If a prime number $p$ divide the number of elements of order $k$ (for some $k\neq p$) in a finite group $G$, then whether we can say that $p$ divide order of $G$?
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No. Consider a group of prime order.
Addition: you can get some more insight into the problem by considering the 3rd Sylow theorem. Take for example a group $G$ of order $p_1p_2\ldots p_r$ where $p_k$ are pairwise different primes. Then the number of elements of order $p_k$ equals $n_k(p_k-1)$, where $n_k$ is the number of $p_k$-Sylow groups. The 3rd Sylow theorem yields that $n_k$ divides $|G|/p_k$. Hence your assertion holds for those primes dividing $n_k$.
The situation becomes more complicated if the primes $p_k$ occur with multiplicities in the group order. Then things depend strongly on the structure of the $p_k$-Sylow groups and their intersections.