I was solving some questions on group theory and I came across a problem something like this:
Let $G$ be a group of order 77. Then the order of the centre of the group is:
$77= 11\times7$. Since $7\nmid(11-1)$, thus we have $G$ to be abelian. Since $G$ is abelian $G=Z(G)$. Therefore $O(G)=O(Z(G))=77$.
But the solution provided a hint saying that "the order of the the centre of the group is order of the largest subgroup of $G$", which is the only normal subgroup of $G$ and hence the order of the centre of the group is $11$.
Can anybody correct me where and what am I missing in this question?