Proof. (1) $\Longrightarrow $ (2). Let G be a subgroup of a slender semigroup S and let h be a homomorphism of $\prod _C$ into G. Then, h is a homomorphism of $\prod _C$ into S. Since S is slender, $(\delta _n)h$ is the identity element of G for almost all positive integers n. Thus, G is a slender group. Thus, assertion $(2)$ is satisfied. Since $(2)$ implies $(3)$ , it remains to prove that $(3)$ implies $(1)$ .

(3) $\Longrightarrow $ (1). Let S be a semigroup containing at least one idempotent element such that every maximal subgroup of S is slender. Let h be a homomorphism of $\prod _C$ into S. Since $(\prod _C)h$ is a subgroup of S, [Reference Clifford and Preston2, Theorem 1.11] implies that there is an idempotent element e of S such that $(\prod _C)h$ is contained by the maximal subgroup $H_e$ of S. Thus, h is a homomorphism of $\prod _C$ into the group $H_e$ . By the assumption, $H_e$ is a slender group and, hence, $(\delta _n)h$ equals the identity element e of $H_e$ for almost all positive integers n. Thus, S is a slender semigroup.

Proof. This is evident by Theorem 3.2, because the one-element group is slender and the elements of a band B are maximal subgroups of B.

Proof. If S is a slender semigroup, then it follows, by the definition of slenderness, that the subsemigroups $S_{\alpha }$ are slender for all $\alpha \in Y$ .

Conversely, assume that the subsemigroups $S_{\alpha }$ are slender for all $\alpha \in Y$ . Let $\eta $ denote the semilattice congruence on S, whose classes are the subsemigroups $S_{\alpha }$ ( $\alpha \in Y$ ). Let h be a homomorphism of $\prod _C$ into S. Then, $(\prod _C)h$ is a subgroup of S. It is clear that $\eta '$ is a semilattice congruence on $(\prod _C)h$ , where $\eta '$ is defined in the following way: for elements $a, b\in (\prod _C)h$ , $(a,b)\in \eta '$ if and only if $(a, b)\in \eta $ . Since every group is semilattice indecomposable, $\eta '$ is the universal relation on $(\prod _C)h$ and so $(\prod _C)h\subseteq S_{\alpha _0}$ for some $\alpha _0\in Y$ . Since $S_{\alpha _0}$ is a slender subsemigroup of S, $(\delta _n)h$ equals an idempotent element of $S_{\alpha _0}$ (and so of S) for almost all indices n. Thus, S is a slender semigroup.

Proof. By [Reference Howie7, Theorem 4.2.1], a semigroup is a Clifford semigroup if and only if it is a semilattice Y of groups $G_{\alpha }$ . Then, by Theorem 3.4, a Clifford semigroup is slender if and only if the subgroups $G_{\alpha }$ are slender for every $\alpha \in Y$ .

Proof. Let ${\mathcal M}(G; I, \Lambda; P)$ be a Rees matrix semigroup over a group G. By [Reference Clifford and Preston2, Lemma 3.2 (iv)], the set $H_{i\lambda }=\{ i\}\times G\times \{ \lambda \}$ ( $i\in I, \lambda \in \Lambda $ ) is a maximal subgroup of ${\mathcal M}(G; I, \Lambda; P)$ . Then, by Theorem 3.2, ${\mathcal M}(G; I, \Lambda; P)$ is slender if and only if all the subgroups $H_{i\lambda }$ ( $i\in I, \lambda \in \Lambda $ ) are slender. Using also [Reference Clifford and Preston2, Lemma 3.2 (iv)], the mapping $a\mapsto (i, ap^{-1}_{\lambda i}, \lambda )$ is an isomorphism of G upon $H_{i\lambda }$ for all $i\in I$ and $\lambda \in \Lambda $ . Thus, the Rees matrix semigroup ${\mathcal M}(G; I, \Lambda; P)$ is slender if and only if G is a slender group.

Proof. By [Reference Howie7, Theorem 3.3.1], a semigroup is completely simple if and only if it is isomorphic to a Rees matrix semigroup over a group. Thus, the assertion follows from Theorem 3.7.

Proof. It is clear that a rectangular group $I\times \Lambda \times G$ is a Rees matrix semigroup ${\mathcal M}(G; I, \Lambda ;P)$ over the group G such that every element of the sandwich matrix P equals the identity element of G. Thus, the assertion of the theorem follows from Theorem 3.7.

Proof. Assume that S is a slender semigroup. Let h be a homomorphism of the rectangular abelian group ${\mathcal RG}^{\infty }_C$ into S. We show that $(\delta _n)h$ is an idempotent element of S for almost all indices n. If $n\neq 3k$ , then $\delta _n$ is an idempotent element of ${\mathcal RG}^{\infty }_C$ and, hence, $(\delta _n)h$ is an idempotent element of S, because a homomorphic image of an idempotent element is idempotent. By the above, it is sufficient to show that $(\delta _{3k})h$ is an idempotent element of S for almost all positive integers k. Let $\xi '$ be the following mapping of $\prod _C$ into ${\mathcal RG}^{\infty }_C$ :

$$ \begin{align*}\xi ': (x_1, x_2, x_3, x_4, x_5, x_6, \ldots )\mapsto (e, e, x_3, e, e, x_6, \ldots ).\end{align*} $$

Since $\xi '$ is an extension of $\xi $ defined by (3.1), $\xi '$ is a homomorphism of $\prod _C$ into ${\mathcal RG}^{\infty }_C$ . Thus, $\xi 'h$ is a homomorphism of $\prod _C$ into S. Since S is a slender semigroup, $(\delta _{3k})\xi 'h$ is an idempotent element of S for almost all positive integers k. Since $(\delta _{3k})\xi '=\delta _{3k}$ ,

$$ \begin{align*}(\delta _{3k})\xi 'h=((\delta _{3k})\xi ')h=(\delta _{3k})h\end{align*} $$

and so $(\delta _{3k})h$ is an idempotent element of S for almost all positive integers k.

To prove the converse, assume that every homomorphism of the rectangular abelian group ${\mathcal RG}^{\infty }_C$ into a semigroup S sends $\delta _n$ into an idempotent of S for almost all indices n. We show that S is a slender semigroup. Let h be a homomorphism of the group $\prod _C$ into S. We show that $(\delta _n)h$ is an idempotent element of S for almost all indices n. Let $\sigma '$ be the following mapping of ${\mathcal RG}^{\infty }_C$ into $\prod _C$ :

$$ \begin{align*}\sigma ': (x_1, x_2, x_3, x_4, x_5, x_6, \ldots )\mapsto (x_3, x_6, \ldots ).\end{align*} $$

Since $\sigma '$ is an extension of $\sigma $ defined by (3.2), $\sigma '$ is a homomorphism of ${\mathcal RG}^{\infty }_C$ into $\prod _C$ , and so $\sigma 'h$ is a homomorphism of ${\mathcal RG}^{\infty }_C$ into S. By the assumption, $(\delta _n)\sigma 'h$ is an idempotent element of S for almost all indices n. Since $(\delta _{3n})\sigma '=\delta _n$ for all positive integers n,

$$ \begin{align*}(\delta _{3n})\sigma 'h=(\delta _n)h,\end{align*} $$

and so $(\delta _n)h$ is an idempotent element of S for almost all indices n. Hence, S is a slender semigroup.

Proof. Assume $f_1, f_2 \in {\mathcal Hom}(H, I\times \Lambda \times G)$ and $x, y\in H$ . Let $(x)f_1=(i, \lambda , a)$ and $(x)f_2=(j, \mu , b)$ , $(y)f_1=(i', \lambda ', a')$ , $(y)f_2=(j', \mu ', b')$ . Then,

$$ \begin{align*} (xy)(f_1f_2) & =(xy)f_1(xy)f_2=(x)f_1(y)f_1(x)f_2(y)f_2 \\ & =(i, \lambda , a)(i', \lambda ', a')(j, \mu , b)(j', \mu ', b')=(i, \mu ', aa'bb') \\ & =(i, \mu ', aba'b')=(i, \lambda , a)(j, \mu , b)(i', \lambda ', a')(j', \mu ', b') \\ & =(x)f_1(x)f_2(y)f_1(y)f_2=(x)(f_1f_2)(y)(f_1f_2). \end{align*} $$

Thus, $f_1f_2\in {\mathcal Hom}(H, I\times \Lambda \times G)$ . Since $(f_1f_2)f_3=f_1(f_2f_3)$ for arbitrary homomorphisms $f_1, f_2, f_3$ of H into $I\times \Lambda \times G$ , it follows that ${\mathcal Hom}(H, I\times \Lambda \times G)$ is a semigroup.

Proof. By Theorem 3.11, ${\mathcal Hom}(H, I\times \Lambda \times G)$ is a semigroup. Since ${\mathcal Hom}(H, G)$ is an abelian group, $I\times \Lambda \times {\mathcal Hom}(H, G)$ is a rectangular abelian group. Let f be an arbitrary element of $I\times \Lambda \times {\cal H}(H,G)$ . Then, there are elements $i\in I$ , $\lambda \in \Lambda $ and $\alpha _f\in {\mathcal Hom}(H, G)$ such that

$$ \begin{align*}f=(i, \lambda , \alpha _f).\end{align*} $$

Let $\Phi _f$ denote the mapping of H into $I\times \Lambda \times G$ defined as follows: for arbitrary $h\in H$ , let

$$ \begin{align*}(h)\Phi _f=(i, \lambda , (h)\alpha _f).\end{align*} $$

Then, for every $h_1, h_2\in H$ ,

$$ \begin{align*} (h_1h_2)\Phi _f & =(i, \lambda , (h_1h_2)\alpha _f)=(i, \lambda , (h_1)\alpha _f(h_2)\alpha _f) \\ & =(i, \lambda , (h_1)\alpha _f)(i, \lambda , (h_2)\alpha _f)=(h_1)\Phi _f(h_2)\Phi _f. \end{align*} $$

Thus,

$$ \begin{align*}\Phi _f\in {\mathcal Hom}(H, I\times \Lambda \times G).\end{align*} $$

We show that $\Phi : f\mapsto \Phi _f$ is an isomorphism which maps $I\times \Lambda \times {\mathcal Hom}(H,G)$ onto ${\mathcal Hom}(H, I\times \Lambda \times G)$ .

First, we show that $\Phi $ is a homomorphism. Let

$$ \begin{align*}f_1=(i, \lambda , \alpha _{f_1}), \ f_2=(j, \mu , \alpha _{f_2})\in I\times \Lambda \times {\mathcal Hom}(H, G)\end{align*} $$

be arbitrary elements. Then, $f_1f_2=(i, \mu , \alpha _{f_1}\alpha _{f_2})$ . For arbitrary $h\in H$ ,

$$ \begin{align*} (h)\Phi _{f_1f_2} & =(h)\Phi _{(i, \mu, \alpha _{f_1}\alpha _{f_2})}=(i, \mu, (h)\alpha _{f_1}\alpha _{f_2}) =(i, \mu, (h)\alpha _{f_1}(h)\alpha _{f_2}) \\ & =(i, \lambda, (h)\alpha _{f_1})(j, \mu, (h)\alpha _{f_2}) =(h)\Phi _{f_1}(h)\Phi _{f_2}=(h)\Phi_{f_1}\Phi _{f_2}, \end{align*} $$

and so

$$ \begin{align*}(f_1f_2)\Phi=(f_1)\Phi (f_2)\Phi.\end{align*} $$

To show that $\Phi $ is injective, assume

$$ \begin{align*}(f_1)\Phi=(f_2)\Phi\end{align*} $$

for elements

$$ \begin{align*}f_1=(i, \lambda , \alpha _{f_1}), \ f_2=(j, \mu , \alpha _{f_2})\in I\times \Lambda \times {\mathcal Hom}(H, G).\end{align*} $$

Then, for all $h\in H$ ,

$$ \begin{align*}(i, \lambda , (h)\alpha _{f_1})=(h)\Phi _{f_1}=(h)\Phi _{f_2}=(j, \mu , (h)\alpha _{f_2})\end{align*} $$

from which it follows that

$$ \begin{align*}i=j, \quad \lambda =\mu, \quad (h)\alpha _{f_1}=(h)\alpha _{f_2},\end{align*} $$

that is,

$$ \begin{align*}f_1=f_2.\end{align*} $$

To show that $\Phi $ is surjective, consider an arbitrary homomorphism $\varphi $ of H into $I\times \Lambda \times G$ . Since $(H)\varphi $ is a subgroup of $I\times \Lambda \times G$ , [Reference Clifford and Preston2, Theorem 1.11] implies that there are elements $i\in I$ and $\lambda \in \Lambda $ such that $(H)\varphi $ is contained by the maximal subgroup $H_{i\lambda }=\{ i\}\times \{ \lambda \}\times G$ of $I\times \Lambda \times G$ . Let

$$ \begin{align*}f=(i, \lambda , \varphi \varphi _{i\lambda}),\end{align*} $$

where $\varphi _{i\lambda }$ is an isomorphism of $\{i\}\times \{\lambda \}\times G$ onto G defined as follows: for arbitrary $(i, \lambda , g)$ ,

$$ \begin{align*}(i, \lambda ,g)\varphi _{i\lambda}=g.\end{align*} $$

Since $\varphi \varphi _{i\lambda }\in {\mathcal Hom}(H,G)$ , we have

$$ \begin{align*} f\in I\times \Lambda \times {\mathcal Hom}(H,G). \end{align*} $$

Let $h\in H$ be an arbitrary element. Then, $(h)\varphi =(i, \lambda , h^*)$ for some $h^*\in G$ . Since

$$ \begin{align*}(h)\Phi _f=(i, \lambda , (h)\varphi \varphi _{i\lambda})=(i, \lambda , (i, \lambda , h^*)\varphi _{i\lambda })=(i, \lambda, h^*)=(h)\varphi ,\end{align*} $$

we see that

$$ \begin{align*}(f)\Phi=\varphi.\end{align*} $$

Thus, $\Phi $ is an isomorphism of the rectangular group $I\times \Lambda \times {\mathcal Hom}(H,G)$ onto ${\mathcal Hom}(H, I\times \Lambda \times G)$ .

Proof. Let $I\times \Lambda \times G$ be a rectangular abelian group and let $G_{\alpha }$ ( $\alpha \in Y$ ) be an arbitrary nonempty set of groups. By Theorem 3.12,

$$ \begin{align*} {\mathcal Hom}\bigg(\!\prod _{\alpha \in Y}G_{\alpha}, I\times \Lambda \times G\bigg)\cong I\times \Lambda \times {\mathcal Hom}\bigg(\!\prod _{\alpha \in Y}G_{\alpha}, G\bigg). \end{align*} $$

If the rectangular abelian group $I\times \Lambda \times G$ is slender, then G is a slender group by Theorem 3.9. It follows from [Reference Göbel6, Theorem 4.3] that

$$ \begin{align*} {\mathcal Hom}\bigg(\!\prod _{\alpha \in Y}G_{\alpha}, G\bigg)=\bigtriangledown {\mathcal Hom}\bigg(\!\prod _{\alpha \in Y}G_{\alpha}, G\bigg). \end{align*} $$

Thus,

$$ \begin{align*} {\mathcal Hom}\bigg(\!\prod _{\alpha \in Y}G_{\alpha}, I\times \Lambda \times G\bigg)\cong I\times \Lambda \times \bigtriangledown {\mathcal Hom}\bigg(\!\prod _{\alpha \in Y}G_{\alpha}, G\bigg). \\[-42pt] \end{align*} $$