I have to find prime number $p\le11$ and $n\in N$, such as multiplicative group of field of $p^n$ elements contains element with order 19. Multiplicative group $F_{p^n}^*$ of field $F_{p^n}$ consists of $p^n-1$ elements. So I found out that for $p=7$ and $n=3$ there can be element of order 19 as $(7^3-1)\div19=18$. But I don't know how to find it. I would appreciate any help
Upd: I have recently asked similar question The element of order 19
By Cauchy's theorem on groups, this means you have to find a prime number $p\le 11$ such that $19\mid p^n- 1$ for some $n$. Indeed, Cauchy's theorem ensures the multiplicative group $\;F_{p^n}^\times$ contains an element of order $19$.
Now any prime $p\le 11$ will meet this condition, since it is equivalent to $p^n\equiv 1\mod19$ for some $n$, and this results from $\,p\in(\mathbf Z/19\mathbf Z)^\times,$ which is a finite group.