This group theory problem has stumped me. I want to prove that if $G=(x)$ is a finite cyclic group that $(x^n) \cap (x^m) = (x^{\operatorname{lcm}(m,n)})$ for all integers $m$ and $n$, where $(x)$ is the group generated by $x$. Thoughts?
Intersection of cyclic subgroups: $(x^m) \cap (x^n) = (x^{lcm(m,n)})$
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5@user52714 Note in particular, on the first of Thomas' links, the following: "Etiquette : Civility is required at all times; rudeness will not be tolerated." β 2012-12-12
3 Answers
HINT: The elements of $\langle x^n\rangle$ are the powers $x^{kn}$ for $k\in\Bbb Z$, and the elements of $\langle x^m\rangle$ are the powers $x^{km}$ for $k\in\Bbb Z$. What integers are in both sets? I.e., what is
$\{kn:k\in\Bbb Z\}\cap\{km:k\in\Bbb Z\}\;?$
And how does that set compare with $\{k\operatorname{lcm}(m,n):k\in\Bbb Z\}$?
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0@user52714: Just use the fact that an integer $r$ is a multiple of both $m$ and $n$ iff itβs a multiple of $\operatorname{lcm}(m,n)$: $\{kn:k\in\Bbb Z\}\cap\{km:k\in\Bbb Z\}=\{k\operatorname{lcm}(m,n):k\in\Bbb Z\}\;.$ β 2012-12-12
Simply put, the intersection of the groups generated by $x^n$ and $x^m$ is the set of $x^r$ where $r$ is divisible by both $m$ and $n$. The element $x^{\hbox{lcm}(m,n)}$ will be contained in this group because $\hbox{lcm}(m,n)$ is such a value of $r$.
Moreover, any number divisible by both $m$ and $n$ will also be divisible by $\hbox{lcm}(m,n)$. That means that $x^{\hbox{lcm}(m,n)}$ generates the group.
This proves that $\left
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0Thanks for catching that. Replaced "dividing" with "divisible by". β 2012-12-12
Let $\ell=\operatorname{lcm}(m,n)$. Then $\langle x^n\rangle=\{x^{kn}:k\in\Bbb Z\}$, $\langle x^m\rangle=\{x^{km}:k\in\Bbb Z\}$, and $\langle x^\ell\rangle=\{x^{k\ell}:k\in\Bbb Z\}$, so $\langle x^n\rangle\cap\langle x^m\rangle=\langle x^{\ell}\rangle$ if and only if $\{kn:n\in\Bbb Z\}\cap\{km:m\in\Bbb Z\}=\{k\ell:k\in\Bbb Z\}$.
Since $\ell=\operatorname{lcm}(m,n)$, there are integers $r$ and $s$ such that $\ell=rm=sn$.
Suppose that $t\in\{k\ell:k\in\Bbb Z\}$, so that $t=k\ell$ for some integer $k$. Then $t=krm=ksn$, where $kr$ and $ks$ are integers, so $t\in\{kn:n\in\Bbb Z\}\cap\{km:m\in\Bbb Z\}$. This shows that
$\{k\ell:k\in\Bbb Z\}\subseteq\{kn:n\in\Bbb Z\}\cap\{km:m\in\Bbb Z\}\;.$
Now suppose that $t\in\{kn:n\in\Bbb Z\}\cap\{km:m\in\Bbb Z\}$, so that there are integers $a$ and $b$ such that $t=am=bn$. Then $t$ is a common multiple of $m$ and $n$, and therefore it is divisible by their least common multiple: $\ell\mid t$, so $t\in\{k\ell:k\in\Bbb Z\}$. This shows that
$\{kn:n\in\Bbb Z\}\cap\{km:m\in\Bbb Z\}\subseteq\{k\ell:k\in\Bbb Z\}$
and hence that
$\{kn:n\in\Bbb Z\}\cap\{km:m\in\Bbb Z\}=\{k\ell:k\in\Bbb Z\}\;.$
And this, as already noted, shows that $\langle x^n\rangle\cap\langle x^m\rangle=\langle x^{\ell}\rangle$.