Question

If F is a field with 49 elements, prove that x^49=x. for all x belongs to F. also find characteristic of F.

Answer 1

Since $F$ is a field, $F\setminus\{0\}$ is a group with the operation of multiplication ( is zero element of the field $F$). Lagrange's Theorem implies that the order of each element of $F\setminus\{0\}$ is a divisor of the order $|F\setminus\{0\}|=48$ of $F\setminus\{0\}$. Let $x\in F\setminus\{0\}$ be arbitrary. If the order of $x$ is equal to $t$, then $x^t=e$ and $48=ts$ for some $s\in\mathbb N$. Then $x^{48}=x^{ts}=(x^{t})^s=e^s=e$. So, $x^{49}=x^{48}x=ex=x$ for any $x\in F\setminus\{0\}$. Taking into account that $0^{49}=0$, we conclude taht $x^{49}=x$ for any $x\in F.$ If $p$ is a prime number, then the characteristic $char F$ of a field $F$ of cardinality $p^n$ is equal to $p$, that is $char F = p.$ Since $49=7^2$, we conclude that $char F = 7.$