**3. Main Results**

First, we will give some useful lemmas.

The *Cartesian product* Γ1✷Γ<sup>2</sup> of graphs Γ<sup>1</sup> and Γ<sup>2</sup> is a graph with vertex set *V*(Γ1) × *V*(Γ2). Two vertices (*u*, *v*) and (*u* , *v* ) are adjacent in Γ1✷Γ<sup>2</sup> if either *u* = *u* and *vv* ∈ *E*(Γ2) or *uu* ∈ *E*(Γ1) and *v* = *v* . Clearly Γ1✷Γ<sup>2</sup> = Γ2✷Γ1.

**Lemma 1.** *Let T be a labeled tree of order n, e any edge of T, and T*1*, T*<sup>2</sup> *two components of T* − *e, where* |*V*(*T*1)| = *r. Furthermore, let S (S*−*, S*1*, S*2*, respectively) be the transposition set on* {1, 2, ... , *n*} *satisfying T*(*S*) = *T (T*(*S*−) = *T* − *e, T*(*S*1) = *T*1*, T*(*S*2) = *T*2*. Then,* Cay(S*n*, *S*−) *has* ( *n <sup>r</sup>*) *components and each component is isomorphic to* Cay(S*r*, *S*1)✷Cay(S*n*−*r*, *S*2)*.*

**Proof.** Without loss of generality, we can assume *r* ≤ *<sup>n</sup>* 2 .

When *r* = 1, *T*<sup>1</sup> is an isolated vertex, *e* is a pendant edge and *S*<sup>1</sup> = ∅. Then, Cay(S1, *S*1)✷Cay(S*n*<sup>−</sup>1, *S*2) = Cay(S*n*<sup>−</sup>1, *S*2). The lemma is true, following from Proposition 3.

When *r* ≥ 2, we relabel *T* as follows: Relabel the vertices of *T*<sup>1</sup> as {1, 2, ... ,*r*} and the vertices of *T*<sup>2</sup> as {*r* + 1,*r* + 2, ... , *n*}. Let *S* , *S*−, *S* <sup>1</sup>, *<sup>S</sup>* <sup>2</sup> be the corresponding transposition sets. Obviously, *S*− = *S* <sup>1</sup> <sup>∪</sup> *<sup>S</sup>* <sup>2</sup>. By Proposition 2, we know that Cay(S*n*, *<sup>S</sup>*) <sup>∼</sup><sup>=</sup> Cay(S*n*, *<sup>S</sup>* ), Cay(S*n*, *S*−) ∼= Cay(S*n*, *S*−), and so on. Thus, we only need to prove the corresponding result on *S* , *S*−, *S* <sup>1</sup> and *<sup>S</sup>* <sup>2</sup>. Since *<sup>T</sup>* − *<sup>e</sup>* is disconnected, Cay(S*n*, *<sup>S</sup>*−) is also disconnected by Proposition 1. Let Γ<sup>1</sup> be the component of Cay(S*n*, *S*−) containing the identity element **1**. Since *T*<sup>1</sup> and *T*<sup>2</sup> are connected, *S* <sup>1</sup> generates <sup>S</sup>*<sup>r</sup>* and *<sup>S</sup>* <sup>2</sup> generates S*n*−*<sup>r</sup>* (let S*n*−*<sup>r</sup>* be symmetric group on {*r* + 1,*r* + 2, ... , *n*}). Then, the vertices in Γ<sup>1</sup> can be represented as **v** = *x*1*x*<sup>2</sup> ... *xrxr*+<sup>1</sup> ... *xn*, where *x*1*x*<sup>2</sup> ... *xr* is a permutation on {1, 2, ... ,*r*} and *xr*+<sup>1</sup> ... *xn* is a permutation on {*r* + 1,*r* + 2, ... , *n*}. Furthermore, let **v** = *x*1*x*<sup>2</sup> ... *xrxr*+<sup>1</sup> ... *xn* and **v** = *x* 1*x* <sup>2</sup> ... *<sup>x</sup> rx <sup>r</sup>*+<sup>1</sup> ... *<sup>x</sup> <sup>n</sup>* be two vertices in Γ1. Then, **v** and **v** are adjacent if and only if for *j*, *k* ≤ *r* and (*jk*) ∈ *S* <sup>1</sup>, *xk* <sup>=</sup> *<sup>x</sup> j* , *xj* = *x <sup>k</sup>* and *xl* <sup>=</sup> *<sup>x</sup> <sup>l</sup>* for other digits, or, for *j*, *k* ≥ *r* + 1 and (*jk*) ∈ *S* <sup>2</sup>, *xk* = *<sup>x</sup> j* , *xj* = *x <sup>k</sup>* and *xl* <sup>=</sup> *<sup>x</sup> <sup>l</sup>* for other digits. Thus, <sup>Γ</sup><sup>1</sup> ∼= Cay(S*r*, *<sup>S</sup>* <sup>1</sup>)✷Cay(S*n*−*r*, *<sup>S</sup>* <sup>2</sup>) and |*V*(Γ1)| = *<sup>r</sup>*!(*<sup>n</sup>* − *<sup>r</sup>*)!. Since Cay(S*n*, *<sup>S</sup>*−) is vertextransitive, all components of Cay(S*n*, *S*−) are isomorphic and there exist *<sup>n</sup>*! *<sup>r</sup>*!(*n*−*r*)! = ( *n r*) components in it.

We need to consider the extendability of the Cartesian product when we investigate the extendability of T*n*(*S*). The following lemmas are used several times in the proof of our theorem.

**Lemma 2** ([25,26])**.** *If* Γ *is a k-extendable graph, then* Γ✷*K*<sup>2</sup> *is* (*k* + 1)*-extendable.*

**Lemma 3** ([25])**.** *If* Γ<sup>1</sup> *and* Γ<sup>2</sup> *are k-extendable and l-extendable graphs, respectively, then their Cartesian product* Γ1✷Γ<sup>2</sup> *is* (*k* + *l* + 1)*-extendable.*

**Lemma 4** ([27])**.** *A bipartite Cayley graph is 2-extendable if and only if it is not a cycle.*

In order to prove the main result, we need other definitions and notations. The *symmetric difference* of two sets *A* and *B* is defined as the set *A*#*B* = (*A* − *B*) ∪ (*B* − *A*). Let Γ be a connected graph. If *e* = *uv* ∈ *E*(Γ), denote *V*(*e*) = {*u*, *v*} and *E*(*v*) = {*e*|*V*(*e*) ∩ {*v*} = ∅}.

Let **x** be a permutation of [*n*]. The smallest positive integer *k* for which **x***<sup>k</sup>* is the identity permutation, this number *k* is called the *order* of **x**, denoted by *o*(**x**) = *k*. *fix*(**x**) denotes the set of points in [*n*] fixed by **x** (see [10]). Let *fix*(**x**)=[*n*] − *fix*(**x**). As we know, there is another way of writing the permutation as products of disjoint cycles which are commutative (see [21]). For example, if **x** ∈ S9, **x** = 324, 158, 967, then **x** = (134)(68)(79) = (68)(134)(79), and further *fix*(**x**) = {2, 5}, *fix*(**x**) = {1, 3, 4, 6, 7, 8, 9}, | *fix*(**x**)| = 7. We say that **x** is a type of (*m*1*m*2*m*3)(*m*4*m*5)(*m*6*m*7) permutation. Clearly **x**<sup>6</sup> = **1** and *o*(**x**) = 6.

**Theorem 1.** *Any Cayley graph* T*n*(*S*) ∈ T*<sup>n</sup> is* (*n* − 2)*-extendable for any integer n* ≥ 3*.*

**Proof.** We prove the theorem by induction on *n*. For *n* = 3, the T3(*S*) is 6-cycle, which is 1-extendable. For *n* = 4, the T4(*S*) is a 3-regular bipartite Cayley graph, which is not a cycle. T4(*S*) is 2-extendable by Lemma 4.

Now we assume the statement is true for all integers smaller than *n* (*n* ≥ 5). Let *S* be a subset of transpositions on [*n*]. The transposition generating graph *T*(*S*) is a tree. We will show that any matching *M* of size (*n* − 2) can be extended to a perfect matching of T*n*(*S*).

Let *M* be a matching with (*n* − 2) edges. There are (*n* − 1) classes of edges in T*n*(*S*) because of |*S*| = *<sup>n</sup>* − 1. We may suppose that *Es*4*t*<sup>4</sup> ∩ *<sup>M</sup>* = <sup>∅</sup>. Let *<sup>S</sup>*<sup>−</sup> = *<sup>S</sup>*\(*s*4*t*4). By Lemma 1, Cay(S*n*, *S*−) has ( *n <sup>r</sup>*) connected components and each component is isomorphic to Cay(S*r*, *S*1)✷Cay(S*n*−*r*, *S*2). We may assume 1 ≤ *r* ≤ *<sup>n</sup>* <sup>2</sup> by the symmetry of Cartesian product. For the convenience, we denote the components of T*n*(*S*)\*Es*4*t*<sup>4</sup> = Cay(S*n*, *<sup>S</sup>*−) by C*i*(*i* = 1, 2, . . . , *l*), where *l* = ( *n r*).

**Claim 1.** C*<sup>i</sup>* is (*n* − 3)-extendable*.*

If *<sup>r</sup>* = 1, the transposition (*s*4*t*4) corresponding to the edge is a leaf of *<sup>T</sup>*(*S*), C*<sup>i</sup>* ∼= T*n*−1(*S* ) by Proposition 3, where *S* = *S*<sup>−</sup> = *S*\(*s*4*t*4), C*<sup>i</sup>* is (*n* − 3)-extendable by the inductive hypothesis.

If *<sup>r</sup>* = 2, T2(*S*) = *<sup>K</sup>*2, C*<sup>i</sup>* ∼= *<sup>K</sup>*2✷Cay(S*n*−2, *<sup>S</sup>*2) = *<sup>K</sup>*2✷T*n*−2(*S*2). T*n*−2(*S*2) is (*<sup>n</sup>* − <sup>4</sup>) extendable by the inductive hypothesis. C*<sup>i</sup>* is (*n* − 3)-extendable by Lemma 2.

If *<sup>r</sup>* ≥ 3, by the inductive hypothesis Cay(S*r*, *<sup>S</sup>*1) ∼= T*r*(*S*1) is (*<sup>r</sup>* − <sup>2</sup>)-extendable and Cay(S*n*−*r*, *<sup>S</sup>*2) ∼= T*n*−*r*(*S*2) is (*<sup>n</sup>* − *<sup>r</sup>* − <sup>2</sup>)-extendable. Hence, Cay(S*r*, *<sup>S</sup>*1)✷Cay(S*n*−*r*, *<sup>S</sup>*2) is (*n* − 3)-extendable by Lemma 3. We get the Claim.

Let *J* = {*i*|*E*(C*i*) ∩ *M* = ∅}. If |*J*| ≥ 2, then |*E*(C*i*) ∩ *M*| ≤ *n* − 3. When *i* ∈ *J*, each edge set *E*(C*i*) ∩ *M* can be extended to a perfect matching of C*i*, which is defined by *M*(C*i*). Clearly, *M* ⊂ *i*∈*J M*(C*i*). When *i* ∈/ *J*, let *M*(C*i*) be an arbitrary perfect matching of C*i*. Then,

$$\bigcup\_{i=1}^{l} M(\mathcal{C}\_{i}) = \left(\bigcup\_{i \in f} M(\mathcal{C}\_{i})\right) \cup \left(\bigcup\_{i \notin f} M(\mathcal{C}\_{i})\right) \text{ is a perfect matching of } \text{Cay}(\mathfrak{S}\_{n}, \mathcal{S}^{-})\text{, which is called a perfect matching of } \mathcal{T}\_{\mathbb{R}}(S).$$

When |*J*| = 1, without loss of generality, we assume that *M* ⊂ *E*(C1) and C<sup>1</sup> contains the identity permutation **1**. If *M* can be extended to a perfect matching of C1, we are done. Suppose that *M* cannot be extended to a perfect matching of C1. Let *e*<sup>2</sup> = *v*1*v*<sup>2</sup> be an edge in *M*. *M*\*e*<sup>2</sup> can be extended to a perfect matching of C<sup>1</sup> (since |*M*\*e*2| = *n* − 3), which is denoted by *M* (C1). Let *E*(*v*1) ∩ *M* (C1) = *e*1, *E*(*v*2) ∩ *M* (C1) = *e*3, *V*(*e*1) = {*v*0, *v*1}, *V*(*e*3) = {*v*2, *v*3} and *e*<sup>4</sup> = *E*(*v*3) ∩ *Es*4*t*<sup>4</sup> . By the transitivity of C<sup>1</sup> and without loss of

generality, we can assume that *<sup>v</sup>*<sup>0</sup> <sup>=</sup> **<sup>1</sup>**. Let *<sup>o</sup>*(*g*(*e*1)*g*(*e*2)*g*(*e*3)*g*(*e*4)) = *<sup>a</sup>*, *vi* <sup>=</sup> *<sup>i</sup>* ∏ *j*=1 *g*(*ej*),

and *e*4*b*+<sup>1</sup> ∈ *Es*1*t*<sup>1</sup> , *e*4*b*+<sup>2</sup> ∈ *Es*2*t*<sup>2</sup> , *e*4*b*+<sup>3</sup> ∈ *Es*3*t*<sup>3</sup> ,*e*4*b*+<sup>4</sup> ∈ *Es*4*t*<sup>4</sup> (*b* = 0, ... , *a* − 1), where {(*s*1*t*1),(*s*2*t*2),(*s*3*t*3),(*s*4*t*4)} ⊂ *S*. It is easy to see *g*(*e*2) = *g*(*ei*) (*i* = 1, 3), *g*(*e*4) = *g*(*ei*) (*i* = 1, 2, 3), *g*(*e*1)*g*(*e*2)*g*(*e*3) = *g*(*e*4), *v*<sup>3</sup> = *g*(*e*1)*g*(*e*2)*g*(*e*3) is an odd permutation and *v*<sup>4</sup> = *g*(*e*1)*g*(*e*2)*g*(*e*3)*g*(*e*4) is an even permutation. The cardinality of *fix*(*v*3) can only be 2, 4, 5 and 6. We discuss these four cases one by one in order to prove that *M* can be extended to a perfect matching of T*n*(*S*).

Case 1. | *fix*(*v*3)| = 2.

*e*4*b*+<sup>1</sup> ∈ *M*

In this case, *v*<sup>3</sup> is a transposition and *o*(*v*3) = 2. There are two subcases for the order of *v*4.

Subcase 1.1. *v*<sup>4</sup> is a type of (*m*1*m*2)(*m*3*m*4) permutation.

We have *<sup>o</sup>*(*v*4) = 2, (*v*4)<sup>2</sup> <sup>=</sup> **<sup>1</sup>**. Note that *vi* <sup>=</sup> *<sup>i</sup>* ∏ *j*=1 *g*(*ej*), where *i* ∈ [8]. Hence, there is an 8-cycle *C*<sup>8</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*7*e*8*v*<sup>8</sup> (*v*<sup>8</sup> = *v*0). The vertex *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) (*i* = 0, 1, 2, 3; *b* = 0, 1). We may take a perfect matching *M* (C2) of C<sup>2</sup> such that *e*<sup>5</sup> ∈ *M* (C2),*e*<sup>7</sup> ∈ *M* (C2) and *e*<sup>6</sup> ∈/ *M* (C2) because of C<sup>2</sup> ∼= C1. Now we take *<sup>M</sup>* = (*M* (C1) *M* (C2))#*E*(*C*8). Clearly *M* ⊂ *M*, *M* is a perfect matching of subgraph T*n*(*S*)[*V*(C1) *V*(C2)]. Let *M*(C*i*) *l*

be a perfect matching of C*<sup>i</sup>* (*i* = 3, ... , *l*). Hence, *i*=3 *M*(C*i*) *M* is a perfect matching of T*n*(*S*).

Subcase 1.2. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3) permutation.

We have *<sup>o</sup>*(*v*4) = 3, (*v*4)<sup>3</sup> <sup>=</sup> **<sup>1</sup>**. Note that *vi* <sup>=</sup> *<sup>i</sup>* ∏ *j*=1 *g*(*ej*), where *i* ∈ [12]. Hence, there is a 12-cycle *C*<sup>12</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*11*e*12*v*<sup>12</sup> (*v*<sup>12</sup> = *v*0). The vertex *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) (*i* = 0, 1, 2, 3; *b* = 0, 1, 2). We may take a perfect matching *M* (C*b*+1) of C*b*+<sup>1</sup> such that (C*b*+1) (*<sup>b</sup>* = 1, 2) because of C*b*+<sup>1</sup> ∼= C1.

(C*b*+1) and *e*4*b*+<sup>2</sup> ∈/ *M*

(C*b*+1),*e*4*b*+<sup>3</sup> ∈ *M*

Now we take *M* = <sup>3</sup> *i*=1 *M* (C*i*) ! #*E*(*C*12). Clearly *M* ⊂ *M*, *M* is a perfect matching of subgraph <sup>T</sup>*n*(*S*)[ <sup>3</sup> *i*=1 *V*(C*i*)]. Let *M*(C*i*) be a perfect matching of C*<sup>i</sup>* (*i* = 4, ... , *l*). Hence, *l i*=4 *M*(C*i*) *M* is a perfect matching of T*n*(*S*).

Case 2. | *fix*(*v*3)| = 4.

In this case, *v*<sup>3</sup> is a type of (*m*1*m*2*m*3*m*4) permutation and *o*(*v*3) = 4. There are two subcases.

Subcase 2.1. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3*m*4)(*m*5*m*6) permutation.

We have *<sup>o</sup>*(*v*4) = 4, (*v*4)<sup>4</sup> <sup>=</sup> **<sup>1</sup>**. Note that *vi* <sup>=</sup> *<sup>i</sup>* ∏ *j*=1 *g*(*ej*), where *i* ∈ [16]. Hence, there is a 16-cycle *C*<sup>16</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*15*e*16*v*<sup>16</sup> (*v*<sup>16</sup> = *v*0). The vertex *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) (*i* = 0, 1, 2, 3; *b* = 0, 1, 2, 3). We may take a perfect matching *M* (C*b*+1) of C*b*+<sup>1</sup> such that *e*4*b*+<sup>1</sup> ∈ *M* (C*b*+1),*e*4*b*+<sup>3</sup> ∈ *M* (C*b*+1) and *e*4*b*+<sup>2</sup> ∈/ *M* (C*b*+1) (*b* = 1, 2, 3) because of C*b*+<sup>1</sup> ∼= C1. Now we take *<sup>M</sup>* = <sup>4</sup> *i*=1 *M* (C*i*) ! #*E*(*C*16). Clearly *M* ⊂ *M*, *M* is a perfect

matching of subgraph <sup>T</sup>*n*(*S*)[ <sup>4</sup> *i*=1 *V*(C*i*)]. Let *M*(C*i*) be a perfect matching of C*<sup>i</sup>* (*i* = 5, ... , *l*). *l*

Hence, *i*=5 *M*(C*i*) *M* is a perfect matching of T*n*(*S*).

Subcase 2.2. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3*m*4*m*5) permutation.

We have *<sup>o</sup>*(*v*4) = 5, (*v*4)<sup>5</sup> <sup>=</sup> **<sup>1</sup>**. Note that *vi* <sup>=</sup> *<sup>i</sup>* ∏ *j*=1 *g*(*ej*), where *i* ∈ [20]. Hence, there is a 20-cycle *C*<sup>20</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*19*e*20*v*<sup>20</sup> (*v*<sup>20</sup> = *v*0). The vertex *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) (*i* = 0, 1, 2, 3; *b* = 0, 1, 2, 3, 4). We may take a perfect matching *M* (C*b*+1) of C*b*+<sup>1</sup> such that *e*4*b*+<sup>1</sup> ∈ *M* (C*b*+1),*e*4*b*+<sup>3</sup> ∈ *M* (C*b*+1) and *e*4*b*+<sup>2</sup> ∈/ *M* (C*b*+1) (*b* = 1, 2, 3, 4) because of C*b*+<sup>1</sup> ∼= C1. Now we take *<sup>M</sup>* = <sup>5</sup> *i*=1 *M* (C*i*) ! #*E*(*C*20). Clearly *M* ⊂ *M*, *M* is a perfect

matching of subgraph <sup>T</sup>*n*(*S*)[ <sup>5</sup> *i*=1 *V*(C*i*)]. Let *M*(C*i*) be a perfect matching of C*<sup>i</sup>* (*i* = 6, ... , *l*). *l*

Hence, *M*(C*i*) *M* is a perfect matching of T*n*(*S*).

*i*=6 Case 3. | *fix*(*v*3)| = 5.

In this case, *v*<sup>3</sup> is a type of (*m*1*m*2*m*3)(*m*4*m*5) permutation and *o*(*v*3) = 6. There are four subcases.

Subcase 3.1. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3)(*m*4*m*5*m*6) permutation.

We have *o*(*v*4) = 3, (*v*4)<sup>3</sup> = **1**. There is a 12-cycle *C*<sup>12</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*11*e*12*v*<sup>12</sup> (*v*<sup>12</sup> <sup>=</sup> *<sup>v</sup>*0) in subgraph <sup>T</sup>*n*(*S*)[ <sup>3</sup> *i*=1 *V*(C*i*)], where *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) for *i* = 0, 1, 2, 3; *b* = 0, 1, 2.

The rest of the proof is similar to Subcase 1.2.

Subcase 3.2. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3*m*4)(*m*5*m*6) permutation.

We have *o*(*v*4) = 4, (*v*4)<sup>4</sup> = **1**. There is a 16-cycle *C*<sup>16</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*15*e*16*v*<sup>16</sup> (*v*<sup>16</sup> <sup>=</sup> *<sup>v</sup>*0) in subgraph <sup>T</sup>*n*(*S*)[ <sup>4</sup> *i*=1 *V*(C*i*)], where *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) for *i* = 0, 1, 2, 3;

*b* = 0, 1, 2, 3. The rest of the proof is similar to Subcase 2.1.

Subcase 3.3. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3*m*4*m*5) permutation.

We have *o*(*v*4) = 5, (*v*4)<sup>5</sup> = **1**. There is a 20-cycle *C*<sup>20</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*19*e*20*v*<sup>20</sup> (*v*<sup>20</sup> <sup>=</sup> *<sup>v</sup>*0) in subgraph <sup>T</sup>*n*(*S*)[ <sup>5</sup> *V*(C*i*)], where *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) for *i* = 0, 1, 2, 3;

*i*=1 *b* = 0, 1, 2, 3, 4. The rest of the proof is similar to Subcase 2.2.

Subcase 3.4. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3)(*m*4*m*5)(*m*6*m*7) permutation.

We have *o*(*v*4) = 6, | *fix*(*v*4)| = 7 and *n* ≥ 7, *l* = ( *n <sup>r</sup>*) <sup>≥</sup> 7, (*v*4)<sup>6</sup> <sup>=</sup> **<sup>1</sup>**. *vi* <sup>=</sup> *<sup>i</sup>* ∏ *j*=1 *g*(*ej*),

where *i* ∈ [24]. Hence, there is a 24-cycle *C*<sup>24</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*23*e*24*v*<sup>24</sup> (*v*<sup>24</sup> = *v*0). The vertex *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) (*i* = 0, 1, 2, 3; *b* = 0, 1, 2, 3, 4, 5). We may take a perfect matching *M* (C*b*+1) of C*b*+<sup>1</sup> such that *e*4*b*+<sup>1</sup> ∈ *M* (C*b*+1),*e*4*b*+<sup>3</sup> ∈ *M* (C*b*+1) and *e*4*b*+<sup>2</sup> ∈/ *M* (C*b*+1) (*<sup>b</sup>* = 1, 2, 3, 4, 5) because of C*b*+<sup>1</sup> ∼= C1. Now we take *<sup>M</sup>* = <sup>6</sup> *i*=1 *M* (C*i*) ! #*E*(*C*24). Clearly,

*<sup>M</sup>* <sup>⊂</sup> *<sup>M</sup>*, *<sup>M</sup>* is a perfect matching of subgraph <sup>T</sup>*n*(*S*)[ <sup>6</sup> *i*=1 *V*(C*i*)]. Let *M*(C*i*) be a perfect

matching of C*<sup>i</sup>* (*i* = 7, . . . , *l*). Hence, *l i*=7 *M*(C*i*) *M* is a perfect matching of T*n*(*S*).

Case 4. | *fix*(*v*3)| = 6.

In this case, *v*<sup>3</sup> is a type of (*m*1*m*2)(*m*3*m*4)(*m*5*m*6) permutation and *o*(*v*3) = 2. There are three subcases.

Subcase 4.1. *v*<sup>4</sup> is a type of (*m*1*m*2)(*m*3*m*4)(*m*5*m*6)(*m*7*m*8) permutation.

We have *o*(*v*4) = 2. There is an 8-cycle *C*<sup>8</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*7*e*8*v*<sup>8</sup> (*v*<sup>8</sup> = *v*0) in subgraph <sup>T</sup>*n*(*S*)[ <sup>2</sup> *i*=1 *V*(C*i*)], where *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) for *i* = 0, 1, 2, 3; *b* = 0, 1. The rest of the proof is

similar to Subcase 1.1.

Subcase 4.2. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3*m*4)(*m*5*m*6) permutation.

We have *o*(*v*4) = 4. There is a 16-cycle *C*<sup>16</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*15*e*16*v*<sup>16</sup> (*v*<sup>16</sup> = *v*0) in subgraph <sup>T</sup>*n*(*S*)[ <sup>4</sup> *i*=1 *V*(C*i*)], where *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) for *i* = 0, 1, 2, 3; *b* = 0, 1, 2, 3. The rest of the proof is similar to Subcase 2.1.

Subcase 4.3. *v*<sup>4</sup> is a type of (*m*1*m*2*m*3)(*m*4*m*5)(*m*6*m*7) permutation.

We have *o*(*v*4) = 6. There is a 24-cycle *C*<sup>24</sup> = *v*0*e*1*v*1*e*<sup>2</sup> ... *v*23*e*24*v*<sup>24</sup> (*v*<sup>24</sup> = *v*0) in subgraph <sup>T</sup>*n*(*S*)[ <sup>6</sup> *i*=1 *V*(C*i*)], where *v*4*b*+*<sup>i</sup>* ∈ *V*(C*b*+1) for *i* = 0, 1, 2, 3; *b* = 0, 1, 2, 3, 4, 5. The rest of the proof is similar to Subcase 3.4.

In conclusion, any matching *M* of size *n* − 2 can be extended to a perfect matching of T*n*(*S*). The proof is complete.

The extendability number of Γ, denoted by *ext*(Γ), is the maximum *k* such that Γ is *k*-extendable. As we know that T*n*(*S*) ∈ T*<sup>n</sup>* is an (*n* − 1)-regular bipartite Cayley graph and not (*n* − 1)-extendable. We can obtain the extendability number of T*n*(*S*) by Theorem 1.

**Corollary 1.** *ext*(T*n*(*S*)) = *n* − 2 *for n* ≥ 3*.*
