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

Question: TIFR 2013 problem 9


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 \(17\)).

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