GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 3, 1996, 293-300

SEMIDIRECT PRODUCTS AND WREATH PRODUCTS OF STRONGLY π-INVERSE MONOIDS

YUFEN ZHANG, SHIZHENG LI, AND DESHENG WANG

Abstract. In this paper we determine the necessary and sufficient conditions for the semidirect products and the wreath products of two monoids to be strongly π-inverse. Furthermore, we determine the least group congruence on a strongly π-inverse monoid, and we give some important isomorphism theorems.

1. Introduction

Our terminology and notation will follow [1] and [2].

Let S and T be two monoids, and let End(T) be the endomorphism monoid of T, and write endomorphisms as exponents to the right of arguments. If α : S → End(T) is a homomorphism, and if s ∈ S and t ∈ T, write t^{s} for t^{α(s)}, since α(s) ∈ End(T) for s ∈ S, then for t_{1}, t_{2} ∈ T, (t_{1}, t_{2})^{s} = t_{1}^{s}t_{2}^{s}. Since α is a homomorphism, (t^{s_{1}})^{s_{2}} = t_{s_{1}s_{2}} for every t ∈ T and s_{1}, s_{2} ∈ S.

The semidirect product S×_{α}T is the monoid with elements {(s, t) : s ∈ S, t ∈ T } and multiplication (s_{1}, t_{1})(s_{2}, t_{2}) = (s_{1}s_{2}, t_{1}^{s_{2}}t_{2}).

In [3], [4] the authors have determined the necessary and sufficient conditions for S ×_{α} T to be regular, inverse, and orthodox. In this paper we determine the necessary and sufficient conditions for S ×_{α} T to be strongly π-inverse and give their applications to the wreath product.

For a monoid S, E(S) and Reg S denote the set of idempotents of S and the set of regular elements of S, respectively.

A semigroup is π-regular if for every s ∈ S there is an m ∈ ℕ such that s^{m} ∈ Reg S. If S is π-regular and E(S) is a commutative subsemigroup, then we call S a strongly π-inverse semigroup. It is easy to see that Reg S is an inverse subsemigroup of a strongly π-inverse semigroup S.

293

1991 Mathematics Subject Classification. 20M18.

Key words and phrases. Wreath product, π-inverse monoid, regular monoid, group congruence.

1072-947X/96/0500-0293$09.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1996 Plenum Publishing Corporation

2. Semidirect Products

Let S and T be two monoids and let S ×_{α} T be the semidirect product of S and T, where α : S → End(T) is a given homomorphism.

Lemma 1. Let S ×_{α} T be a strongly π-inverse monoid; then

(1) both S and T are strongly π-inverse monoids;

(2) u^{e} = u for every e ∈ E(S) and every u ∈ E(T);

(3) if t^{e}t = t for t ∈ T and e ∈ E(S), then t^{e} = t;

(4) t^{e} = t for every t ∈ Reg T and every e ∈ E(S);

(5) for every s ∈ S and t ∈ T, there exists m ∈ ℕ such that s^{m} ∈ Reg S and t^{s(m)} ∈ Reg T, where t^{s(m)} = t^{s^{m−1}}t^{s^{m−2}} ⋯ t^{s}t.

Proof. (1) For arbitrary s ∈ S, there exist m ∈ ℕ and (s_{1}, t_{1}) ∈ T such that (s, 1)^{m}(s_{1}, t_{1})(s, 1)^{m} = (s, 1)^{m}. Hence (s^{m}s_{1}s^{m}, t_{1}^{s^{m}}) = (s^{m}, 1), s^{m}s_{1}s^{m} = s^{m} and then S is π-regular.

Since (e, 1), (f, a) ∈ E(S ×_{α} T) for e, f ∈ E(S), we have (e, 1)(f, 1) = (f, 1)(e, 1), and then ef = fe. Hence S is strongly π-inverse monoid.

For arbitrary t ∈ T, there exist m ∈ ℕ and (s_{2}, t_{2}) ∈ S ×_{α} T such that (1, t)^{m}(s_{2}, t_{2})(1, t)^{m} = (1, t)^{m}, that is, (s_{2}, (t^{m})^{s_{2}}t_{2}t^{m}) = (1, t^{m}). Then s_{2} = 1 and t^{m}t_{2}t^{m} = (t^{m})^{s_{2}}t_{2}t^{m} = t^{m}. Thus T is π-regular.

Since (1, u), (1, v) ∈ E(S ×_{α} T) for u, v ∈ E(T) and S ×_{α} T is strongly π-inverse, we have (1, u)(1, v) = (1, v)(1, u), so that uv = vu, which implies that T is strongly π-inverse.

(2) Let e ∈ E(S) and u ∈ E(T). Then (e, 1), (1, u) ∈ E(S ×_{α} T) and (e, 1)(1, u) = (1, u)(e, 1), which implies u^{e} = u.

(3) If t^{e}t = t, then (e, t) ∈ E(S ×_{α} T) and (e, t)(e, 1) = (e, 1)(e, t) since (e, 1) ∈ E(S ×_{α} T). Hence t^{e} = t.

(4) From (1), for every t ∈ Reg T, there exists a unique t_{1} ∈ T such that tt_{1}t = t, t_{1}tt_{1} = t_{1}. Then t^{e}t_{i}^{e}t^{e} = t^{e}, further, (t^{e}t_{1})^{e}t^{e}t_{1} = t^{e}t_{1}. From (3) we have (t^{e}t_{1})^{e} = t^{e}t_{1}, that is, (tt_{1})^{e} = t^{e}t_{1}. Since tt_{1} ∈ E(T), from (2) we have (tt_{1})^{e} = tt_{1} = t^{e}t_{1}, and then t_{1}tt_{1} = t_{1}t^{e}t_{1} = t_{1}, t^{e}t_{1}t^{e} = t^{e}. Thus both t and t^{e} are inverses of t_{1}, and then t^{e} = t.

(5) Since S ×_{α} T is a strongly π-inverse monoid, for every (s, t) ∈ S ×_{α} T there exist m ∈ ℕ and (s_{1}, t_{1}) ∈ S ×_{α} T such that (s, t)^{m}(s_{1}, t_{1})(s, t)^{m} = (s, t)^{m}. Then

(s^{m}s_{1}s^{m}, (t^{s(m)})^{s_{1}s^{m}}t_{1}^{s^{m}}t^{s(m)}) = (s^{m}, t^{s(m)}),

so that s^{m}s_{1}s^{m} =s^{m}, (t^{s(m)})^{s_{1}s^{m}}t_{1}^{s^{m}}t^{s(m)} =t^{s(m)}. Then

(t^{s(m)}t_{1}^{s^{m}})^{s_{1}s^{m}}t^{s(m)}t_{1}^{s^{m}} = t^{s(m)}t_{1}^{s^{m}}.

From (3) we have (t^{s(m)}t_{1}^{s^{m}})^{s_{1}s^{m}} = t^{s(m)}t_{1}^{s^{m}}. Thus t^{s(m)}t_{1}^{s^{m}}t^{s(m)} = t^{s(m)}, and then s^{m} ∈ Reg S and t^{s(m)} ∈ Reg T.

STRONGLY π-INVERSE MONOIDS 295

Theorem 2. Let S and T be two monoids and let α : S → End(T) be the given homomorphism, and let S ×_{α} T be the semidirect product of S and T. Then S ×_{α} T is a strongly π-inverse monoid iff

(1) both S and T are strongly π-inverse monoids,

(2) t^{e} = t for every t ∈ Reg T and every e ∈ E(S), and

(3) for every s ∈ S and t ∈ T there exists m ∈ ℕ such that s^{m} ∈ Reg S and t^{s(m)} ∈ Reg T, where t^{s(m)} = t^{s^{m−1}}t^{s^{m−2}} ⋯ t^{s}t.

Proof. The necessity of the assertion is obvious by Lemma 1. We only prove the sufficient part.

For every (s, t) ∈ S ×_{α} T, from (3) there exist m ∈ ℕ, s_{1} ∈ S and t_{1} ∈ T such that

s^{m}s_{1}s^{m} = s^{m},
t^{s(m)}t_{1}t^{s(m)} = t^{s(m)}.

From (2) we have (t^{s(m)}t_{1})^{s_{1}s^{m}} = t^{s(m)}t_{1}. Hence (t^{s(m)})^{s_{1}s^{m}}t_{1}^{s_{1}s^{m}}t^{s(m)} = t^{s(m)}, and then (s, t)^{m}(s_{1}, t_{1}^{s_{1}})(s, t)^{m} = (s, t)^{m}. This means that S ×_{α} T is π-regular.

For arbitrary (e, u) ∈ E(S ×_{α} T) we prove that e ∈ E(S) and u ∈ E(T). In fact, if (e, u)^{2} = (e, u), then e^{2} = e, u^{e}u = u. Thus u^{e} ∈ E(T), and then, from (3) there exists m ∈ ℕ such that u^{e^{m−1}} ⋯ u^{e}u = u^{e}u ∈ Reg T. From (2), (u^{e}u)^{e} = u^{e}u = u^{e}u^{e} = u^{e} = u. So that u^{2} = u.

Now, for (e, u), (f, v) ∈ E(S ×_{α} T), we have e, f ∈ E(S) and u, v ∈ E(T). By (1) and (2) we have

(e, u)(f, v) = (ef, u^{f}v) = fe, v^{e}u) = (f, v)(e, u).

Therefore S ×_{α} T is strongly π-inverse.

Theorem 3. Let S and T be two monoids and let S ×_{α} T be a strongly π-inverse monoid.

(1) (e, u) ∈ E(S ×_{α} T) iff e ∈ E(S) and u ∈ E(T).

(2) For every e ∈ E(S), let α^{∗}(e) be the restriction of α(e) on E(T); then α^{∗}(e) ∈ End(E(T)).

(3) Let α^{∗} : E(S) → End(E(T)) such that e → α^{∗}(e); then α^{∗} is a homomorphism from E(S) to End(T).

(4) E(S ×_{α} T) ≅ E(S) × E(T) ≅ E(S) ×_{α^{∗}} E(T).

Proof. It is a immediate consequence of Lemma 1 and Theorem 2.

Theorem 4. Let S and T be two monoids and let S ×_{α} T be a strongly π-inverse monoid.

(1) (s, t) ∈ Reg(S ×_{α} T) iff s ∈ Reg S and t ∈ Reg T.

(2) For every s ∈ Reg S, let α^{∗}(s) be the restriction of α(s) on Reg T; then α^{∗}(s) ∈ End(Reg T).

(3) Define α^{∗} : Reg S → End(Reg T) by s → α^{∗}(s); then α^{∗} is a homomorphism from Reg S to End(Reg T).

(4) Reg(S ×_{α} T) ≅ Reg(S) ×_{α^{∗}} Reg(T).

Proof. (1) Let (s, t) ∈ Reg(S ×_{α} T); then there exists (s_{1}, t_{1}) ∈ S ×_{α} T such that (s, t)(s_{1}, t_{1})(s, t) = (s, t), and then ss_{1}s = s, t^{s_{1}s}t_{1}^{s}t = t. From the latter equation we have (tt_{1}^{s})^{s_{1}s}tt_{1}^{s} = tt_{1}^{s}, and from Lemma 1 (3), (tt_{1}^{s})^{s_{1}s} = tt_{1}^{s}, that is, tt_{1}^{s}t = t. So that s ∈ Reg S and t ∈ Reg T.

Conversely, let s ∈ Reg S and t ∈ Reg T; then there exist s_{1} ∈ S and t_{1} ∈ T such that ss_{1}s = s, tt_{1}t = t. Hence

(s, t)(s_{1}, t_{1}^{s_{1}})(s, t) = (ss_{1}s, t_{1}^{s_{1}s}t_{1}^{s_{1}s}t) = (s, tt_{1}t) − (s, t).

Therefore (s, t) ∈ Reg(S ×_{α} T).

(2) For every s ∈ Reg S and t ∈ Reg T, t^{s} ∈ Reg T; then α^{∗}(s) ∈ End(Reg T).

(3) and (4) are obvious.

3. Least Group Congruence on a Strongly π-Inverse Monoid

Theorem 5. Let S be a strongly π-inverse monoid; then the relation

δ = {(s_{1}, s_{2}) ∈ S × S : s_{1}e = s_{2}e for some e ∈ E(S)}

is the least group congruence on S.

Proof. It is obvious that δ is a left compatible equivalent relation on S. Let xe = ye, for x, y ∈ S and e ∈ E(S). For any z ∈ S, since S is a strongly π-inverse monoid, there exist m ∈ ℕ and s ∈ S such that z^{m}sz^{m} = z^{m}, sz^{m}s = s; and we have

xz(z^{m−1}sez) = xez^{m}sez = yez^{m}sez = yz(z^{m−1}sez)

and

(z^{m−1}sez)^{2} = z^{m−1}sez^{m}sez = z^{m−1}sz^{m}sez = z^{m−1}sz^{m}sez = z^{m−1}sez.

Thus (xz, yz) ∈ δ.

It is obvious that eδ = fδ = 1 is the identity of S/δ for every e, f ∈ E(S). Now, for sδ ∈ S/δ there exist m ∈ ℕ, s_{1} ∈ S such that s^{m}s_{1}, s_{1}s^{m} ∈ E(S). Thus (s^{m}s_{1})δ = sδ(s^{m−1}s_{1})δ = 1 and (s_{1}s^{m})δ = (s_{1}s^{m−1})δ(sδ) = 1, so that sδ has an inverse element. This means that S/δ is a group.

Let ρ be an arbitrary group congruence on S. If (x, y) ∈ δ, then there exists e ∈ E(S) such that xe = ye, so that (xe)ρ = (ye)ρ. Since eρ = 1 ∈ S/ρ, we have xρ = yρ. Hence δ ⊂ ρ.

Theorem 6. Let S and T be two monoids, let S ×_{α} T be a strongly πinverse monoid, and let δ_{S×_{α}T}, δ_{S} and δ_{T} be the least group congruences on S ×_{α} T, S and T, respectively. Then

(1) for every sδ_{S} ∈ S/δ_{S} define α^{∗}(sδ_{S}) : T/δ_{T} → T/δ_{T} by tδ_{T} → t^{s}δ_{T}; then α^{∗}(sδ_{S}) ∈ End(T/δ_{T});

(2) define α^{∗} : S/δ_{S} → End(T/δ_{T}) by sδ_{S} → α^{∗}(sδ_{S}); then α^{∗} is a homomorphism;

(3) S/δ_{S} ×_{α^{∗}} T/δ_{T} ≅ (×_{α}T)/δ_{S×_{α}T}.

Proof. (1) If t_{1}δ_{T} = t_{1}δ_{T} for t_{1}, t_{2} ∈ T, then there exists u ∈ E(T) such that t_{1}u = t_{2}u and then t_{1}^{s}u^{s} = t_{2}^{s}u^{s} for s ∈ S. Since u^{s} ∈ E(T), we have t_{1}^{s}δ_{T} = t_{2}^{s}δ_{T}. Thus α^{∗}(sδ_{S}) is well defined for every sδ_{S} ∈ S/δ_{S}. It is easy to see that α^{∗}(sδ_{S}) is a homomorphism.

(2) If s_{1}δ_{S} = s_{2}δ_{S} for s_{1}, s_{2} ∈ S, then s_{1}e = s_{2}e for e ∈ E(S). For arbitrary tδ_{T} ∈ T/δ_{T}, there exists t_{1}δ_{T} ∈ T/δ_{T} such that (t_{1}t)δ_{T} = 1 ∈ T/δ_{T}, and then t_{1}tu = u for some u ∈ E(T). Thus, tut_{1}tu = tu, hence tu ∈ Reg T. From Lemma 1, t^{e}u = tu for every e ∈ E(S) and then t^{e}δ_{T} = tδ_{T}, so α^{∗}(eδ_{S}) is an identity mapping on T/δ_{T}. Thus t^{s_{1}}δ_{T} = t^{s_{1}e}δ_{T} = t^{s_{2}e}δ_{T} = t^{s_{2}}δ_{T}, and then α^{∗}(s_{1}δ_{S}) = α^{∗}(s_{2}δ_{S}), so that α^{∗} is well defined.

For arbitrary s_{1}δ_{S}, s_{2}δ_{S} ∈ S/δ_{S} and arbitrary tδ_{T} ∈ T/δ_{T}, we have t^{(s_{1}s_{2})}δ_{T} = (t^{s_{1}})^{s_{2}}δ_{T}, that is, α^{∗}(s_{1}s_{2}) = α^{∗}(s_{1}δ_{S})α^{∗}(s_{2}δ_{S}). Thus α^{∗} is a homomorphism from S/δ_{S} to End(T/δ_{T}).

(3) Define φ : (S ×_{α} T)/δ_{S×_{α}T} ⟶ S/δ_{S} × T/δ_{T} by (s, t)δ_{S×_{α}T} ⟶ (sδ_{S}, tδ_{T}). Suppose (s_{1}, t_{1})δ_{S×_{α}T} = (s_{1}, t_{2})δ_{S×_{α}T}. Then there exist (e, u) ∈ E(S ×_{α} T) such that (s_{1}, t_{1})(e, u) = (s_{2}, t_{2})(e, u); then s_{1}e = s_{2}e, t_{1}^{e}u = t_{2}^{e}u. So s_{1}δ_{S} = s_{2}δ_{S}, t_{1}δ_{T} = t_{2}δ_{T}, and then (s_{1}δ_{S}, t_{1}δ_{T}) = (s_{2}δ_{S}, t_{2}δ_{T}). Thus φ is well defined. φ is obviously surjective.

If (s_{1}δ_{S}, t_{1}δ_{T}) = (s_{2}δ_{S}, t_{1}δ_{T}), then s_{1}δ_{S} = s_{2}δ_{S}, t_{1}δ_{T} = t_{2}δ_{T}, and then there exist e ∈ E(S) and u ∈ E(T) such that s_{1}e = s_{2}e, t_{1}u = t_{2}u. From Lemma 1 (2), t_{1}^{e}u = t_{1}^{e}u^{e} = t_{2}^{e}u^{e} = t_{2}^{e}u. Hence (s_{1}, t_{1})(e, u) = (s_{2}, t_{2})(e, u), and then (s_{1}, t_{1})δ_{S×_{α}T} = (s_{2}, t_{2})δ_{S×_{α}T}. Thus φ is one-to-one.

It is easy to see that φ is a homomorphism. Thus (S ×_{α} T)/δ_{S×_{α^{∗}} T} ≅ δ_{S} ×_{α^{∗}} T/δ_{T}.

Corollary 7. Let S be a strongly π-inverse monoid. Then for every s ∈ S there exist e, f ∈ E(S) such that se, fs ∈ Reg S.

Proof. From the proof of Theorem 6 we know that se ∈ Reg S for some e ∈ E(S). Using a similar way, we can prove that the binary relation defined on S by

σ = {(s_{1}, s_{2}) : fs_{1} = fs_{2} for some f ∈ E(S)}

is also the least group congruence on S. Then there exists f ∈ E(S) such that fs ∈ Reg S, using the same method as in the proof of Theorem 6.

4. Wreath Products

Let S be a monoid, S acts on a set X from the left, that is, sx ∈ X, 1x = x and s(rx) = (sr)x for every s, r ∈ S and x ∈ X. Let T also be a monoid, then the wreath product Sw_{x}T = S ×_{α} T^{X}, where T^{X} = {f : X ⟶ T is a function} is the Cartesian power of T, that is, fg(x) = f(x)g(x) for every f, g ∈ T^{X} and every x ∈ X, and where the homomorphism α : S ⟶ End(T^{X}) is defined by (f^{s})(x) = f(sx) for every s ∈ S, f ∈ T^{X} and x ∈ X.

Lemma 8. Let T be a monoid and let R = {T′ ⊂ T | |T′| ≤ |X|}. Then T^{X} is a strongly π-inverse monoid iff

(1) T is a strongly π-inverse monoid, and

(2) for every T′ ∈ R there exists m ∈ ℕ such that (t′)^{m} ∈ Reg T for all t′ ∈ T′.

Proof. Suppose that T^{X} is strongly π-inverse and T′ ∈ R. Then there exists g ∈ T^{X} such that g(X) = T′. Let m ∈ ℕ such that g^{m} ∈ Reg T^{X}; then (t′)^{m} = (g(x))^{m} = g^{m}(x) ∈ Reg T for all t′ ∈ T′.

Now, for each t ∈ T, let T′ = {t}; then there exists m ∈ ℕ such that t^{m} ∈ Reg T. Thus T is π-regular.

For every u, v ∈ E(T), define g : X ⟶ T by g(x) = u and h : X ⟶ T by h(x) = v for all x ∈ X. Then, g, h ∈ E(T^{X}) and

uv = g(x)h(x) = gh(x) = h(x)g(x) = vu.

Thus T is a strongly π-inverse monoid.

Conversely, for any g ∈ T^{X}, we have g(x) ∈ R, and then there exists m ∈ ℕ such that g^{m}(x) = (g(x))^{m} ∈ Reg T for all x ∈ X, so that g^{m} ∈ Reg T^{X}. Thus T^{X} is π-regular.

For each g, h ∈ E(T^{X}) we have g(x), h(x) ∈ E(T) for all x ∈ X. Then

gh(x) = g(x)h(x) = h(x)g(x) = (hg)(x)

for all x ∈ X, and then gh = hg. Thus T^{X} is strongly π-inverse.

Lemma 9. Let S and T be two monoids; S acts on a set X from the left. Then the following conditions are equivalent:

(1) for each e ∈ E(S) and g ∈ Reg T^{X}, g^{e} = g;

(2) |T | = 1 or ex = x for each e ∈ E(S) and x ∈ X.

Proof. Suppose that (1) holds. If there exist e ∈ E(S) and x ∈ X such that ex ≠ x, then for t′ ∈ Reg T define g : X ⟶ T by

g(y) = \begin{cases}1, & if y = ex, \\ t′, & if y ≠ ex. \end{cases}

We have g ∈ Reg T^{X}. Hence g^{e} = g, and then t′ = g(x) = g^{e}(x) = g(ex) = 1. Thus Reg T = {1}. But for each t ∈ T there exists m ∈ ℕ such that t^{m} ∈ Reg T. So t^{m} = 1, and then t ∈ Reg T. Hence t = 1, therefore |T | = 1.

Conversely, assume (2) holds. Let e ∈ E(S) and g ∈ Reg T^{X}. If |T | = 1, then (1) holds. If |T | ≠ 1, then ex = x for e ∈ E(S) and all x ∈ X. So g^{e}(x) = g(x) for all x ∈ X, which means that g^{e} = g.

Theorem 10. Let S and T be two monoids; S acts on a set X from the left. Then the wreath product Sw_{x}T is a strongly π-inverse monoid iff

(1) S and T are strongly π-inverse monoids,

(2) for each subset T′ of T with |T′| ≤ |X| there exists m ∈ ℕ such that (t′)^{m} ∈ Reg T′ for all t′ ∈ T′,

(3) |T | = 1 or ex = x for every e ∈ E(S) and all x ∈ X, and

(4) for each x ∈ S and g ∈ T^{X} there exists m ∈ ℕ such that s^{m} ∈ Reg S and g^{s(m)}(x) ∈ Reg T for all x ∈ X, where g^{s(m)} = g^{s^{m−1}} ⋯ g^{s}g ∈ T^{X}.

Proof. It is easy to see that g^{s(m)} ∈ Reg T^{X} iff g^{s(m)}(x) ∈ Reg T for all x ∈ X. Thus Theorem 10 is an immediate consequence of Lemma 8, Lemma 9, and Theorem 2.

Recall that the standard wreath product SwT of two monoids is formed by the left regular representation of S on itself, and we have

Theorem 11. The standard wreath product SwT of two monoids S and T is a strongly π-inverse monoid iff

(1) both S and T are strongly π-inverse monoids,

(2) for each subset T′ of T with |T′| ≤ |S| there exists m ∈ ℕ such that (t′)^{m} ∈ Reg T for all t′ ∈ T′,

(3) S is a group or |T | = 1, and

(4) for every s ∈ S and g ∈ T^{S} there exists m ∈ ℕ such that g^{s(m)}(x) ∈ Reg T for all x ∈ S.

Acknowledgement

The authors would like to thank Professor T. Saito for his careful reading of the paper and his constructive suggestions.

References

1. S. Bogdanowić, Semigroups with a system of subsemigroups. University of Novy Sad, Institute of Mathematics, Novy Sad, 1985.

2. J. M. Howie, Introduction to semigroup theory. Academic Press, London, New York, San Francisco, 1972.

3. W. R. Nico, On the regularity of semidirect products. J. Algebra 80(1983), 29–36.

4. T. Saito, Orthodox semidirect products and wreath products of monoids. Semigroup Forum 38(1989), 347–354.

(Received 17.05.1994)

Authors’ address: Department of Mathematics Shandong Normal University Jinan, P. R. China