Can a proper subgroup of the multiplicative group of a finite field form an arithmetic progression. To rule out special cases, here the proper subgroup should not be $1$, and the length of arithmetic progression is finite and at least $3$. If we can permute a subgroup to form an AP, the it also meets the requirement.
Arithmetic progression in this context is just $\{a+nd| n\in \mathbb{N}， n\le N\}\subset \mathbb{F}$, where $\mathbb{F}$ is the given finite field. I tried some cases like $\mathbb{Z}/(p)$, and didn't find any.
So my question is whether we have infinitely many finite fields qualified or the proposition is generally false?
 A: The only such progressions are the "obvious" ones, namely:
$$S = \{1\}, \quad \{-1,1\}  \ \text{when $\mathrm{char}(k) > 2$}, \quad \{1,2,3,4,\ldots,p-1\}.$$
The first coincides with the last when $\mathrm{char}(k) = 2$, and the second coincides with the last when $\mathrm{char}(k) = 3$.
Let $p$ denote the characteristic of $k$, let $n$ denote the length of the subgroup/arithmetic progression $S$, and let $q = |k|$. The multiplicative subgroup $k^{\times}$ of a finite field is cyclic, so if $|S| = n$, then
$$S = \{1,\zeta,\zeta^2,\ldots,\zeta^{n-1}\}$$
where $\zeta$ is a primitive $n$th root of unity.
Some Reductions: If $k$ has characteristic $p$, then in any arithmetic progression $a, a+d,a+2d,\ldots$, the initial term $a$ and the $p+1$st term $a+pd=a$ coincide, and thus $n = |S| \le p$. Moreover, the order $|S|$ divides the order of $k^{\times}$ which is $q-1$ where $q$ a power of $p$. Thus $(n,p) = 1$, and hence $n < p$. It follows that if $p = 2$ then $n = 1$ and so $S = \{1\}$, and if $p = 3$ then either $n = 1$ and $S = \{1\}$ or $n = 2$ and $S = \{1,-1\}$.
The argument: We may assume that $p > 3$, $2 < n \le p - 1$, and $(n,p) = 1$.
Suppose that the common difference of the arithmetic progression is $d \in k$. Since $1 \in S$, the set $S$ also must have the form:
$$S = \{1+id \ | \ i \in [-a,n-1-a]\}$$
for some $0 \le a \le n-1$. But now we find that
$$\sum_{x \in S} x = \sum_{i=-a}^{n-1-a} 1 + i d
= \frac{n(2-d-2ad+dn)}{2}.$$
but we also have
$$\sum_{x \in S} x = 1 + \zeta + \ldots + \zeta^{n-1} = \frac{\zeta^n-1}{\zeta - 1} = 0$$
since $\zeta^n = 1$. From these two evaluations we deduce (noting that $n \ne 0$ because $(n,p) = 1$) that
$$(1+2a-n)d = 2.$$
Since $k$ has characteristic different from $2$, it follows that $d$ and $(1+2a-n)$ are both units, and hence that
$$d = \frac{2}{1+2a-n}$$
But now we compute that
$$\sum_{x \in S} x^2 = 1 + \zeta^2 + \ldots + \zeta^{2(n-1)} = \frac{\zeta^{2n} - 1}{\zeta^2 - 1} = 0$$
where we assume that $n > 2$ so $\zeta^2 - 1 \ne 0$, and
$$\sum_{x \in S} x^2 =  \sum_{n=-a}^{n-1-a} (1 + i d)^2
= \frac{n (n^2 - 1)}{3 (1 + 2 a - n)^2},$$
where we use that $k$ has characteristic $p > 3$ so there are no issues with the RHS. Now combining these two formulae, we deduce that $n(n+1)(n-1) = 0$. Since $1< n < p$ and $p$ is prime, this can happen only when $n = p-1$. The elements of order $p-1$ in $k$ are precisely the non-zero roots of $x^p - x = 0$, or equivalently the primitive subfield $\mathbf{F}_p$. Thus we have found all such examples.
