# Normal Subgroups

• Nov 18th 2009, 10:03 AM
elninio
Normal Subgroups
Let H and K be normal subgroups of G such that H∩K = {e} and HK = G.

Prove that G ∼= H × K
• Nov 18th 2009, 10:07 AM
aman_cc
Quote:

Originally Posted by elninio
Let H and K be normal subgroups of G such that H∩K = {e} and HK = G.

Prove that G ∼= H × K

Hint - Can you define an isomorphism from G->HxK. (Use the fact G=HK and H∩K={e})
• Nov 18th 2009, 10:08 AM
tonio
Quote:

Originally Posted by elninio
Let H and K be normal subgroups of G such that H∩K = {e} and HK = G.

Prove that G ∼= H × K

First show that if $\displaystyle h_1\,,\,h_2\in H\,,\,\,k_1\,,\,k_2\in K\,,\,\,then\,\,h_1k_1=h_2k_2\Longleftrightarrow h_1=h_2\,,\,k_1=k_2$ , next define $\displaystyle f: G\rightarrow H\times K \,\,\,as\,\,\,f(g=hk)=(h,k)$ and show f is a isomorphism of groups.

Tonio
• Nov 18th 2009, 05:31 PM
elninio
Tonio, I'm not sure if this is the same thing as what you've posted but i've worked on this problem for over an hour and I know now that I can simply show that f is in isomorphism in a map defined f: H x K -> G by f((h,k)) = hk (Where hk is in G).

Does this seem correct so far? This is where I am having my trouble. I worked on one isomorphism problem similar to this before but I cant seem to apply any of the same methods or logic to this one.

How do I show f is an isomorphism in this problem?
• Nov 18th 2009, 06:06 PM
tonio
Quote:

Originally Posted by elninio
Tonio, I'm not sure if this is the same thing as what you've posted but i've worked on this problem for over an hour and I know now that I can simply show that f is in isomorphism in a map defined f: H x K -> G by f((h,k)) = hk (Where hk is in G).

Does this seem correct so far? This is where I am having my trouble. I worked on one isomorphism problem similar to this before but I cant seem to apply any of the same methods or logic to this one.

How do I show f is an isomorphism in this problem?

After you show what I proposed in my prior post, you have:
$\displaystyle f\left((h_1,k_1)+(h_2,k_2)\right)=f(h_1h_2,k_1k_2) =h_1h_2k_1k_2=h_1k_1h_2k_2=f(h_1,k_1)f(h_2,k_2)$ since by now it must be clear H, K commute POINTWISE.
Also, $\displaystyle f(h,k)=hk=1=1\cdot 1 \Longleftrightarrow h=1=k\Longrightarrow\,Ker(f)=1$ , and surjectivity is immediate.

Tonio