Exponent of noncyclic finite abelian group

Question

Prove that a finite abelian group $G$ of order $n$ which is not cyclic has an exponent $m$ with $m < n$.

Here is what I have so far:

Suppose that $G$ is a finite abelian group which is not cyclic. We know that for all $g \in G$, the order $o(g)$ of $g$ is a divisor of $n$ and is strictly less than $n$, since otherwise $g$ would generate the entire group $G$ and then $G$ would be cyclic. We also know that $o(g)$ is the smallest positive integer $k$ such that $g^k = e$. Since $G$ is finite, write $G = \{g_1, g_2, ..., g_n\}$ and let $n = p_1^{\alpha_1}p_2^{\alpha_2}...p_j^{\alpha_j}$ be the prime factorization of $n$.

Since $o(g)$ divides $n$ for each $g \in G$, we can write

$$ o(g_1) = p_1^{\beta_{1,1}}p_2^{\beta_{2,1}}...p_j^{\beta_{j,1}}, \quad o(g_2) = p_1^{\beta_{1,2}}p_2^{\beta_{2,2}}...p_j^{\beta_{j,2}}, \quad ..., \quad o(g_n) = p_1^{\beta_{1,n}}p_2^{\beta_{2,n}}...p_j^{\beta_{j,n}}, $$

where $\beta_{l,k} \leq \alpha_l$ for all $l,k$ and where we don't have equality in all of the exponents (this would give that the order of some group element was $n$). I know that if I choose $m = \text{lcm}(o(g_1), o(g_2), ..., o(g_n))$ then $m$ will have the property that $g^m = e$, and since each of $o(g_i)$ divide $n$ we will have $m \leq n$.

My problem is that I can't seem to prove that there is a strict inequality $m < n$, since in taking the least common multiple we consider the maximum exponent for each of the primes $p_1, p_2, ..., p_j$ in the prime factorization of $n$. So, it seems very plausible that we could have $m = n$. What am I missing?

Answer 1

Hint

One of the $p_i$ has to be repeated in the expression for $G$ as a product of cyclic groups of order $p_i^{\alpha _i}$ (by the fundamental theorem of finite abelian groups).

Otherwise we get cyclicity by the Chinese remainder theorem. Thus the exponent is strictly smaller than $n$.

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At least $p^2-1$ of the $g_i$ have order $p$, for some prime $p$.