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How can I prove that if $G$ is an Abelian group with elements $a$ and $b$ with orders $m$ and $n$, respectively, then $G$ contains an element whose order is the least common multiple of $m$ and $n$?

It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when $m$ and $n$ are coprimes. I did this part. But I have no idea how to solve the general case.

Thanks.

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5 Answers 5

up vote 11 down vote accepted

Below is an inductive proof excerpted from my post in this prior thread here.

LEMMA $\ $ A finite abelian group $\rm\:G\:$ has an lcm-closed order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$

$\rm\quad\quad\quad\quad\quad\quad\ X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G:\ o(Z) = lcm(o(X),o(Y))$

Proof $\ \ $ By induction on $\rm o(X)\ o(Y)\:.\ $ If it's $\:1\:$ then trivially $\rm\:Z = 1\:$. $\ $ Otherwise

write $\rm\ o(X)\ =\ AP,\: \ \ o(Y) = BP',\ \ \ P'|P = p^m > 1\:,\ \ $ prime $\rm\: p\:$ coprime to $\rm\: A,B$

Then $\rm\: o(X^P) = A,\ \ o(Y^{P'}) = B\:.\ $ By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$

so $\rm\ \ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y))\:$.

NOTE $\ \ $ This is the element-wise form of what's known as "Herstein's hardest problem", viz. 2.5.11, p. 41 in the first edition of Herstein's popular textbook "Topics in Algebra". In the 2nd edition Herstein added the following note (problem 2.5.26, p.48)

Don't be discouraged if you don't get this problem with what you know of group theory up to this stage. I don't know anybody, including myself, who has done it subject to the restriction of using material developed so far in the text. But it is fun to try. I've had more correspondence about this problem than about any other point in the whole book."

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@Bill: I believe the (in)famous problem was about subgroups, not elements; that is, if $G$ is abelian and $H$ and $K$ are finite subgroups, then $G$ has a subgroup of order $\mathrm{lcm}(|H|,|K|)$. In my Spanish translation of the first edition, this is problem 2.5.11. –  Arturo Magidin Nov 16 '10 at 22:39
    
@Arturo: Indeed, but essentially the same proof works for the subgroup case. –  Bill Dubuque Nov 16 '10 at 23:11
    
@Bill: if you know that if $X$ has order $ab$ and $\gcd(a,b)=1$, then $X^a$ has order $b$; that would require at least a lemma or two with the material known up to that point... But in any case, your text does not make clear what "Herstein's hardest problem" was, just that it is something of which this is the "element-wise form" –  Arturo Magidin Nov 17 '10 at 1:25
    
@Arturo: No, it's trivial and gcd(a,b)=1 isn't needed: if o(X) = ab then o(X^a) = b (it can't be c < b else X^(ac) = 1 for ac < ab, contra to o(X) = ab). The above proof is for the OP, not for Herstein's exercise (I don't recall the constraints imposed by order of presentation of material in Herstein's textbook). –  Bill Dubuque Nov 17 '10 at 2:24
    
@Bill: Sorry! I was commenting on the subgroup version, sorry I did not make that clear; should have used "H" instead of "X"; my apologies about the confusion. I guess I think you just may want to mention in the body what Herstein's problem is. –  Arturo Magidin Nov 17 '10 at 2:27

Let $n = \prod_i p_i^{n_i}$ and $m = \prod_i p_i^{m_i}$, where the $p_i$s are the same. Thus $\mathrm{lcm}(n,m) = \prod_i p_i^{\max(n_i,m_i)}$. Let $N = \{ i : n_i \geq m_i \}$ and $M = \{ i : n_i < m_i \}$. Define $n' = \prod_{i \in N} p_i^{n_i}$ and $m' = \prod_{i \in M} p_i^{m_i}$. Now $a' = a^{n/n'}$ is of order $n'$, and $b' = b^{m/m'}$ is of order $m'$. Since $n',m'$ are coprime, we know that there is an element of order $n'm' = \mathrm{lcm}(n,m)$.


You can also prove it using the structure theorem for abelian groups, but that's an overkill.

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Thanks Yuval. That's a nice proof. But how would you do this using the structure theorem? Here the abelian group isn't necessarily finitely generated. So would you pick a finitely generated subgroup containing a and b? –  Álvaro Garcia Nov 17 '10 at 13:59
1  
Indeed, you can take the subgroup generated by $a,b$, which is going to contain the element you're looking for anyhow. –  Yuval Filmus Nov 17 '10 at 14:55

Both responses you have received provide correct answers. In fact, you don't need to go so far as to obtain elements of relatively prime orders; all you need to worry about are primes with the property that they divide $nm$, and the largest power of $p$ that divides $n$ is equal to the largest power of $p$ that divides $m$.

Lemma. Let $x$ and $y$ be commuting elements of a group $G$, and assume $x$ and $y$ are of finite orders $n$ and $m$, respectively. Suppose that for every prime $p$ that divides $nm$, the largest power of $p$ that divides $n$ is different from the largest power of $p$ that divides $m$. Then the order of $xy$ in $G$ is $\mathrm{lcm}(n,m)$.

Proof. Let $\ell=\mathrm{lcm}(n,m)$. It is easy to verify that $(xy)^{\ell}=1$, so we just need to show that $\ell$ is the smallest positive integer $k$ such that $(xy)^k=1$. Suppose that $(xy)^k = 1$, with $0\lt k\leq \ell$.

Since $(xy)^k = x^ky^k = 1$, then $x^k = y^{-k}$, and in particular $x^k$ has the same order as $y^k$. The order of $x^k$ is $n/\gcd(n,k)$, and the order of $y^k$ is $m/\gcd(n,k)$. Let $p$ be a prime that divides $\ell$, and let $a=\mathrm{ord}_p(n)$, $b=\mathrm{ord}_p(m)$, and $c=\mathrm{ord}_p(k)$. Assume $b\lt a$. If $c\lt a$, then $\mathrm{ord}_p(n/\gcd(n,k)) = a-c$, and $\mathrm{ord}_p(m/\gcd(m,k))=\max(0,b-c)\lt a-c$, which is impossible. Thus, $c=a$. A symmetric argument shows that if $a\lt b$, then $c=b$. That is, for all primes $p$ that divide $\ell$, we have $\mathrm{ord}_p(k) = \max(\mathrm{ord}_p(n),\mathrm{ord}_p(n)) = \mathrm{ord}_p(\ell)$. Hence $k=\ell$. QED

So now assume that $a$ is an element of order $n$ in an abelian group, and let $b$ be an element of order $m$. Let $p_1,\ldots,p_k$ be the primes that divide $nm$ and for which the $p$-order of $p_i$ in $m$ and in $n$ are equal. Then $$\mathrm{lcm}(n,m) = \mathrm{lcm}\left(n, \frac{m}{p_1\cdots p_k}\right).$$ Since $m/(p_1\cdots p_k)$ is the order of $b^{p_1\cdots p_k}$, then it follows from the lemma that $ab^{p_1\cdots p_k}$ has order $\mathrm{lcm}(n,m)$, as desired.

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You ended $$ by ] due to which math after that wasn't displaying properly. I have edited it. –  anonymous Nov 16 '10 at 23:02
    
Thanks; I was on my way out and in a bit of a hurry. –  Arturo Magidin Nov 17 '10 at 0:40
    
Very nice Arturo. It's good to know that we can do this assuming only that the elements commute. –  Álvaro Garcia Nov 17 '10 at 13:59
    
@Álvaro Garcia: Well, that much is really a very simple observation: just work inside $\langle x,y\rangle$. Of course, the result is horribly false if $x$ and $y$ don't commute. All sorts of weird things can happen then. –  Arturo Magidin Nov 17 '10 at 17:23

This is also one of the starred problems in Herstein's: Topics in Algebra book. Take a look into page 3 of this file:

In case you aren't able to view the solution i type it down as i am not able to download this file, but able to view it in Google Docs.

Solution: Let $a$ and $b$ be elements of order $m$ and $n$ respectively. Write $\text{lcm} (m,n)= \prod\limits_{i=1}^{t}p_{i}^{\alpha_{i}}$, where the $p_{i}$'s are distinct primes, and $\alpha_{i} > 0$ for each $i$. For any $i \in [1,2, \cdots,t ]$, $p_{i}^{\alpha_{i}}$ divides, either $m$ or $n$. If say, $p_{i}^{\alpha_{i}} \mid m$, then, $a^{m/p_{i}^{\alpha_{i}}}$ is of order $p_{i}^{\alpha_{i}}$. In this way we obtain elements $x_{1},\cdots,x_{t}$ of orders $p_{1}^{\alpha_{1}}, \cdots, p_{t}^{\alpha_{t}}$ respectively. Now consider elements in $G$ of the form $$x_{1}^{\beta_{1}}, \cdots, x_{t}^{\beta_{t}} \quad \cdots (3)$$ where $\beta_{i} \in \mathbb{N}$, and $i \in [1,2,\cdots,t]$.

Our next goal, is to show that, the single elements, $(x_{1},\cdots,x_{t})$ generates all elements of the form $(3)$. For this purpose, we prove that for all $\beta_{i} \in \mathbb{N}$, $i \in [1,2,\cdots t]$, the system $$ k = \beta_{1} \ (\text{mod} \ p_{1}^{\alpha_{1}})$$ $$ \cdot $$ $$\cdot $$ $$\cdot $$ $$k = \beta_{t} \ (\text{mod} \ p_{t}^{\alpha_{t}}$$ has a solution in $k$. This follows by setting $$ k =\sum\limits_{i=1}^{t} \beta_{i}c_{i} p_{1}^{\alpha_{1}} \cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}} \cdots p_{t}^{\alpha_{t}}$$ where $c_{i}$ satisfies $$ c_{i} p_{1}^{\alpha_{1}} \cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}} \cdots p_{t}^{\alpha_{t}} \equiv 1 \ (\text{mod} \ p_{i}^{\alpha_{i}})$$

It's not hard to see that the elements of the form in Eq.(3) form a subgroup $H$ of $G$. Hence for each $i$ the order $p_{i}^{\alpha_{i}}$ of the subgroup generated by $x_{i}$ divides $|H|$ by Lagrange's theorem. This gives $|H| \geq \prod\limits_{i=1}^{t} p_{i}^{\alpha_{i}}$. Clearly $|H| \leq \prod\limits_{i=1}^{t} p_{i}^{\alpha_{i}}$ also holds. Since the element, $(x_{1}, \cdots, x_{t})$ generates exactly $H$ it has order $|H|$

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Thanks Chandru1. –  Álvaro Garcia Nov 17 '10 at 13:58

Let G be an abelian group. Let a and b belong to G with order m and n respectively. 3 Cases arise: Case 1: If m|n or n|m then lcm(m,n)=m or n in the 2 cases respectively. Hence our condition holds trivially.

Case 2: Suppose m and n are co-primes.Hence lcm(m,n) = m*n By abelian property of G (a.b)^mn = a^mn . b^mn =e.e=e. if o(a.b) Hence o(b^k) | o(a)(=m) but o(b^k)|o(b)(=n) Hence m and n have a common factor. Contradiction. Hence o(a.b)=m.n

Case 3: Let m and n not be coprimes let d= gcd(m,n) Hence n/d is co-prme with m. so o(a)=m and o(b^d)=n/d.

But as m and n/d are coprimes , Hence by Case 2 we get o(a.(b^d))=(m.n)/d=m.n/gcd(m,n)=lcm(m,n). Hence we have an element which has order lcm(m,n)

Hence Proved.

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