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July 12, 2017

TIFR 2013 Problem 10 Solutions - Normal Subgroup of Order 2


TIFR 2013 Problem 10 Solutions is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Contemporary Abstract Algebra by Joseph A. Gallian. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program


Problem:True/False?


Any normal subgroup of order 2 is contained in the center of the group.


Discussion:


If \(N\) is a normal subgroup of a group \(G\) and \(|N|=2\), then \(N=\left\{e,a\right \}\) where \(a^2=e\).

For all \(g\in G\) we have \(gag^{-1}\in N\).

Can \(gag^{-1}=e\)? No. Since that would imply \(a=e\).

Therefore, for all \(g\in G\), \(gag^{-1}=a\).

Which proves that a is in the center of the group.


Helpdesk

  • What is this topic:Abstract Algebra
  • What are some of the associated concept: Normal Subgroup, Center of a Group
  • Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian

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