 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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## Question:

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

## Discussion:

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, $$H=S_3$$.

## Helpdesk

• 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