# Exponent of noncyclic finite abelian group

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?

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 $$n$$.

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

• Using the FT of finite abelian groups I get $G \cong A_1 \times ... \times A_j$, where $|A_i| = p_i^{\alpha_i}$ for each $i$, and each of these $A_i$ can be expressed as the direct product of groups of the form $$\mathbb{Z}_{p_i^{\beta_i]}$$, where the exponents $\beta_i$ must sum to $\alpha_i$. When you say one of the $p_i$ have to be repeated in the expression of $G$, this simply means that one of the $p_i$ must have exponent $\alpha_i$ greater than one, is that correct? Sep 23, 2022 at 4:50
• Well it does. But also, it means we have at least two cyclic factors for prime power $p$. Sep 23, 2022 at 5:04
• So, $G$ has a subgroup isomorphic to $\Bbb Z_p×\Bbb Z_p$. That forces the conclusion because now the contribution to the exponent from those two factors is less than the power of $p$ in the prime factorization. Sep 23, 2022 at 5:07
• Oops, correction, none of the $\alpha _i$ has to be greater than one. But in the prime factorization of $n$ it does. Sep 23, 2022 at 5:15