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# Understand the problem

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]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 . [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]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 [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]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 . [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]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.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]