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If $G$ is a finite abelian group, $a, b\in G$ such that $|a|=m>1, |b|=n>1$ where $m, n$ are not necessarily relatively prime, then prove or disprove that $G$ has an element $ab$ of order lcm$(m, n).$

My claim: It is not true in general. For example consider $G=\langle x\rangle $ with $|x|=4, a=x$ and $b=x^3$. Then we can see that $|a|=4=|b|$ but $|ab|=1\neq lcm(4, 4)$.

Am I correct in my logic? Please help me. Thanks in advance

amWhy
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KON3
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  • I assume you meant that $|G|=4$ so that $x^4=e \in G$ for all $x \in G$. But then notice that $|a|=|x|=4$ and $|b|=|x^3|=4$ so lcm is $4$ but then $ab=xx^3=x^4=e$. So $|ab|=|e|=1 \neq 4$. This still doesn't produce a counterexample, you need to show that given $|a|$ and $|b|$ that there would be no element having order of their lcm, if you believe that the conjecture is false. – mathematics2x2life Dec 03 '13 at 18:04
  • @mathematics2x2life I apologize. It was not in my mind. Yes you are correct. I did the edit. Thanks for reminding me. – KON3 Dec 03 '13 at 18:45
  • Note, Anjan, that if every group $G$, the following holds: $\exists e \in G$, so put $a = b = e$. Then $ab = e^2 = e$ and $|ab| = |e| = 1 = \operatorname{lcm}(|a|, |b|)$. So trivially, yes, for every group, there is some $(a, b)$ such that $|ab| = \operatorname{lcm}(|a|, |b|)$. – amWhy Dec 03 '13 at 19:00
  • @amWhy I understood. The edit has been made. So we shall now have to deal with non identity elements only. In that case, I got your answer already. – KON3 Dec 03 '13 at 19:07
  • I felt already :-D @amWhy – KON3 Dec 03 '13 at 19:15

3 Answers3

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You are correct that the proposition is false, if it's a proposition that is being claimed for all products $ab$ of arbitrary $a\,b, \in G$. And your counterexample is a very good one: provided you specify that $|G| = |\langle x\rangle| = 4$ (so you need to make that explicit).

For the sake of transparency, perhaps make explicit what you want to denote as $a$, as $b$, and $G$. For example, we could put $G = \mathbb Z_4$ (the additive group of integers modulo $4$, with specific elements $a = 1$, $b = 3$ such that $|a| = |b| = 4$.

Then as you rightly claim, $|a| = m = 4 = n = |b|$, but $|ab| = 1 \neq \operatorname{lcm}(m, n) = 4$.

Unless $\gcd(|a|, |b|)=1$ is given we can't say that order of $ab$ is same as lcm$(|a|, |b|)$. We can only make such a claim provided $G$ is abelian and $\gcd(|a|, |b|) = 1$, i.e., if the orders of $a$ and $b$ are relatively prime.

ADDED:

Given the OP's edited question: Existence is easily established.

In every group $G$, the following holds: $\exists e \in G$, so put $a = b = e$. Then $ab = e^2 = e$ and $|ab| = |e| = 1 = \operatorname{lcm}(|a|, |b|)$.

So trivially, yes, for every group, there exists $(a, b)$ such that $|ab| = \operatorname{lcm}(|a|, |b|)$.

amWhy
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  • How is the OP's argument a counterexample to anything? – Igor Rivin Dec 03 '13 at 18:15
  • It's a rough counterexample to the proposition that given $a, b \in G$, $|ab| = \operatorname{lcm}(|a|, |b|)$. It would be better if the user gave a specific group and specific elements: E.g. $\mathbb Z_4$ as $G$, $a = 1, b = 3$. – amWhy Dec 03 '13 at 18:22
  • but that is not his question, which seems to be asking about the existence of an element of order $lcm(|a|, |b|).$ – Igor Rivin Dec 03 '13 at 18:24
  • @amWhy But in your (and the OP's case), $|ab|=|1 \cdot 3|=|3|=4$. But $|a|=|1|=1$ and $|b|=|3|=4$. So the lcm is $4$ and there is an element of order $=$lcm, precisely $3$! – mathematics2x2life Dec 03 '13 at 18:34
  • @mathematics2x2life Would you please note that the group I suggest is the additive group of integers, modulo $4$, generated by $1$ and by $3$.: so $|ab| = |1 + 3| = 1$. $|a| = 4, $|b| = 4$. – amWhy Dec 03 '13 at 18:38
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Let $G$ be a finite abelian group. Then if $a,b$ such that $ab \neq e_G \in G$ with $|a|=m$ and $|b|=n$, we know that the order of $|ab|=\text{lcm}(m,n)$ (or is a multiple of the order). But since $G$ is closed under its operation, we know that $ab \in G$. But since $ab \in G$ and $|ab|=\text{lcm}(m,n)$, the result holds (if it is a multiple, we can still construct such an element, I encourage you to see why). It is for you show that that in any abelian group that $|ab|=r \mid \text{lcm}(|a|,|b|)$, their orders not necessarily being finite.

EDIT. Note that I am only stating the statement is true if $ab \neq e_G \in G$. If we allow this, then indeed the result is false. Otherwise, the group you are suggesting, $\mathbb{Z}_4$, the integers under addition modulo $4$, with $a=1$ and $b=3$ works as counterexample (note that $1+3=4\equiv 0 \mod 4$ so that $ab=0=e_G \in G$).

mathematics2x2life
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  • You can't assume what is to be proven: that is, you can not assume that we know $|ab| = \operatorname{lcm}(m, n)$. That's precisely what needs to be established, or proven false.! – amWhy Dec 03 '13 at 18:39
  • @amWhy Which is precisely what I told the OP! Since we are in an abelian group, we know that this is true. But unless they can use that fact as a given, the OP need show it, as I mentioned in my answer. – mathematics2x2life Dec 03 '13 at 18:43
  • The OP shows gives an example $G, a, b $ such that $|ab| = 1 \neq 4 = \text{lcm}(a, b)$. (Hence the claim fails to hold in always in any abelian group.) And the specific group I mentioned is precisely such an abelian group in which the counterexample holds: disproving the claim you are asking the OP to prove! – amWhy Dec 03 '13 at 18:46
  • @amWhy Note that my solution excludes the case you mention, where $ab=e_G \in G$. Then the result does hold in all cases but the one mentioned. Perhaps I should have been more clear about this in my solution. I will add this as an edit to make it clearer. Thank you. – mathematics2x2life Dec 03 '13 at 18:49
  • Your case excludes $a = e, b= e$. It says nothing about $ab$. – amWhy Dec 03 '13 at 18:50
  • @amWhy Thank you for catching that! I had intended $ab \neq e_G \in G$ not $a,b \neq e_G$. This has been fixed. – mathematics2x2life Dec 03 '13 at 18:51
  • @AnjanDebnath I would focus on the answer given by amWhy. The result is in general false. If you excluded the case where $ab=e_G$, then we could show that the statement was true! However, since that is not part of our assumption, then the conjecture is false, as you demonstrated (just clean up your notation--you want an additive group not multiplicative--as suggested by amWhy and I). – mathematics2x2life Dec 03 '13 at 19:01
  • @mathematics2x2life With due respect, why is that required? I mean by Fundamental theorem of finite abelian group, every finite abelian group is isomorphic some external direct sum of $\mathbb{Z}_n$ ( I am sorry, I couldn't write the complete statement; I refer "Contemporary Abstract Algebra" by Gallian ) Isn't it? So in either way, the finite Ableian group whether it is additive or not, it does not matter. Does it? After all, it is being made in to Additive group – KON3 Dec 03 '13 at 19:12
  • @mathematics2x2life I will of course do as you said. But I just want to be convinced about it. Nothing else – KON3 Dec 03 '13 at 19:14
  • Try it! Compare $\mathbb{Z}_4$ multiplicatively versus in addition $\mod 4$. You don't quite get what the theorem actually says. It tells you that any finite abelian group breaks down into external direct products of $\mathbb{Z}_n$, not that if $|G|=|H|=n$ are both abelian groups that their external direct product decomposition has to be the same. For example, $\mathbb{Z}_4$ and $V_4$, the Klein-4 group, both are abelian groups of order $4$ but are not isomorphic and do not have the same decomposition. – mathematics2x2life Dec 03 '13 at 19:16
  • @mathematics2x2life I agree with you about the theorem. The only part I would like to know, are you asking me to consider $\mathbb{Z}_4$ as a gorup under multiplication modulo 4? I am sorry, actually I couldn't figure out this part from your comment. Please specify. Thank you – KON3 Dec 03 '13 at 19:19
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    @AnjanDebnath No, just think of $\mathbb{Z}$ under addition modulo $4$. Let $a=1$ and $b=3$. Then notice $|a|=1$, $|b|=4$, so their lcm is $4$ but $|ab|=|1+3|=|4|$. But $4\equiv 0 \mod 4$ so $|4|=1$. Then we have found a counterexample. – mathematics2x2life Dec 03 '13 at 19:23
  • My final query. What if we are supposed to find an element of order lcm$(|a|, |b|)$ in $G$ with all the same properties given above? – KON3 Dec 04 '13 at 01:52
  • @AnjanDebnath You mean if given $|a|=n$ and $|b|=m$ then can we find an element of order $|nm|$? That is in my solution. So long as $ab\neq e_G$, then $ab$ has order $\text{lcm}(|a|,|b|)$. It is easy enough to show that this is true for any abelian group (not necessarily finite). – mathematics2x2life Dec 04 '13 at 01:56
  • @mathematics2x2life but in your example you have already assumed that $|ab|=lcm(|a|, |b|)$ which was supposed to be 'proved'. How can we do that? My confusion is because of that only. – KON3 Dec 04 '13 at 01:59
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    See any of the following: http://math.stackexchange.com/questions/10616/order-of-elements-in-abelian-groups

    http://math.stackexchange.com/questions/78544/if-orda-m-ordb-n-then-does-there-exist-c-such-that-ord-c-lcmm-n

     – mathematics2x2life Dec 04 '13 at 02:06
  • @mathematics2x2life Excellent! Thank you so much. My all doubts are clear now. – KON3 Dec 04 '13 at 02:09
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Thanks to all of you for giving such helpful response. Your valuable discussion has helped me a lot. Based on whatever the information I received, here is my solution.

Let $a, b\in G$ with $|a|=m>1, |b|=n>1$. Assume that $|ab|=\alpha$. Let $\gcd(m, n)=d$ and $lcm(m, n)=L$. Then $mn=dL$ i.e. $\frac{mn}{d}=L$.

Now $gcd(m, n)=d\Rightarrow gcd(\dfrac m d, \dfrac n d)=1$.

Consider the element $a^d b^d$. Then $(a^d b^d)^{\frac{mn}{d}}=(a^d)^{\frac{mn}{d}} (b^d)^{\frac{mn}{d}}$ because we know that in an abelian group $G$, $(ab)^t=a^t b^t$ for every integer $t$.

From the orders of $a$ and $b$, so it follows that $(a^d b^d)^{\frac{mn}{d}}=e$ where $e$ is identity element in $G$.

Thus $\alpha | \frac{mn}{d}=L$ i.e. $\alpha|L$. Once we prove that $L|\alpha$ we are done. So let $L\nmid \alpha$. Note that here now we have $L<\alpha$. By division algorithm we then write $\alpha = Lq+r, 0\leq r < L$. If $r=0$ then $L|\alpha$. So let $0<r<L$.

then $(a^d b^d)^r=(a^d b^d)^{\alpha-Lq}=(ab)^{d(\alpha-Lq)}=e$ (WHY?) which is contradiction as $r<L<\alpha$ where $\alpha=|ab|$. Thus we must conclude $L|\alpha$.

Hence $L=\alpha$ viz $|a^d b^d|= lcm(m, n)$.

Hence the proof.

Am I correct?

KON3
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