INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

July 13, 2017

TIFR 2013 problem 9 | Existence of element of order 51

Try this problem from TIFR 2013 problem 9 based on the existence of an element of order 51.

Question: TIFR 2013 problem 9

True/False?

There is an element of order \(51\) in \(\mathbb{Z_{103}^*}\)

Hint: 103 is prime.

Discussion: Since for \(p\)-prime, \(\mathbb{Z_p}\) is a field with addition modulo p and multiplication modulo p as the first and second binary operations respectively, and for a finite field \(\mathbb{F}\), \(\mathbb{F}^*\) forms a cyclic group with respect to multiplication. Hence \(\mathbb{Z_p}^*\) forms a cyclic group.

We have \(\mathbb{Z_{103}^*}\), a cyclic group or order \(103-1=102\).

For a cyclic group of order \(n\), if \(d|n\) then there exists an element of order d.

Since \(51|102\), there exists an element of order \(51\) in \(\mathbb{Z_{103}^*}\).

Alternative: By Sylow's first theorem, there exists elements of order 3&17 in \(\mathbb{Z_{103}^*}\). ( \(|\mathbb{Z_{103}^*}|=102=2*3*17\))

Since the group \(\mathbb{Z_{103}^*}\) is abelian, the group and 3,17 are co-prime we get an element of order \(3*17=51\) (by simply multiplying the element of order \(3\) to the element of order \(17\)).

Some Useful Links:

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com