If the players are symmetric and the core is nonempty, then $x_i = v(N)/n$ for all $i$ is a Core element

Question

Consider a TU game $\langle N,v \rangle$ with $N = \{1,2,\ldots,n\}$ being the set of players and $v : 2^N \to \mathbb R$, $v(\emptyset) = 0$ the characteristic function. The Core of $v$ is defined by undominated allocations $x \in \mathbb R^n$ \begin{align} \mathcal C(v) = \left\{x \in \mathbb R^n ~ \bigg| ~ \sum_{i \in S}{x_i} \geq v(S) ~ \forall S \subseteq N\right\}. \end{align} Let $m_i(S) = v(S \cup \{i\}) - v(S)$ denote the marginal contribution of $i$ to coalition $S$. We say that the players are symmetric if $m_i(S) = m_j(S)$ for all $i,j \in N$ and $S \subset N$. The marginal contribution of any agent to any coalition is thus identical.

I am wondering whether the following Proposition is true and how one proves it formally.

Proposition: If the players are symmetric and the core is nonempty $\mathcal C(v) \neq \emptyset$, then the equal division of the worth of the grand coalition is an element of the core $(v(N)/n,\ldots,v(N)/n) \in \mathcal C(v)$.

Answer 1

The center allocation $z_{i}:=(v(N)/|N|)$ for all $i \in N$ is in the core of the symmetric game $v$, whenever $z(S) = \sum_{i \in S} z_{i} \ge v(S) $ holds for all $S \subseteq N$. Now, if $v(N)/|N| \ge v(S)/|S|$ for all $S \subseteq N$, then it holds $z(S) = \sum_{i \in S} z_{i} = |S|\cdot v(N)/|N| \ge v(S)$. Thus, $\mathbf{z} \in C(v)$.

We observe that for a symmetric game $v$, the core is non-empty if $v(N)/|N| \ge v(S)/S$ holds for all $S \subseteq N$. Otherwise, the core is empty.