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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.

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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).

**What is this topic:**Abstract Algebra**What are some of the associated concept:**Order of a Subgroup**Book Suggestions:**Contemporary Abstract Algebra by Joseph A. Gallian

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