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.
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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.
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- 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
Center of group and normal subgroup of order 2
Any normal subgroup of order 2 is contained in the center of the group.
True
Discussion: Center of a group Z(G) is the sub group of elements that commute with all members of the group. A subgroup of order two has two elements: identity element and another element, say x, which is self inverse. Since Z(G) is a subgroup it contains the identity element. We show that the other element x also is in Z(G).
Suppose H is the normal subgroup of order 2 and H={1, x}. If g is an arbitrary element of G, then gH = Hg as H is normal. That is {g, gx} = {g, xg}. Since the two sets are equal and x is not identity, hence gx = xg. This implies that x is in the Center of the Group (as it commutes with an arbitrary element of the group).
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Cyclic Group - TIFR Problem - Video
There is an element of order 51 in the multiplicative group (Z/103Z)
True
Discussion: First note that (Z/103Z) has 102 elements as 103 is a prime (in fact one of the twin primes of 101, 103 pair). Also 102 = 2317. So it has Sylow-3 subgroup of order 3 (prime order hence it is cyclic too) and a Sylow-17 subgroup (which is similarly cyclic). Since (Z/103Z) is abelian all it's subgroups are normal. Thus product of Sylow-3 and Sylow-17 subgroups is a subgroup (direct product of normal subgroups is a subgroup) containing 51 elements which is again cyclic. Hence there is an element of order 51 (generator of this subgroup).