CSIR JUNE 2011 PART B QUESTION 40 SOLUTION (Center of groups of order $77$)

Let $G$ be a group of order $77$, then the center of $G$ is isomorphic to
1)$\Bbb Z_1$,
2)$\Bbb Z_7$,
3)$\Bbb Z_{11}$,
4)$\Bbb Z_{77}$.
Solution:
Result: Let $G$ be a finite group of order $pq$, where $p,q$ are prime numbers such that $p<q$. If $p \nmid q-1$, then $G$ is cyclic. In particular isomorphic to $\Bbb Z_{pq}$. If $p \mid q-1$, then there exists a non-abelian group of order $pq$.
Now, in our problem, it is given that order of $G$ is $77 = 7 \times 11$ and $7 \nmid 10 = 11-1$. Hence by the above result, $G$ should be cyclic and isomorphic to $\Bbb Z_{77}$. Since this group is abelian, the center is the full group $\Bbb Z_{77}$.
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