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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
Group Theory
Medium
Suggested Book
Abstract Algebra Dummit and Foote

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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