that yet

That's lame. I wrote up a proof. Here's the basic idea. Break it into two cases

1. If $g\in N$ it's obvious what to do.

2. If $g\notin N$ then it has to be in the other coset, call it $hN$. And, so if $g^2\notin N\implies\cdots g\notin g^{-1}N\implies\cdots$
• Apr 28th 2010, 07:28 AM
rainyice
Quote:

Originally Posted by Drexel28
That's lame. I wrote up a proof. Here's the basic idea. Break it into two cases

1. If $g\in N$ it's obvious what to do.

2. If $g\notin N$ then it has to be in the other coset, call it $hN$. And, so if $g^2\notin N\implies\cdots g\notin g^{-1}N\implies\cdots$

i don't get it ....
• Apr 29th 2010, 08:12 AM
MissMousey
Quote:

Originally Posted by glasssocks

if g is not an element of H, G = eH U gH because there are only 2 cosets.
also G = He U Hg.

Doesn't this only work if G is a normal group?