In the ring of integers, the only maximal ideals are those generated by the prime elements. Is the same true for the group of integers? Are the only maximal subgroups of integers the ones generated by the primes?
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Yes: this is true, and the proof goes exactly as for the case of the ring. – Crostul Mar 06 '16 at 10:25
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Any ideal is an additive subgroup; in the case of the integers, also the converse is true. – egreg Mar 06 '16 at 10:55
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Any subgroup of $\mathbb{Z}$ is a copy of $\mathbb{Z}$, of the form $n\mathbb{Z}$ for $n \in \mathbb{N}$. Since $n \mathbb{Z} \subseteq m \mathbb{Z}$ if and only if $m$ divides $n$, and primes are minimal among the natural numbers (excluding $1$) ordered by divisibility, the answer to your question is yes.
mad_algebraist
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The answer is yes. Since every subgroup of $\mathbb{Z}$ looks like $a \mathbb{Z}$ with $a$ integer, we only need to consider groups generated by a single element.
Proof :
Suppose $G=p\mathbb{Z}$ generated by $p$, a prime. Suppose we have an other group $H$ such that $G \varsubsetneq H$, then $\exists m \notin G, m \in H$.
But $gcd(m,p)=1$ since p is prime and $m \notin p\mathbb{Z}$. So by Bézout identity :
$$\exists u,v \in \mathbb{Z}, pu+mv=1$$
Hence $H=\mathbb{Z}$. So $G$ is maximal.
Now let $G$ being a maximal subgroup of $\mathbb{Z}$, suppose $G$ is not generated by a prime. We know that $\exists a \in \mathbb{Z}, G=a \mathbb{Z}$ because all the subgroups of $\mathbb{Z}$ have this form. But then, $a$ can be rewritten $a=pq$ with $p$ a prime. Then $a \mathbb{Z} \varsubsetneq p \mathbb{Z} \neq \mathbb{Z}$, contradiction.
nicomezi
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