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Let $G$ be a finite abelian group in which there are at most $n$ solutions of the equation $x^n = e$ for each posivite integer $n$. How to determine if $G$ is cyclic or not?

Saaqib Mahmood
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    For general case-it is not necessary abelian, you can see http://math.stackexchange.com/questions/346936/finite-group-for-which-xxm-e-leq-m-for-all-m-is-cyclic. – Bobby Apr 02 '14 at 14:25

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By the classification of finite abelian groups we know $$ G \cong \mathbb{Z}/(n_1) \oplus \mathbb{Z}/(n_2) \oplus \cdots \oplus \mathbb{Z}/(n_k)$$

where $n_i$ divides $n_{i+1}.$ If $k\geq 2$ i.e. there are at least two summands in this decomposition, then the condition fails: $x^{n_1}=e$ has $n_1$ solutions from $\mathbb{Z}/(n_1) \oplus 0 \oplus \cdots \oplus 0,$ as well as some from inside $0\oplus \mathbb{Z}/(n_2) \oplus 0 \oplus \cdots \oplus 0.$ So if the condition holds then $k=1$ i.e. G is cyclic.

Ragib Zaman
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