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Here is my attempted proof of the title (in response to an exercise I randomly found on the web) and I was wondering if anyone could verify it.

Firstly, in a finite field $F$, if $a,b\in F$ then $a+1=b+1$ only if $a=b$. This follows from the existence of the additive inverse.

Next, if $a_i$ are the elements of $F$ indexed from $1$ up to $|F|$, then we can use what we just found to show that $$a_1+a_2+a_3+...+a_{|F|}=(a_1+1)+(a_2+1)+(a_3+1)+...+(a_{|F|}+1)$$

So:

$$a_1+a_2+a_3+...+a_{|F|}=a_1+a_2+a_3+...+a_{|F|}+\overbrace{1+1+1+...+1}^{|F| \ times}$$ $$0=\overbrace{1+1+1+...+1}^{|F| \ times}$$

The last thing to show is that the statement: $$0=\overbrace{1+1+1+...+1}^{n \ times}$$

Implies that the characteristic divides $n$. But if it did not, then one could take the difference between 1 added $n$ times and 1 added the largest multiple of the characteristic less than $n$ times. This would also be equal to 0 but has less 1's than the characteristic which would contradict its minimality. The result follows (I think).

I'm not super confident that I haven't missed something in my argument, and while I do enjoy taking inspiration from Fermat's Little Theorem I suspect there would be a much easier way which I would like to hear.

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    $\begingroup$ If you want to write out the rest of your proof more formally you could do this. Let $p$ be the characteristic. Write $|F|=n=mp+r$ with $0\leq r<p$. Then $0=n\cdot 1 = (mp+r)\cdot 1 = m(p\cdot 1)+r\cdot 1 = r\cdot 1$. So $r$ has to be $0$ since $r<p$. So $p$ divides $n$. Btw, although the result can also be done by quoting Lagrange's Theorem, yours is a nice proof too. $\endgroup$ Aug 31, 2020 at 11:29

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This is a simple consequence of Lagrange's theorem for finite groups.

Consider the finite additive group $(F,+)$. The characteristic of $F$ is simply the order of the subgroup generated by $1$, thus it divides the order of $F$.

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