# For a group whose order is the product of two distinct primes, What is the order of the center of the group?

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?

• @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.) – user1729 Jun 5 at 13:52
• 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? – 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 $$|Z(G)|=77$$.