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)$\Bbb Z_{77}$.

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,

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 $\Bbb Z_{77}$**.**Share to your groups:**

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