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July 14, 2017

TIFR 2013 problem 5 | Non-Cyclic Subgroup of \(\mathbb{R}\)

Try this problem from TIFR 2013 problem 5 based on Non-Cyclic Subgroup of \(\mathbb{R}\).

Question: TIFR 2013 problem 5

True/False?

All non-trivial proper subgroups of \((\mathbb{R},+)\) are cyclic.

Hint: What subgroups comes to our mind immediately?

Discussion: \((\mathbb{Q},+)\) is a subgroup of \((\mathbb{R},+)\). Is \((\mathbb{Q},+)\) a cyclic group?

Suppose \((\mathbb{Q},+)\) is cyclic. Then there exists a generator say \(\frac{a}{b}\). Note that, we are only allowed to use addition (and subtraction) to create \(\mathbb{Q})\

Therefore, we can create $$ \frac{a}{b}+\frac{a}{b}+...+\frac{a}{b}=n\frac{a}{b}=\frac{na}{b} $$

Also, we can create  $$ (-\frac{a}{b})+(-\frac{a}{b})+...+(-\frac{a}{b})=n(-\frac{a}{b})=-\frac{na}{b} $$

Notice that we can increase the magnitude of the numerator, but not the magnitude of the denominator.

For example, we cannot create \(\frac{a}{2b}\) using \(\frac{a}{b}\) and the binary operation +.

Therefore, \((\mathbb{Q},+)\) is not cyclic.

Remark: There is one result which states that subgroups of \((\mathbb{R},+)\) are either cyclic or dense. Notice that although \((\mathbb{Q},+)\) is not cyclic it is dense.

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