Let G be a graph of order n whose chromatic polynomial is...

Let $G$ be a graph of order $n$ whose chromatic polynomial is $p_{G}(k)=k(k-1)^{n-1}$ (i.e., the chromatic polynomial of $G$ is the same as that of a tree of order $n$ ). Prove that $G$ is a tree.

Key Concepts

Graph Connectivity

Connectivity in a graph means that there is a path between any pair of vertices. This property is crucial for trees, as they are defined as connected graphs with no cycles. Connectivity ensures that the entire graph is a single component and influences the form of the chromatic polynomial. A disconnected graph would yield a chromatic polynomial that factors over its components, differing from the form k(k - 1)^(n - 1).

Acyclicity and Cycles

A graph that contains a cycle will have additional constraints on its vertex colorings, causing a deviation in the form of its chromatic polynomial from that of a tree. The absence of cycles (acyclicity) ensures that the number of colorings follows the simple multiplication of available choices for each vertex, as seen in trees. Hence, if a graph has a chromatic polynomial identical to that of a tree, it indicates the graph must be acyclic.

Chromatic Polynomial

The chromatic polynomial of a graph is a function that counts the number of ways to properly color its vertices using a given number of colors, k, while ensuring adjacent vertices receive different colors. It encapsulates key combinatorial properties of a graph and, through its form, can indicate structural features such as the presence or absence of cycles.

Tree

A tree is an acyclic connected graph and is characterized by having exactly n - 1 edges when there are n vertices. Trees have a particularly simple chromatic polynomial, k(k - 1)^(n - 1), which arises from their acyclic nature, meaning that each additional vertex beyond the first is connected by exactly one eligible edge that does not form a cycle.

Please give Ace some feedback