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August 31, 2019

Are juniors countable if seniors are?: TIFR GS 2018 Part A Problem 21

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]A countable group can have only countably many distinct subgroups.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part A Problem 21[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Group Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Abstract Algebra, Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][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="1" _address="0.1.0.1"]

Start with hints

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This is a really artistic problem.
Pre-Solution Thoughts:
Lemma: A group is finite iff the number of subgroups of a group is finite.(Check!)
The lemma is a way to understand that an infinite group will have infinite number of subgroups.
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  • We are considering here set of all subsets of a countable group G, which is a subset of the Power Set of G.
  • Given G is countable the Power Set of G is uncountable.
  • Now we know that \(2^d\), where d is the set of natural numbers denotes the cardinality of the Power Set of G which is an uncountable set as it is bijective to the Real Numbers [Here d is not finite by the way, d is countably infinite].
  • So it is kinda intuitive that it may be uncountable.
  • First I took the group (Z,+), but we all know that the subgroups of  Z are nZ only. This doesn’t solve our purpose.
  • So naturally the next choice was the group (Q.+) whose subgroup is (Z,+).
  • While understanding the subgroups of (Q.+), the question is solved.
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  • We need to understand the subgroups of (Q.+).
  • Consider any rational number q and the subgroup qZ of (Q.+) generated by q.
  • What if we take two rational numbers?
  • For simplicity check the subgroup generated by {1/2 , 1/3}.
  • Prove that the subgroup generated by {1/2 , 1/3} is (1/2.3)Z.{Observe that co-prime property of 2 and 3 is playing an important role}.
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  • Now what if we take three mutually coprime natural numbers say a,b,c and see the subgroup generated by { 1/a , 1/b ,1/c}. In fact for simplicity take a,b,c to be primes.
  • Observe that it is of the form (1/a.b.c)Z.
  • Hence for every finite subset of primes, we generate a distinct subgroup of (Q,+).
  • Hence the total number of subgroups contained all the subsets of primes, which has a bijection with the Real Numbers as mentioned above.
  • So the answer is False.
Exercise
  • Find all the subgroups of (Q,+) with proof.
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Watch the video

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