 Solution

Use the Chinese remainder theorem to separate relatively prime factors:These five cyclic prime-power summands are the elementary-divisor form. Leave one $\mathbb{Z}/3\mathbb{Z}$ summand aside and combine the factors of orders $4,3,5,7$, which are pairwise coprime. Their direct sum of modules is cyclic of order $420$. The invariant-factor canonical form is thereforeIts group order is $3\cdot 420=1260=60\cdot 21$, and its exponent of a finite group is $420=\mathrm{lcm}(60,21)$, providing independent checks on the decomposition. Since its exponent is smaller than its order, $A$ itself is not cyclic.