Understand the problem

Let G, H be finite groups. Then any subgroup of G × H is equal to A × B for some subgroups A<G and B<H
Source of the problem
TATA INSTITUTE OF FUNDAMENTAL RESEARCH -GS-2018 (Mathematics)  -Part A -Question 9
Topic
Group Theory 
Difficulty Level
Medium 
Suggested Book
Abstract Algebra Dummit and Foote

Start with hints

Do you really need a hint? Try it first!

Consider K \(\leq\) G x H . Now if ( a , b ) \(\to\) K then  \( a^{-1} , b^{-1}\)  \(\to\)  K . Also if ( a , b) , ( p,q )  \(\in\) K then ( ap , bq ) , ( pa , qb )  \(\in\) K  So , G \(|_k\) = { g \(\in\) G | ( g , h ) \(\in\) K  for some h \(\in\) H }  is a subgroup of G . Similarly,  H \(|_k\) \(\leq\) H .
We have arrived to the conclusion that G\(|_k\) \(\leq\) G &  H\(|_k\) \(\leq\) H . Now use this as a fact  to guess the answer . Are you sure that  G\(|_k\)  x  H\(|_k\) = K ?  I mean that who confirms that K can be within as A X B 
Yes this is true that for every g \(\in\) G\(|_K\)  \(\exists\) k s.t  ( g , h )  \(\in\) K . But  G\(|_K\) x  H\(|_K\) contain  ( g , h )  \(\forall h\)    \(\in\)   H\(|_K\) .  This is certainly having a bigger expectation .
So , here is a nice counter example . Take G = \(\mathbb{Z_2}\)  &  H = \(\mathbb{Z_4}\) Consider a cyclic subgroup of G x H ; < ( 1 , 1 ) > = { ( 1 , 1 ) , ( 0, 2 ) , ( 1, 3) , (0,0) } Observe that G\(|_k\) = G &  H\(|_k\) = H But  < ( 1 , 1 ) > \(\neq\)  G x H  Hence , the answer is false.

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