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 $17$).