Jtdylktuy

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

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

No. For instance if $G=mathbb Z_2oplusmathbb Z_2$, then $|G|=4$, but the least common multiple of the order of it's group elements is $2$.