 TIFR 2015 Problem 4 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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## PROBLEM:

Let $$S$$ be the collection of isomorphism classes of groups $$G$$ such that every element of G commutes with only the identity element and itself. Then what is $$|S|$$?

## Discussion:

Given any $$g\in G$$, it commutes with few obvious elements: $$e,g,g^2,…$$ , i.e, all integral powers of $$g$$.

So by given condition, this whole set {$$e,g,g^2,…$$} is same as the set {$$e,g$$}. So any element in $$G$$ must have order $$\le 2$$.

Now let us look at $$e$$. The identity commutes with every element. But by given condition, $$e$$ commutes with $$e$$ only. That implies there is no other element in $$G$$.

So, $$G=(e)$$.

So, $$|S|=1$$.

## HELPDESK

• What is this topic:Modern Algebra
• What are some of the associated concept: Isomorphism
• Book Suggestions: Topics in Algebra by I.N.Herstein