# Abstract Algebra-Isomorphism

• Dec 18th 2009, 11:12 AM
WannaBe
Abstract Algebra-Isomorphism
Let A,B be normal sub-groups of a group G.
G=AB.

Prove that:
G/AnB is isomorphic to G/A*G/B

Have no idea how to start...Maybe the second isom. theorem can help us...

TNX!
• Dec 18th 2009, 11:16 AM
tonio
Quote:

Originally Posted by WannaBe
Let A,B be normal sub-groups of a group G.
G=AB.

Prove that:
G/AnB is isomorphic to G/A*G/B

Have no idea how to start...Maybe the second isom. theorem can help us...

TNX!

Define $\displaystyle \phi:G\rightarrow G\slash A\times G\slash B\,\,\,by\,\,\,\phi(g):=(gA,gB)$ and apply the first isomorphism theorem.

Hint to show the above map is onto: use that every $\displaystyle g\in G$ can be written as $\displaystyle g=ab\,,\,\,a\in A\,,\,\,b\in B$.

Tonio
• Dec 18th 2009, 12:06 PM
WannaBe
Well... So after we define this map and show that G is homomorphic to G/AXG/B , by the first iso. theorem, we know that G/AXG/B is isomorphic to G/kerPhi
The only thing that is missing is to show toat kerPhi=AnB...
We can show that this way: (I'll use the symbol p for phi)
If k is in kerp then p(k)=(A,B), which means that: (kA, kB) = (A,B) hence:
kA=A , kB=B which means k is in A and k is in B which means k is in AnB and we're done...

Am I right?

• Dec 18th 2009, 02:46 PM
tonio
Quote:

Originally Posted by WannaBe
Well... So after we define this map and show that G is homomorphic to G/AXG/B , by the first iso. theorem, we know that G/AXG/B is isomorphic to G/kerPhi
The only thing that is missing is to show toat kerPhi=AnB...
We can show that this way: (I'll use the symbol p for phi)
If k is in kerp then p(k)=(A,B), which means that: (kA, kB) = (A,B) hence:
kA=A , kB=B which means k is in A and k is in B which means k is in AnB and we're done...

Am I right?

Yes, that's right....but please do note that showing $\displaystyle \phi(G)=G\slash A\times G\slash B$ (i.e., that $\displaystyle \phi$ is onto) is NOT trivial: you MUST prove this.

Tonio
• Dec 19th 2009, 12:12 AM
WannaBe
Ofcourse it's not trivial...But according to the hint you gave me:
Each g in G can be written as ab when a is in A and b is in B...
So we can write: p(g)=p(a)*p(b)=(A,aB)*(bA,B)...So if we have (kA,kB) in G/AXG/B, we only need to prove that (kA,kB)=(A,aB)*(bA,B) i.e (kA,kB)=(A*bA, aB*B)=(abA, abB)...
But it's pretty much obvious, because if kA is in G/A then k is in G hence k=ab for some a in A and b in B... Am I right?

TNX again
• Dec 19th 2009, 02:55 AM
tonio
Quote:

Originally Posted by WannaBe
Ofcourse it's not trivial...But according to the hint you gave me:
Each g in G can be written as ab when a is in A and b is in B...
So we can write: p(g)=p(a)*p(b)=(A,aB)*(bA,B)...So if we have (kA,kB) in G/AXG/B, we only need to prove that (kA,kB)=(A,aB)*(bA,B) i.e (kA,kB)=(A*bA, aB*B)=(abA, abB)...
But it's pretty much obvious, because if kA is in G/A then k is in G hence k=ab for some a in A and b in B... Am I right?

TNX again

Yeppers!

Tonio
• Dec 19th 2009, 04:33 AM
WannaBe
Tnx a lot man! I realy need your help in a homework assigment I need to send tommorow...I'll be delighted if you'll be able to look at my post in Differential Equations forum...I'm not sure what I need to do...

TNX a lot!