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The problem is to find the maximal subgroups of $Z$ and $Z/nZ$.

I did the first part and I found the maximal subgroups of $Z$ are:

$nZ$ with n as a prime integer.

I'm a little bit confused with second part.

I know the structure of a subgroup of $Z/nZ$ is : $mZ/nZ$ with $nZ⊆mZ$ which means $m|n$

Here is my answer but I'm not sure if this is correct or not?

Let $H$ a maximal subgroup of $Z/nZ$: $∃m$ integer :$H$=$mZ/nZ$.

Let $K$ a subgroup of $Z/nZ$: $∃m'$ integer : $K$=$m'Z/nZ$

$H$$\subset$$K$ => $mZ/nZ$ $\subset$ $m'Z/nZ$ => $mZ$ $\subset$ $m'Z$ <=> $m'|m$.

H maximal and $H$$\subset$$K$ => $K$=$Z/nZ$ => $m'=1$ =>

the divisors of m are only m and 1 => m is prime. so Are the maximal subgroups of $Z/nZ$ : $mZ/nZ$ with $m$ a prime integer?!

user281002
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  • You can add that $m$ is a prime divisor of $n$. To shorten the proof, you can use the bijection between subgroup/ideals of $\mathbf Z/n\mathbf Z$ and subgroup/ideals of $\mathbf Z/n\mathbf Z$ which contain $n$ (i.e. divisors of $n$). – Bernard Jan 04 '18 at 12:53

1 Answers1

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Your solution is essentially right but can be better rendered as follows:

  • The maximal subgroups of $\mathbb Z$ are exactly $p\mathbb Z$ where $p$ is prime.

  • By the canonical correspondence $\mathbb Z \to \mathbb Z/n\mathbb Z$, the maximal subgroups of $\mathbb Z/n\mathbb Z$ correspond exactly to the maximal subgroups of $\mathbb Z$ that contain $n\mathbb Z$.

  • The subgroups of $\mathbb Z$ that contain $n\mathbb Z$ are exactly $m\mathbb Z$ where $m$ is a divisor of $n$.

  • The maximal subgroups of $\mathbb Z/n\mathbb Z$ are exactly $p\mathbb Z/n\mathbb Z$ where $p$ is prime divisor of $n$.

lhf
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