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
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?
TNX a lot for the answer and for your verification!
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