On finite groups Let $(G,*)$ is a finite group, whose order is $n$. Also let $p|n$ and $p\in P$. Can you bring an example of such group which doesn't have an element of order $p$.
Sincerely,

As Ragib said, Cauchy's Theorem assures that if $p$ is a prime divisor of the order of the finite group $G$, then there is an element $g\in G$ of order $p$, hence at least $p-1$ of them. One may note that the proof of Cauchy Theorem's relies on a counting elements argument, thus it is not constructive.
In the comments, the questioner extended his question to elements of power prime order. Sylow's Theorems say that if $p^k$ is the highest power of the prime $p$ dividing the order of $G$, then there exists a subgroup $P<G$ with $|P|=p^k$ (such subgroups are called $p$-sylows). One can then apply the theory of $p$-grous to conclude that $G$ has subgroups of order any power $p^\ell$ with $\ell\leq k$. Again, the proof is not constructive.
Now, in order to decide about the existence of elements of order $p^k$ is enough to consider just one $p$-sylow and decide whether it needs to be cyclic (because by the Sylow's Theorems again all $p$-sylows are conjugated inside $G$, hence isomorphic one to another).
To answer (in the negative) to the general question it is enough to observe that in the permutation group on $10$ elements, a $5$-sylow has $25$ elements and one such is the subgroup 
$$
P=\langle(1\ 2\ 3\ 4\ 5)\rangle\times\langle(6\ 7\ 8\ 9\ 10)\rangle
$$
which is visibly non-cyclic.
Of course, one can come up with many more counterexamples, some of which much simpler than the one described above.
