Let $G_1, G_2, \ldots, G_t$ be finite cyclic groups, and define $G = G_1 \times G_2 \times \ldots \times G_t$. Let $n_j = |G_j|$, such that $|G| = \prod_{j=1}^{t} n_j$.
For part (a) of the problem, we are asked to suppose that $\gcd(n_i, n_j) > 1$ for some $i \neq j$ and show that $G$ is not cyclic.
I understand that a group $G$ is cyclic if there exists an element $g \in G$ such that every element of $G$ can be written as $g^k$ for some integer $k$. My attempt to solve this part is based on the hint given:
Hint: Show that the order of any element is less than $n$, where $n = \prod_{j=1}^{t} n_j$.
Here is my incomplete attempt:
I started by considering an arbitrary element $(g_1, g_2, \ldots, g_t) \in G$, where each $g_j$ is a generator of $G_j$. Then, the order of this element would be the least common multiple of the orders of $g_j$, which is $\operatorname{lcm}(|g_1|, |g_2|, \ldots, |g_t|)$.
Since $G_j$ are all cyclic, $|g_j| = n_j$. However, I'm stuck on how to proceed from here. I understand that the $\gcd(n_i, n_j) > 1$ condition should imply that the $\operatorname{lcm}$ of the $n_j$'s is less than the product of the $n_j$'s, but I'm not sure how to show this formally. Can someone help me complete this proof?
Note that for all $(g_m)_{m=1}^{t} ∈G$ $$ \text{lcm}(|g_m|) ≤ \text{lcm}(|g_{m ≠ i, j}|)×\text{lcm}(|g_i|, |g_j|) = \text{lcm}(|g_{m ≠ i, j}|)\frac{n_{i}n_{j}}{\text{gcd}(n_i, n_j)} \lt n_jn_i\prod_{m \ne i,j} n_m = |G|$$ as $\text{gcd}(n_i, n_j) > 1$. Thus the order of every element is less than the order of the group. Alternatively, you could prove that the product $G_i \times G_j$ is non-cyclic, thus the entire product is non-cyclic.