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So I want to show that $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle$. My approach to this problem was to show a double containment, i.e. to show that $\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle$ and $ \langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle$.

I would like to see a full proof for this, specifically $\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle$. (I tried it with the approach of breaking it down into to cases; $a$ has infinite order and $a$ has finite order, the latter of which i would appreciate the most help on.)

My approach to solving the whole problem: (I would appreciate any feedback on anything that is wrong, or a different approach to the proof.)

To show the easier containment, $\langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle$, I did the following: Let $l = \operatorname{lcm}(m, n)$. Let $j \in \langle a^l\rangle$, so $j = (a^l)^k = a^{lk}$ for some $k \in \mathbb Z$. Since $l$ is a multiple of $m$ and $n$ by definition, we can say $l = ms = nt$ for some $s, t \in \mathbb Z$. Now $j = a^{kl} = a^{kms} = (a^m)^{ks} \in \langle a^m\rangle$. Similarly, $j = a^{kl} = a^{knt} = (a^n)^{kt} \in \langle a^n\rangle$. Now, since $j \in \langle a^m\rangle$ and $j \in \langle a^n\rangle$, it follows that $j \in \langle a^m \rangle \cap \langle a^n \rangle$. Thus, by definition, $\langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle$.

For the second containment, the one which I'm having more problems with, I attempted the following:

Case in which $a$ is infinite: Suppose that $\vert a \vert = \infty$. Let $c \in \langle a^m \rangle \cap \langle a^n \rangle$. Then $c = a^{mx} = a^{ny}$ where $x, y \in \mathbb Z$. It follows that $a^{mx - ny} = e$ and so $mx = ny$ because if $mx > ny$ then the difference would not be zero, and we would have an element that was finite, contradicting our hypothesis. And since $mx = ny$ we know $\operatorname{lcm}(mx, ny) = mx = ny$ and $\operatorname{lcm}(mx, ny)$ is a multiple of $\operatorname{lcm}(m, n)$. Hence $c \in \langle a^{\operatorname{lcm}(m, n)}\rangle$ and thus $\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle$.

Case in which $a$ is finite: I tried starting it out the same as the previous case, but i could never reach my conclusion :(

Shaun
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  • Robert:May you show where this problem came from? – Victor Apr 15 '12 at 21:09
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    @Victor This problem does not require context to the level of source of the problem. Robert has elaborated enough for us to believe he has worked through this problem a bit and not merely worked his fingers on the ctrl keys. Further, this is not a problem that specialists' skills are required to track. This is a typical problem in a first course in group theory. Regards, –  Apr 15 '12 at 21:12
  • @Victor In any case 'Abstract Algebra: Theory and Applications' by Thomas Judson, and Yes, this is a problem from an introductory course on the subject. – Robert Cardona Apr 15 '12 at 21:16
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    The $\gcd$ and $\operatorname{lcm}$ have dual definitions through their universal properties as follows: $$\forall n:\quad\begin{array}{c c}n|\gcd(a,b)\iff n|a~\wedge~n|b \ a|n~\wedge~b|n\iff\operatorname{lcm}(a,b)|n.\end{array}$$ I saw these in a Bill D answer, and they make arguments for propositions like this pretty slick. – anon Apr 15 '12 at 21:24
  • repeat: http://math.stackexchange.com/q/82899/24430 , Also, it may help to try small examples. Maybe where $m$ and $n$ are relatively prime and then one where they are not. – john w. Apr 15 '12 at 21:29
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    @johnw.: Not quite a repeat; it's a "repeat" only in the case where $a$ is of infinite order, not in the case where it is of finite order. – Arturo Magidin Apr 15 '12 at 21:38
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    In fancy terms, the lattice of subgroups of a cyclic group is isomorphic to the lattice of divisors of the order of the group. – lhf Apr 16 '12 at 00:01
  • In a similar vein, $<a^m,a^n>$ = $<a^{gcd(m,n)}>$. – Nicky Hekster Apr 17 '12 at 22:07

2 Answers2

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$\def\lcm{\mathrm{lcm}}$Let $k$ be the order of $a$, with $k=0$ if $a$ is of infinite order. Note that $x\equiv y \pmod{0}$ is equivalent to $x=y$ (since $0|x-y$ if and only if $x-y=0$). Also, $\mathrm{lcm}(n,0) = \gcd(n,0) = n$.

Consider first the case with $n|k$ and $m|k$.

To show $\langle a^m\rangle\cap\langle a^n\rangle \subseteq \langle a^{\lcm(m,n)}\rangle$, let $x$ lie in the intersection. Then there exist $r,s\in\mathbb{Z}$ such that $x=a^{mr} = a^{ns}$, so $mr\equiv ns\pmod{k}$. Thus, $k|mr-ns$, so $n|mr-ns$ and $m|mr-ns$. Therefore, $n|mr$ and $m|ns$. Since $n|mr$, then $mr$ is a common multiple of $n$ and $m$, hence is a multiple of $\lcm(m,n)$ (similarly with $ns$), so we can find $q$ such that $mr=q\lcm(m,n)$. Thus, $x = a^{mr} = a^{q\lcm(m,n)}\in\langle a^{\lcm(m,n)}\rangle$, as desired.

Now for the general case, note that $\langle a^n\rangle = \langle a^{\gcd(n,k)}\rangle$; thus, we are reduced to showing that $\lcm(n,m) \equiv \lcm(\gcd(n,k),\gcd(m,k))\pmod{k}$. This is true, but you may not have it in your arsenal yet, so here's a proof (that no doubt Bill Dubuque can prove in a much slicker way):

Lemma. $\lcm(\gcd(n,k),\gcd(m,k)) = \gcd(\lcm(n,m),k)$.

Proof. Indeed, $\gcd(n,k)$ and $\gcd(m,k)$ both divide $k$, hence their least common multiple divides $k$; and $\lcm(n,m)$ is a common multiple of $\gcd(n,k)$ and $\gcd(m,k)$, so $\lcm(\gcd(n,k),\gcd(m,k))$ divides $\lcm(n,m)$. Thus, the left hand side divides the right hand side. Conversely, let $p$ be a prime and suppose that $p^a$ is the largest power of $p$ that divides $\gcd(\lcm(n,m),k)$. Then $p^a$ divides both $\lcm(n,m)$ and $k$, but $p^{a+1}$ does not divide at least one of them. Since $p^a$ divides $\lcm(n,m)$, then $p^a$ divides either $n$ or $m$, and we also know $p^a$ divides $k$. Either $p^{a+1}$ does not divide $k$, or it does not divide either $n$ or $m$. Thus, $p^a$ divides either $\gcd(n,k)$ or $\gcd(m,k)$, but $p^{a+1}$ divides neither. Therefore, $p^a$ is the largest power of $p$ that divides $\lcm(\gcd(n,k),\gcd(m,k))$. $\Box$

Now the result follows: note that $\gcd(\lcm(n,m),k) \equiv \lcm(n,m)\pmod{k}$ (express $\gcd(\lcm(m,n),k)$ as a linear combination of $\lcm(m,,n)$ and $k$), so we are done.

Arturo Magidin
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  • It took me a while to work it through. I followed it for the most part, however I don't think I would have figured that out myself :( Thank You though! – Robert Cardona Apr 15 '12 at 22:40
  • @Robert: The difficulty is when you are not picking the "distinguished" generators for $\langle a^n\rangle$ and $\langle a^m\rangle$ (with $n|k$ and $m|k$); as you can see, the case where $n|k$ and $m|k$ follows pretty much like it does in the infinite case. – Arturo Magidin Apr 15 '12 at 22:54
  • To Arturo Magidin: i have confusing about the difference between general case and the case where n|k and m|k, please help. – Victor Apr 16 '12 at 03:37
  • What is confusing about the difference? In one of them we assume $n|k$ and $m|k$; in the other we don't. For example, if $a$ has order $16$, we could have $n=6$ and $m=8$. What is it you are confused about, really? – Arturo Magidin Apr 16 '12 at 03:41
  • @ArturoMagidin: Sorry to bring up such an old question, but what is a proof for p^(a)|lcm(m,n) implies p^(a)|m or p^(a)|n? Intuitively it seems to make sense, but i havent found a proof yet (can't apply Euclid's Lemma if a>1). – greycatbird Aug 27 '20 at 20:52
  • @greycatbird: A proof depends on what you know/can apply. How do you define the least common multiple? How do you define the greatest common divisor? What properties about them do you know? Absent context, it’s hard to answer your question in a way that doesn’t end up wasting my time and yours by talking at cross purposes. – Arturo Magidin Aug 27 '20 at 21:26
  • @ArturoMagidin: Sorry. Adv. Undergrad w/ Intro. Elem. NT and Intro. Abstract Algebra courses. My attempt at proof: p^(a)|lcm(m,n)=mn/(m,n) implies p^(a)|mn (multiply both sides by g.c.d.). Can't necessarily use Euclid's lemma (if a>1), so i tried showing either (p^(a),m)=1 or (p^(a),n)=1 to show p^(a)|m or p^(a)|n. No luck yet. – greycatbird Aug 27 '20 at 23:44
  • @greycatbird: But how do you define the gcd and the lcm? Via their universal properties? With unique factorization? Have you established the connection between unique factorization and the gcd/lcm? – Arturo Magidin Aug 27 '20 at 23:54
  • @Greycatbird: You can’t prove $(p^a,m)=1$ or $(p^a,n)=1$ because they don’t have be true: $p$ could divide both $m$ and $n$. But the highest power of $p$ that divides lcm$(a,b)$ can’t be larger than the greater of largest power dividing $n$ and largest power dividing $m$. – Arturo Magidin Aug 27 '20 at 23:55
  • @ArturoMagidin: I was using lcm(m,n)=mn/((m,n), but using the prime factorization definition, I think I get it. Certainly p^(b)|m and p^(c)|n for some b,c in {0,1,2,...} (it may even be that b=0 or c=0). It cannot be, however, that a>b and a>c, since p^(a)|lcm(m,n)=...p^max{b,c}... . If this were true, then p^(a-max{b,c})|lcm(m,n)/p^max{b,c}, which is impossible. Hence a<b or a<c. Therefore p^(a)|m or p^(a)|n. – greycatbird Aug 28 '20 at 13:20
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    @greycatbird: The prime factorization definition tells you that the power of $p$ that divides the gcd is the smallest of the largest power dividing $m$ and the largest power dividing $n$; and the power dividing the lcm is the largest of those two. But you can also deduce it from $mn=\gcd(m,n)\mathrm{lcm}(m,n)$ by showing what is the largest power that divides $\gcd(m,n)$. – Arturo Magidin Aug 28 '20 at 14:25
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    Upon trying to prove $\gcd(\text{lcm }(n,m), k)\equiv \text{ lcm }(n,m) \text{ mod }k$, as suggested, writing said gcd as a linear combination $\text{ lcm }(n,m)\cdot s +k\cdot t$, and reducing modulo $k$ leaves $s\cdot\text{lcm }(n,m)$. How is $s\cdot\text{lcm }(n,m)\equiv\text{ lcm }(n,m)\text{ mod }k$? –  Jan 05 '23 at 23:37
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    @ArturoMagidin Coming here from upanddownintegrate's question. Note that $\operatorname{lcm}(n,m) \equiv \operatorname{lcm}(\gcd(n,k),\gcd(m,k))\pmod{k}$ is not generally true (an example from that question's comments: $n=2, m=3, k=11$ gives $6\equiv 1 \pmod{11}$). However, you never needed to show that; instead you only needed to show exactly what you showed in the lemma. – Brian Moehring Jan 07 '23 at 02:29
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Note $ $ The equality in the lemma in Arturo's answer - that gcd distributes over lcm - can be derived simply via lcm, gcd laws. Rewriting lcms to gcds using $\rm\,[a,b] := lcm(a,b) =\: ab/(a,b)\,$ and clearing denominators, we deduce $\,\rm (k,[n,m]) = [(k,n),(k,m)]\,$ is equivalent to

$$\rm (k,m,n)\:(km,kn,mn)\ =\ (k,m)\:(k,n)\:(m,n)$$

whic is true, by both $\rm = (kkm,kkn,kmn,kmm,knn,mmn,mnn)\:$ by the distributive law. $ $ QED
For further details on the above gcd arithmetic this answer.

Dually, we could rewrite gcds to lcms, reducing it to the lcm dual of the above. And, of course, the dual identity is true: lcm distributes over gcd. These are well-known identities that come to the fore in the lattice viewpoint, e.g. see Birkhoff, Lattice Theory.

Though it is no longer needed after the above, it is instructive to note that Arturo's proof of lhs|rhs follows nicely from a useful lemma. First, recall the universal properties of lcm, gcd

$$\rm\ [a,b]\:|\:n \;\iff\; a,b\:|\:n\quad where\quad\: [a,b] := lcm(a,b) $$

$$\rm n\:|\:(c,d) \;\iff\; n\:|\:c,d\quad where\quad (c,d) := gcd(c,d) $$

Employing these we immediately deduce the following "lcm divides gcd" lemma

$$\rm [a,b]\ |\ (c,d)\iff a,b\ |\ (c,d)\iff a,b\ |\ c,d $$

Now, applying this to the particular values in Arturo's lemma we deduce

$$\rm [(n,k),(m,k)]\ |\ (k,[n,m])\iff (n,k),(m,k)\ |\ k,[n,m]$$

which is true: $\rm\:(n,k),(m,k)\ |\ k\:$ and $\rm\:(n,k)\ |\ n\ |\ [n,m],\:$ and $\rm\: (m,k)\ |\ m\ |\ [n,m]\quad$ QED

Bill Dubuque
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