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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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