-2
$\begingroup$

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.

$\endgroup$

closed as off-topic by Derek Holt, Davide Giraudo, Arnaldo, Namaste, Claude Leibovici Jun 23 '17 at 13:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Davide Giraudo, Arnaldo, Namaste, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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 $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.