Understanding Converse of Lagrange's theorem! Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)

The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$.                 ......         (2)

If my statement is $p\to q$ then converse is $q\to p$.

i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $p\to q$ then converse is $q\to p$
Actually, you need to write the Lagrange theorem in $p \rightarrow q$ form. So, the Lagrange theorem is actually,

If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.

Here $p:$ $H$ is a subgroup of $G$

$q;$ Order of $H$ divides order of $G$.

Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.

So, the converse will be

If a number $m$ divides order of $G$, then there is a subgroup of order $m$.

I hope this gives you insights.
Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!