# Group of Order $pq$ is Either Abelian or the Center is Trivial ## Problem 30

Let $G$ be a group of order $|G|=pq$, where $p$ and $q$ are (not necessarily distinct) prime numbers.

Then show that $G$ is either abelian group or the center $Z(G)=1$. Add to solve later

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## Hint.

Use the result of the problem “If the Quotient by the Center is Cyclic, then the Group is Abelian”.

## Proof.

Since the center $Z(G)$ is a (normal) subgroup  of $G$, the order of $Z(G)$ divides the order of $G$ by Lagrange’s theorem.
Thus the order of $Z(G)$ is one of $1,p,q,pq$.

Suppose that $Z(G)\neq 1$.
Then the order of the quotient group $G/Z(G)$ is one of $1,p,q$.
Hence the group $G/Z(G)$ is a cyclic group.

We conclude that $G$ is abelian group by Problem “If the Quotient by the Center is Cyclic, then the Group is Abelian”.

Therefore, either $Z(G)=1$ or $G$ is abelian. Add to solve later

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