A strange problem in Dummit Foote Abstract Algebra Recall the following fact.

If $G$ is a group of order $pq$ where $p,q$ are primes such that $p<q$ and $q\not\equiv 1\bmod p$ then $G$ is cyclic.

The following is Exercise 4.3.6 in Dummit Foote Abstract algebra.

Assume $G$ is a non-abelian group of order 15. Prove that $Z(G)=1$.

Since $|G| = 3\cdot 5$ and $5\not\equiv 1\bmod 3$ so from the above fact, $G$ is cyclic. In particular, it's abelian. Hence, such $G$ does not exist. Am I misunderstanding something here?
 A: No, you are not. And it follows from what you wrote that it is indeed true that the center of non-abelian group of order $15$ is trivial (since there is no such group).
A: What you wrote is of course correct, and because of this, the implication is vacuously true. However, probably the author expected something like this:
Lemma: Let $G$ be a group. If $G / Z(G)$ is cyclic, then $G$ is Abelian.
Proof: Let $\bar{x}$ the generator of $G / Z(G)$. Now, let $y, z \in G$. There exists $n, m \in \mathbb{Z}$ and $a, b \in Z(G)$ such that $y = x^n a$ and $z = x^m b$. It is easy to see that $y z = z y$, so $G$ is indeed Abelian. $\square$
Now, let's go back to our problem. By Lagrange's theorem, possible orders for the center of a non-Abelian group of order 15 are 1, 3, and 5. However, 3 and 5 are also impossible from the previous result, so the order of the center is indeed 1.
A: Given those assumptions, the result follows from the class equation. Since $G$ is non-abelian, $|Z(G)|=1$, or $3$, or $5$. If $|Z(G)|=3$, then every noncentral element has centralizer of order $3$, and hence (class equation): $3.5=$ $3+k\frac{3.5}{3}=$ $3+5k$, for some nonnegative integer $k$: contradiction, because $5\nmid 3$. Likewise, $|Z(G)|\ne 5$, because $3\nmid 5$. Therefore, $|Z(G)|=1$.
I guess you have incurred in a case of vacuous truth.
