Prove that k-cube graph is a Cayley graph

Question

Define Cayley graph as following:

G is finite group. C $\subseteq$ G such that C does not contain identity element of G and g^-1 $\in$ C for all g $\in$ C. Cayley graph X(G,C) is formed with vertices V(X)=G, edges E(X)={(a,b): a,b $\in$ G, ab^-1 $\in$ C}.

The k-cube is the graph with vertex set {0,1}^k such that any two vertices x,y $\in$ {0,1}^k are joined by an edge if and only if x and y differ in exactly one coordinate. Show that the k-cube is a Cayley graph.

Answer 1

Hint: Try $G=\mathbb Z_2^k$ and let $C$ be the set of $k$-tuples with $1$ in one coordinate and $0$ in every other.

Answer 2

I work this out with your hint.

I have g={0,1}^k. The operation in G is addtion modulo 2.

g^-1=g, e={0}^k

So the vertices are a,b $\in$ G.

There is edge (a,b) if a*b=e_j=(x₁,_x2,...,x_k) where x_j=1, 1 $\leq$ j $\leq$ k and x_i=0 for all i$\neq$j

In this case a*b^-1=a*b

So C={e_j, j=1,2,...,k}