Let be G an abelian finite group. We suppose that G is not cyclic. We have to show that there exists a subgroup of G that is isomorphic to $Z/pZ \times Z/pZ $ for a prime number p.


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By the classification theorem for finitely-generated abelian groups, there are primes $p_1,\dots,p_r$ and positive integers $k_1,\dots,k_r$ such that $$ G\simeq \mathbb{Z}/p_1^{k_1}\mathbb{Z}\times\dots\times \mathbb{Z}/p_r^{k_r}\mathbb{Z}$$

If the primes $p_i$ were distinct then $G$ would be cyclic, so since $G$ is not cyclic at least two of the $p_i$ are equal, say $p_1$ and $p_2$.

Setting $p=p_1=p_2$, it follows that $G$ contains a subgroup isomorphic to $\mathbb{Z}/p^{k_1}\mathbb{Z}\times\mathbb{Z}/p^{k_2}\mathbb{Z}$, hence contains a subgroup isomorphic to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$.

  • $\begingroup$ Thank you for your explanations it is pretty clear i didnt think this way $\endgroup$ – Farouk Deutsch Jun 21 '17 at 1:30

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