# 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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