Let $F_p$ be a finite field of order $p$ where $p$ is a prime number. Let $\{ \alpha_1, \ldots, \alpha_{p-1} \}$ be a multi-set with $\alpha_i \in F_p$ and each $\alpha_i$ non-zero.
I want to show that $$\sum_{i\in K} \alpha_i = -1$$ for some subset $K \subseteq \{1,\ldots, p-1\}$.
I am stuck at this. Any hint(s) would be appreciated.
On
Here are a few hints, without giving away the complete solution:
Hint 1. Note that it is not really a question about fields at all; the same question could be stated for any finite cyclic group $\mathbb{Z} / n\mathbb{Z}$. However, I'm telling you now that the result crucially depends on the assumption that $p$ is prime. (For instance, in $\mathbb{Z} / 4\mathbb{Z}$ no subset of the multiset $\{2,2,2\}$ sums up to $-1$.) Which group-theoretic property does $\mathbb{Z} / p\mathbb{Z}$ have that a general $\mathbb{Z} / n\mathbb{Z}$ doesn't?
Hint 2. There is nothing special about $-1$; the same conclusion holds for any other non-zero element of $F_p$.
Hint 3. For every $k \in \{0,\ldots,p-1\}$, let $S_k \subseteq F_p$ denote the set of all partial sums using only the first $k$ numbers. In other words, define $$ S_k := \left\{\sum_{i\in K} a_i \, : \, K \subseteq \{1,\ldots,k\}\right\}, $$ so we have $S_0 = \{0\}$, $S_1 = \{0,a_1\}$, $S_2 = \{0,a_1,a_2,a_1+a_2\}$, etcetera. Prove that these sets grow sufficiently quickly. (Of course the cardinality of $S_k$ is at most $2^k$, but typically less since different expressions may sum up to the same value.)
Since any element $\alpha \in \mathbf{F}_p^*$ generates $(\mathbf{F}_p,+)$, the only subset $S \subset \mathbf{F}_p$ such that $S= S \cup (\alpha+S) \bmod p$ is $\mathbf{F}_p$.
If all the $\alpha_i$ are the same element $\alpha$, take $b \equiv -\alpha^{-1} \bmod p$ so that $\sum_{i=1}^b \alpha_i = -1$.
Otherwise wlog. $\alpha_1 \ne \alpha_2$ so that $S_2 = \{\alpha_1,\alpha_2,\alpha_1+\alpha_2\}$ contains $3$ elements, and $S_{i+1} = S_i \cup ( \alpha_{i+1}+S_i) $ contains at least $i+2$ elements, therefore $S_{p-1}$ contains $-1$.