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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?

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marked as duplicate by amWhy, MathOverview, TMM, Vedran Šego, egreg Dec 6 '13 at 14:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Can I asked why you posted the same question you posted an hour ago? – amWhy Dec 12 '12 at 2:56
@user52714: please read the faq about how this site works: Also please read:… – Thomas Dec 12 '12 at 2:59
@user52714 Note in particular, on the first of Thomas' links, the following: "Etiquette : Civility is required at all times; rudeness will not be tolerated." – John Martin Dec 12 '12 at 3:13

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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I've looked at it from that angle, but I'm just not reaching my conclusion. I need only show that the intersection is a subset of (x^lcm(m,n)) but this proving more challenging than it appears – Zack C Dec 12 '12 at 1:51
@user52714: Just use the fact that an integer $r$ is a multiple of both $m$ and $n$ iff it’s a multiple of $\operatorname{lcm}(m,n)$: $$\{kn:k\in\Bbb Z\}\cap\{km:k\in\Bbb Z\}=\{k\operatorname{lcm}(m,n):k\in\Bbb Z\}\;.$$ – Brian M. Scott Dec 12 '12 at 1:54

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 $\left<x^m\right>\cap \left<x^n\right>$ is in fact $\left< x^{\hbox{lcm}(m,n)}\right>$.

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Thanks for catching that. Replaced "dividing" with "divisible by". – orlandpm Dec 12 '12 at 3:09

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 $\langle x^n\rangle\cap\langle x^m\rangle=\langle x^{\ell}\rangle$.

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