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College Mathematics

# Direct product of groups: TIFR 2018 Part A, Problem 9

This problem is a cute and simple application of the direct product of groups in the abstract algebra section. It appeared in TIFR GS 2018.

# 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.

# Connected Program at Cheenta

#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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