# ord $(b)|\max\{\text{ord}(g)|g\in G\}$ for all $b\in G\,$ a finite abelian group

Let $a$ be an element of maximum order from a finite Abelian group $G$. Prove that for any element $b$, $|b|$ divides $|a|$ (order of $b$ divides order of $a$).

• Note that this is exercise 46 in additional exercises to chapters 1-4 in Gallian's Contemporary Abstract Algebra. – a student Dec 16 '15 at 2:58

## 3 Answers

By the structure theorem for finite(ly-generated) abelian groups, there exist $d_1, \cdots, d_k \in \mathbb{Z}^+$ such that

$$G \cong \dfrac{\mathbb{Z}}{d_1 \mathbb{Z}} \oplus \cdots \oplus \dfrac{\mathbb{Z}}{d_k \mathbb{Z}}$$

and $d_i$ divides $d_{i+1}$ for each $1 \le i < k$.

But then each element has order dividing $d_k$, and $d_k$ is the maximum order of any element of $G$.

• Thanks so much, but I'm wondering whether there is any way to prove this without using the structure of Albelian group?This problem is from some first chapters of the book, which just about group, subgroup and cyclic group. – le duc quang Sep 6 '12 at 16:25
• @leducquang: Looks like Bill Dubuque beat me to it ;) An elementary proof relies on the idea that if $a$ and $b$ are elements of an abelian group then there is some element of the group whose order is the lcm of the order of $a$ and the order of $b$. If $a$ is maximal then you need $\text{lcm}(|a|,|b|) \le |a|$, and for this, you need $|b|$ to divide $|a|$. – Clive Newstead Sep 6 '12 at 17:04
• I think indeed one needs to show a result like that of the question (though it could be in disguise, e.g. if one is doing modules over a PID) in order to prove the structure theorem mentioned. So the question for a direct proof is quite reasonable. – Marc van Leeuwen Sep 7 '12 at 6:54

Lemma: If the orders $|x|,|y|$ of $x,y\in G$ are coprime then the order of $xy$ is $|x| |y|$.

Proof: If $(xy)^m = 1$ then $x^{m|y|} = 1$, so $|x|$ divides $m|y|$. Since $|x|$ and $|y|$ are coprime, this implies $|x|$ divides $m$. Similarly $|y|$ divides $m$, so by coprimality their product divides $m$.

Now let $a$ be your element of maximum order, and $b$ any other element. Suppose $p$ is a prime dividing $|b|$ to a higher power than $|a|$. Write $|a| = p^i m$ and $|b| = p^j n$, where $j>i$ and $p$ divides neither $m$ nor $n$. Then $a^{p^i}$ and $b^{n}$ have coprime orders, so $a^{p^i} b^n$ has order $p^j m > |a|$, a contradiction.

• Thank you, this really helped me! What a great answer! – a student Dec 16 '15 at 2:59
• It is maybe good to note that in general if in a group $G$ to lements $x$ and $y$ commute, then $order(xy)=\text{ lcm}(order(x),order(y))$. – Nicky Hekster Aug 21 '16 at 11:21
• @NickyHekster That's not true. (Take, like, $x=y=1$ in $\mathbf{Z}/2\mathbf{Z}$.) – Sean Eberhard Aug 22 '16 at 9:36
• Sean, you are right is has to be "divides" $|xy| \mid lcm(|x|,|y|)$ whenever $[x,y]=1$. – Nicky Hekster Aug 22 '16 at 10:44
• Can you tell me why $(xy)^m=1$ then $x^{m|y|}=1$ holds? – scarface Jan 13 at 16:15

Below is a simple proof that does not employ the high-powered structure theorem.

Theorem $\rm\quad maxord(G)\ =\ expt(G)\$ for a finite abelian group $\rm\: G,\$ i.e.

$\rm\quad\ \ max\ \{ ord(g) : \: g \in G\}\ =\ min\ \{ n>0 : \: g^n = 1\ \:\forall\ g \in G\}$

Proof $\ \:$ By the lemma below, $\rm\: S\: =\: \{ ord(g) : \:g \in G \}$ is a finite set of naturals closed under$\rm\ lcm\,.$

Therefore every $\rm\ s \in S\:$ is a divisor of the max elt $\rm\, m\:$ [else $\rm\: lcm(s,m) > m\:$],$\$ so $\rm\ m = expt(G).$

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

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

Proof $\ \$ By induction on $\rm\, o(X)\: o(Y).\$ If it is $\:1\:$ 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)).\ \$ QED

• Sean Eberhard provides an answer that not only doesn't employ a high-powered structure theorem, but also avoids hight-density cryptic formalism that makes the essence of the argument hard to locate, and avoids bludgeoning MathJax into using roman fonts only – Marc van Leeuwen Sep 7 '12 at 7:09
• @Marc I'm afraid you misread. The above proof does not use the structure theorem, as I emphasized. Moreover, the point of the above presentation is to bring to the fore arithmetical essence of the matter, viz. that the result holds true for any set of naturals closed under lcm - something that is not made clear in the other answers. – Bill Dubuque Sep 7 '12 at 14:26
• I'm afraid it's you who misread my comment, which says "not only doesn't". I don't deny you didn't use the structure theorem, but neither does that of Sean Eberhard, and in addition it avoids a bit more. Also if "the essence of the matter" is that for a finite set closed under lcm every number divides the maximal one, I'm unimpressed. But maybe I'm misreading here. – Marc van Leeuwen Sep 7 '12 at 14:32
• @Marc Said arithmetical structure of the order set is the essence of the matter to me. My presentation was explicitly designed to highlight that structure. While that may seem trivial to you or I, it is nontrivial to most students first learning this material. But, of course, we may disagree on subjective matters like pedagogy - just as we apparently disagree on other subjective matters, such as typesetting (for the record, the proof was composed (long ago) with Roman fonts because early versions of MathJax rendered italic fonts horribly on some platforms). – Bill Dubuque Sep 7 '12 at 14:45
• Sincerly, I don't want to polemicize. But doesn't $\operatorname{lcm}(a,b)=a*(b/\gcd(a,b))$ with $a$ the maximum (the parenthesised factor being an integer${}\geq1$ that is${}>1$ unless $b$ divides $a$) prove "said arithmetical structure"? – Marc van Leeuwen Sep 7 '12 at 15:20