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:

My attempt:

$77= 11\times7$. 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?

  • 1
    $\begingroup$ @AndreasCaranti If the hint is just "the order of the centre of the group is order of the largest subgroup of $G$", then this is fine. As the largest subgroup of $G$ is $G$ itself. (I mean, it's a rubbish hint. But its correct.) $\endgroup$ – user1729 Jun 5 at 13:52
  • $\begingroup$ Thanks: But is there any relation between center of the group and largest subgroup(not itself) or the normal subgroup for any group not specifically needed to be abelian? $\endgroup$ – MB17 Jun 5 at 13:55

Prove the following lemma: if $G/Z(G)$ is cyclic, then $G$ is abelian.

In particular this shows that $Z(G)$ never has prime index. So if $|G|= pq$ this rules out the cases $|Z(G)| \in \{p,q\}$. So either $G$ is abelian or its center is trivial.

Both cases are possible. Here seems to be a construction of a nonabelian group of order $pq$, where $p< q$ and $q = 1 \pmod{p}$.


Theorem. A group of order $pq$ is either abelian or has trivial centre.

Proof. Suppose the theorem does not hold. Then there exists a group $G$ of order $pq$ whose centre has order $p$, say. Therefore, $G/Z(G)$ is cyclic of order $q$. However, if $G/Z(G)$ is cyclic then $G$ is abelian (see this old question for a proof), a contradiction. QED

On the other hand, this is overkill for the question at hand. As the OP points out, because $7\nmid(11-1)$ the group is abelian (essentially by the classification of groups of order $pq$). Therefore, if $|G|=77$ then $|Z(G)|=77$.


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