I was reading the answer to this question: Explicit descriptions of groups of order 45 and the accepted answer says the Sylow $3$-subgroup is either isomorphic to $\mathbb Z_9$ or $\mathbb Z_3 \times \mathbb Z_3$. But now my question is: Why isn't $\mathbb Z_9 \cong \mathbb Z_3 \times \mathbb Z_3$ by the fundamental theorem of finite abelian groups?
The fundamental theorem of finite abelian groups says: Every finite abelian group is the direct product of cyclic groups. (Herstein)
So isn't $\mathbb Z_9 \cong \mathbb Z_3 \times \mathbb Z_3$ by this theorem?
Best Answer
In general, one needs $\mathrm{gcd}(m,n) = 1$ for $\mathbb{Z}_{mn}$ to be isomorphic to $\mathbb{Z}_m \times \mathbb{Z}_n$.