# Order of Group and LCM of group elements. [closed]

Is order of Group equal to lowest common multiples of order of group elements.?If yes, what are the conditions?

No. For instance if $$G=\mathbb Z_2\oplus\mathbb Z_2$$, then $$|G|=4$$, but the least common multiple of the order of it's group elements is $$2$$.

No, for a finite abelian group the $$\rm lcm$$ of the orders is the expt (not order) of the group, namely

$$\begin{eqnarray}\rm{\bf Proposition}\quad maxord(G) \!&\,=\,&\rm expt(G)\ \text{ for a finite abelian group}\ G,\ i.e.\\ \\ \rm max\ \{ ord(g) : \: g \in G\} \!&=&\rm min\ \{ n>0 : \: g^n = 1\ \ \forall\ g \in G\}\end{eqnarray}$$

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

Hence 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$$

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

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