Chapter 3 Cayley Graphs

Monday, January 25, 1993

Given graphs $\Gamma = (X, E)$ and $\Gamma' = (X', E')$.

Definition 3.1 A map $\sigma: X \to X'$ is an isomorphism of graphs whenever;

1. $\sigma$ is one-to-one and onto,
2. $xy\in E$ if and only if $\sigma x \sigma y \in E'$ for all $x, y\in X$.

We do not distinguish between isomorphic graphs.

Definition 3.2 Suppose $\Gamma = \Gamma'$. Above isomorphism $\sigma$ is called an automorphism of $\Gamma$. Then set $\mathrm{Aut}(\Gamma)$ of all automorphisms of $\Gamma$ becomes a finite group under composition.

Definition 3.3 If $\mathrm{Aut}(\Gamma)$ acts transitive on $X$, $\Gamma$ is called vertex transitive.

Definition 3.4 (Cayley Graphs) Let $G$ be any finite group, and $\Delta$ any generating set for $G$ such that $1_G \not\in \Delta$ and $g\in \Delta \to g^{-1}\in \Delta$. Then Cayley graph $\Gamma = \Gamma(G, \Delta)$ is defined on the vetex set $X = G$ with the edge set $E$ define by the following.

$$E = \{(h_1,h_2)\mid h_1, h_2\in G, h_1^{-1}h_2\in \Delta\} = \{(h, hg) \mid h\in G, g\in \Delta\}.$$

Example 3.1 $G = \langle a \mid a^6 = 1\rangle$, $\Delta = \{a, a^{-1}\}$.

Example 3.2 $G = \langle a \mid a^6 = 1\rangle$, $\Delta = \{a, a^{-1}, a^2, a^{-2}\}$.

Example 3.3 $G = \langle a, b \mid a^6 = 1 = b^2, ab = ba\rangle$, $\Delta = \{a, a^{-1}, b\}$.

HS MEMO

$\mathrm{Aut}(\Gamma) \simeq D_6\times \mathbb{Z}_2$ contains two regular subgroups isomorphic to $D_6$ and $\mathbb{Z}_6 \times \mathbb{Z}_2$ and $\Gamma$ is obtained as Cayley graphs in two ways.

Cayley graphs are vertex transitive, indeed.

Theorem 3.1 The following hold.

$(i)$ For any Cayley graph $\Gamma = \Gamma(G, \Delta)$, the map

$$G \to \mathrm{Aut}(\Gamma) \; (g\mapsto \hat{g})$$ is an injective homomorphism of groups, where $$\hat{g}(x) = gx \quad \textrm{for all }\: g\in G \textrm{ and for all } x\in X (= G).$$ Also, the image $\hat{G}$ is regular on $X$. i.e., the image $\hat{G}$ acts transitively on $X$ with trivial vertex stabilizers.

$(ii)$ For any graph $\Gamma = (X, E)$, suppose there exists a subgroup $G \subseteq \mathrm{Aut}(\Gamma)$ that is regular on $X$. Pick $x\in X$, and let

$$\Delta = \{g\in G\mid \langle x, g(x)\in E\}.$$ Then $1\not\in \Delta$, $g\in \Delta \to g^{-1}\in \Delta$, and $\Delta$ generates $G$. Moreover, $\Gamma \simeq \Gamma(G, \Delta)$.

Proof. $(i)$ Let $g\in G$. We want to show that $\hat{g}\in \mathrm{Aut}(\Gamma)$. Let $h_1, h_2\in X = G$. Then, $$\begin{align} (h_1, h_2)\in E & \to h_1^{-1}h_2\in \Delta\\ & \to (gh_1)^{-1}(gh_2)\in \Delta \\ & \to (gh_1, gh_2)\in E\\ & \to (\hat{g}(h_1), \hat{g}(h_2)) \in E. \end{align}$$ Hence, $\hat{g}\in \mathrm{Aut}(\Gamma)$.

Observe: $g \mapsto \hat{g}$ is a homomorphism of groups: $$\hat{1}_G = 1, \; \widehat{g_1g_2} = \widehat{g_1}\widehat{g_2}.$$

Observe: $g \mapsto \hat{g}$ is one-to-one: $$\widehat{g_1} = \widehat{g_2} \to g_1 = \widehat{g_1}(1_G) = \widehat{g_2}(1_G) = g_2.$$

Observe: $\hat{G}$ is regular on $X$: Clear by construction.

$(ii)$ $1_G\not\in \Delta$: Since $\Gamma$ has not loops, $(x, 1_Gx) \not\in E$.

$g\in \Delta \to g^{-1}\in \Delta$: $$g\in \Delta \to (x, g(x))\in E \to E \ni (g^{-1}(x), g^{-1}(g(x))) = (g^{-1}(x), x).$$

$\Delta$ generates $G$: Suppose $\langle \Delta \rangle \subsetneq G$. Let $\hat{X} = \{g(x)\mid g\in \langle \Delta\rangle\} \subsetneq X$. ($\hat{X} \subsetneq X$ as $G$ acts regularly on $X$.)

Since $\Gamma$ is connected, there exists $y\in \hat{X}$ and $z\in X\setminus \hat{X}$ with $yz\in E$.

Let $y = g(x)$, $g\in \langle \Delta\rangle$, $z\in h(x)$, $h\in G\setminus \langle \Delta\rangle$. Then $$(y,z)=(g(x),h(x))\in E \to (x,g^{-1}h(x))\in E \to g^{-1}h\in \langle \Delta \rangle \to h\in \langle \Delta \rangle.$$ This is a contradition. Therefore, $\Delta$ generates $G$.

Let $\Gamma' = (X', E')$ denote $\Gamma(G, \Delta)$. We shall show that $$\theta: X' \to X \; (g\mapsto g(x))$$ is an isomorphism of graphs.

$\theta$ is one-to-one: For $h_1, h_2\in X' = G$, $$\theta(h_1)=\theta(h_2) \to h_1(x) = h_2(x) \to h_2^{-1}h_1(x)=x \to h_2^{-1}h_1\in \mathrm{Stab}_G(x) = \{1_G\} \to h_1 = h_2.$$ ($\mathrm{Stab}_G(x) = \{g\in G\mid g(x) = x\}$.)

$\theta$ is onto: Since $G$ is transitive, $$X = \{g(x)\mid g\in G\} = \theta(X') = \theta(G).$$

$\theta$ respects adjacency: For $h_1, h_2\in X' = G$, $$(h_1,h_2)\in E' \leftrightarrow h_1^{-1}h_2\in \Delta \leftrightarrow (x, h_1^{-1}h_2(x))\in E \leftrightarrow (h_1(x),h_2(x))\in E \leftrightarrow (\theta(h_1), \theta(h_2))\in E.$$ Therefore $\theta$ is an isomorphism between graphs $\Gamma(G, \Delta)$ and $\Gamma(X, E)$.

How to compute the eigenvalues of the Cayley graph of and abelian group.

Let $G$ be any finite abelian group. Let $\mathbb{C}^*$ be the multiplicative group on $\mathbb{C}\setminus \{0\}$.

Definition 3.5 A (linear) $G$-character is any group homomorphism $\theta: G \to \mathbb{C}^*$.

Example 3.4 $G = \langle a\mid a^3 =1\rangle$ has three characters, $\theta_0, \theta_1, \theta_2$. $$\begin{array}{c|ccc} \theta_i(a^j) & 1 & a & a^2 \\ \hline \theta_0 & 1 & 1 & 1\\ \theta_1 & 1 & \omega & \omega^2\\ \theta_2 & 1 & \omega^2 & \omega \end{array}, \quad \textrm{with }\; \omega = \frac{-1+\sqrt{-3}}{2}.$$ Here $\omega$ is a primitive cube root of $q$ in $\mathbb{C}^*$, i.e., $1+\omega + \omega^2 = 0$.

For arbitraty group $G$, let $X(G)$ be the set of all characters of $G$.

Observe: For $\theta_1, \theta_2\in X(G)$, one can define product $\theta_1\theta_2$: $$\theta_1\theta_2(g) = \theta_1(g)\theta_2(g) \quad \textrm{for all }\; g\in G.$$ Then $\theta_1\theta_2\in X(G)$.

Observe: $X(G)$ with this product is an (abelian) group.

Lemma 3.1 The groups $G$ and $X(G)$ are isomorphic for all finite abelian groups $G$.

Proof. $G$ is a direct sum of cyclic groups; $$G = G_1\oplus G_2 \oplus \cdots \oplus G_m, \quad \textrm{where } \; G_i = \langle a_i\mid a_i^{d_i} = 1\rangle \quad (1\leq i\leq m).$$ Pick any element $\omega_i$ of order $d_i$ in $\mathbb{C}^*$, i.e., a primitive $d_i$-the root of $1$. Define $$\theta_i: G \to \mathbb{C}^* \quad (a_1^{\varepsilon_1}\cdots a_m^{\varepsilon_m} \mapsto \omega_i^{\varepsilon_i} \quad \textrm{where }\; 0\leq \varepsilon_i < d_i, 1\leq i\leq m).$$ Then $\theta_i\in X(G)$. (Exercise)

Claim: There exists an isomorphism of groups $G \to X(G)$ that sends $a_i$ to $\theta_i$.

Observe: $\theta_i^{d_i} = 1$. For every $g = a_1^{\varepsilon_1}\cdots a_m^{\varepsilon_m} \in G$, $$\theta_i^{d_i}(g) = (\theta_i(g))^{d_i} = (\omega_i^{\varepsilon_i})^{d_i} = (\omega_i^{d_i})^{\varepsilon_i} = 1.$$ Observe: If $\theta_1^{\varepsilon_1}\theta_2^{\varepsilon_2}\cdots \theta_m^{\varepsilon_m} = 1$ for some $0\leq \varepsilon_i < d_i, 1\leq i\leq m$. Then $\varepsilon_1 = \varepsilon_2 = \cdots = \varepsilon_m = 0$.

Pf. $1 = \theta_1^{\varepsilon_1}\theta_2^{\varepsilon_2}\cdots \theta_m^{\varepsilon_m}(a_i) = \omega_i^{\varepsilon_i}$, Since $\omega_i$ is a primitive $d_i$-th root of $1$, $\varepsilon_i = 0$ for $1\leq i\leq m$.

Observe: $\theta_1, \ldots, \theta_m$ generate $X(G)$. Pick $\theta\in X(G)$. Since $a_i^{d_i} = 1$, $1 = \theta(a_i^{d_i}) = \theta(a_i)^{d_i}$.

Hence $\theta(a_i) = \omega^{\varepsilon_i}$ for some $\varepsilon_i$ with $0\leq \varepsilon_i < d_i$.

Now $\theta = \theta_1^{\varepsilon_1}\cdots \theta_m^{\varepsilon_m}$, since these are both equal to $\omega_i^{\varepsilon_i}$ at $a_i$ for $1\leq i \leq m$.

Therefore, $$G \to X(G) \quad (a_i \mapsto \theta_i)$$ is an isomorphism of groups.

Note. The correspondence above is clearly a group homomorphism.