Order of Group and LCM of group elements. Is order of Group equal to lowest common multiples of order of group elements.?If yes, what are the conditions?
 A: 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 $2$.
A: 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  $\rm\ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).$
