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I know that a group $G$ satisfies the maximal condition if and only if every subgroup of $G$ is finitely generated. So I think that if $G$ satisfies the maximal condition then it is finitely generated because $G$ is a subgroup of itself. Am I right or not?

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  • $\begingroup$ Correctamundo. ${}{}$ $\endgroup$
    – Pedro
    Dec 25, 2015 at 21:54
  • $\begingroup$ i can't understand what does amundo mean? $\endgroup$
    – Lauren
    Dec 25, 2015 at 21:57
  • $\begingroup$ Try this. $\endgroup$ Dec 25, 2015 at 21:58
  • $\begingroup$ @MerieMarissa It's a way of saying "correct". $\endgroup$
    – Pedro
    Dec 25, 2015 at 21:59

2 Answers 2

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Yes, indeed, $G$ is a subgroup of itself.

On the other hand, you could have groups so that every ${\it proper}$ subgroup is finitely generated, while $G$ itself is not. An example would be to consider a prime number and the $G$ the quotient group $\mathbb{Q}_p/ \mathbb{Z}$, where $\mathbb{Q}_p$ consists of all rational numbers with denominator a power of $p$.

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As said by Pedro Tamaroff in the comments, you are correct.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Dec 25, 2015 at 23:05
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    $\begingroup$ @StellaBiderman it's a community wiki... Spend a few seconds reading, while reviewing. $\endgroup$ Dec 25, 2015 at 23:06
  • $\begingroup$ Ah, sorry. I didn't see that when going through the review queue. $\endgroup$ Dec 25, 2015 at 23:07

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