1

I'm citing below the definition of and a theorem about cyclic groups, as it is written in my book (Algebra, by Thomas Hungerford):

Definition:

Let $ G $ be a group (notation is multiplicative in here.) For every $ a \in G, $ a cyclic group is: $ \langle a \rangle = \{a^n : n\in \mathbb{Z}\} $

and,

Theorem:

Every subgroup $ H $ of the additive group $ ( \mathbb{Z},+ ) $ is cyclic. Either $ H = \langle 0 \rangle, $ or $ H = \langle m \rangle, $ where $ m $ is the least positive integer in $ H $.

My question is;

I know that, and I would, express $ H $ as the union of all $ H_m = \{ m \mathbb{Z} : m \in H \} $.

The reason why I'm specifying this notation is that I don't understand the fact that $ m $ should be the least integer there (if I get it right of course.)

Thank you for giving advice.

Community
  • 1
freehumorist
  • 571

3 Answers3

2

You can use division with remainder to show that $H = m\mathbb{Z}$ for some $m$. For let $m$ be the smallest positive element of $H$. (If $H=\{0\}$, there is nothing to prove.) Then for any $k \in H$, we have

$$ k = q \cdot m + r \qquad \text{for some }q\text{, with } 0 \leq r < m.$$ Since both $k$ and $m$ lie in $H$ and $H$ is a subgroup, this means $r \in H$. But $m$ was the smallest positive element, so $r = 0$.

SvanN
  • 2,307
2

$ H$ closed under subtraction $\Rightarrow$ $H$ closed under remainder $= \bmod $ (via repeated subtraction) hence $H$ is closed under $\gcd,\,$ since gcds are computable by repeated $\!\bmod$ (or subtraction) by Euclid. Now it's easy to show $H$ is generated by the gcd of its elements - its least positive element (for $H\neq 0)$

This is the intuition behind the proof. You might find it enlightening to see how it is employed in this proof of Bezout's GCD identity. These ideas will be clarified when one studies (principal) ideals in rings, and the result that Euclidean domains are PIDs (a generalization of Bezout)

Bill Dubuque
  • 272,048
  • Thank you Bill. To specify, my trouble wasn't about the method, but the idea behind. – freehumorist Dec 21 '18 at 20:32
  • @bugrahaskan The idea is as I explained, i.e. being closed under (repeated) subtraction implies $H$ is closed under gcds, so $H$ is generated by the gcd of all its elements, necessarily the least positive element. Algorithmically it amounts to applying the Euclidean algorithm to compute $,\gcd(h_1,h_2,h_3,\ldots)\ $ – Bill Dubuque Dec 21 '18 at 20:38
  • See here and here for further conceptual elaboration. – Bill Dubuque Nov 04 '23 at 07:22
0

Suppose $H\neq \{0\}=\langle 0\rangle$. Then $\exists h\in H$ such that $h\neq 0$. Let $$m=\min \{g\in H\mid 0\lt g\leq |h|\}$$ where $|h|=h$ if $h\gt 0$, and $|h|=-h$ if $h\lt 0$. Note $m$ exists by well-ordering principle. We claim $H= \langle m\rangle=\{km\mid k\in \Bbb{Z}\}$. Since $m\in H$ and $H$ is closed under $+$, we have $km\in H$ for all $k\in \Bbb{Z}$. Thus $\langle m\rangle\subseteq H$. Assume towards contradiction, $\exists a\in H$ such that $a\notin \langle m\rangle$. That is $m \nmid a$. By division algorithm, $a=mq+r$ where $q\in \Bbb{Z}$ and $0\lt r\lt |m|=m$. Since $-mq=m(-q)\in \langle m\rangle$, we have $-mq\in H$. So $a-mq=r\in H$. So $\exists r\in H$ such that $0\lt r\lt m$. Thus we reach contradiction. Hence $H\subseteq \langle m\rangle$ and $H= \langle m\rangle$.

user264745
  • 4,143