TIFR 2014 Problem 24 Solution 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
Let (H_1, H_2) be two distinct subgroups of a finite group G, each of order 2. Let (H) be the smallest subgroup containing (H_1) and (H_2). Then order of (H) is
A. always 2
B. always 4
C. always 8
D. none of the above
Let's check out the simplest example where we suspect things might get wrong.
Take (S_3) as the group. It has 3 order 2 subgroups. If we take any two of those then the subgroup generated by (i.e the smallest subgroup containing) should have order greater than 2; so it must have order 3 or 6 and that itself leads us to conclude that the answer is none of the above.
For example take (<(1 2)>=H_1,<(13)>=H_2). Then (H) contains ( (1 2), (1 3), (1 2)(1 3)=(1 3 2), (1 3)(1 2)=(1 2 3)). Therefore (H) has atleast 4 elements, so it must have 6 elements i.e, (H=S_3).
TIFR 2014 Problem 24 Solution 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
Let (H_1, H_2) be two distinct subgroups of a finite group G, each of order 2. Let (H) be the smallest subgroup containing (H_1) and (H_2). Then order of (H) is
A. always 2
B. always 4
C. always 8
D. none of the above
Let's check out the simplest example where we suspect things might get wrong.
Take (S_3) as the group. It has 3 order 2 subgroups. If we take any two of those then the subgroup generated by (i.e the smallest subgroup containing) should have order greater than 2; so it must have order 3 or 6 and that itself leads us to conclude that the answer is none of the above.
For example take (<(1 2)>=H_1,<(13)>=H_2). Then (H) contains ( (1 2), (1 3), (1 2)(1 3)=(1 3 2), (1 3)(1 2)=(1 2 3)). Therefore (H) has atleast 4 elements, so it must have 6 elements i.e, (H=S_3).
