An algebra $A$ is called hopfian if it is not isomorphic to a proper homomorphic image of itself (an equivalent definition: every surjective endomorphism $\alpha :A\to A$ is injective (and therefore it is automorphism)). A conception of hopfness (and dual notion of co-hopfness) appeared in the group theory (see ( [1] , v. 2, Section 15), [2] - [4] ). Abelian groups which are hopfian were also investigated in [5] . There are a lot of articles where hopfness and co-hopfness in modules over rings and other algebraic structures are considered.

The hopfness is a finiteness condition, since all the finite objects are hopfian. The hopfness is a weaker condition as the noetherness (maximal condition on congruences). Immediate position between hopfian and noetherian algebras occupy the strongly hopfian algebras. An algebra $A$ is called strongly hopfian if, for every endomorphism $\phi :A\to A$, a chain of kernels $\ker \phi \subseteq \ker {\phi }^2\subseteq \cdots$ is stabilized, i.e. $\ker {\phi }^{n+1}=\ker {\phi }^n$ for some $n$.

Let $S$ be a semigroup. A set $X$ is called an act over a semigroup $S$ if a mapping $X\times S\to X$ is given such that $x\left(st\right)=\left(xs\right)t$ for any $x\in X$ and $s,t\in S$ [6] [7] . The act is an algebraic model of automaton (there $X$ is a set of states and $S$ is a semigroup of input signals). Besides, an act is an unary algebra, i.e. an algebra with unary operations.

The hopfian acts over semigroups were considered in [8] - [11] . V. K. Kartashov proved ( [12] , Thm. 2) that every finitely generated commutative act is hopfian (it is some reformulation).

In the present work, we find the necessary and sufficient conditions to be strongly hopfian for a coproduct of the connected acts over arbitrary semigroup. Using this result we characterize strongly hopfian unitary acts over a group. As a consequence we obtain the fact that a coproduct of a finite number of acts is strongly hopfian if and only if each of them is strongly hopfian. Authors do not know whether this statement is true for the hopfness.

Non-defined terms from the theory of acts can be found in [6] [7] , semigroup theory in [13] , universal algebra in [14] .

For a mapping $\phi :A\to B$, a relation $\ker \phi =\left\{\left(a,a^{\prime }\right)|\phi \left(a\right)=\phi \left(a^{\prime }\right)\right\}$ is called a kernel of $\phi$. If $\phi$ is a homomorphism of algebras then $\ker \phi$ is a congruence. For any set $X$ we put ${\Delta }_X=\left\{\left(x,x\right)|[\in X\right\}$ (the equality relation on $X$).

An act $X$ over a semigroup $S$ is called unitary if $S$ has a unity $e$ and $xe=x$ for any $x\in X$.

A coproduct of universal algebras is a notion dual to the direct product. In case of acts, a coproduct ${\displaystyle {\coprod }_{i\in I}{X}_i}$ of acts $X_i$ ( $i\in I$) is a disjoint union of this acts (or isomorphic copies of them).

For an act $X$ over a semigroup $S$, we can consider a graph where $X$ is a set of vertices and $\left\{\left(x,xs\right)|x\in X,s\in S\right\}$ is a set of edges. An act $X$ is called connected if the graph is connected (in the usual sense, i.e. as a non-oriented graph). It is clear that any graph is a coproduct of connected acts (its connected components).

Let $G$ be a group and $H$ be its subgroup, not necessarily normal. By $G/H$ we will denote a set of right cosets $Hg$ for $g\in G$. Remark that $G/H$ is not necessarily a group but it is an act over the group $G$ where a multiplication is defined by the rule $Hg\cdot g^{\prime }=Hg{g}^{\prime }$. The act $G/H$ is unitary cyclic (even a simple) act over $G$. The following facts can be established straightforwardly.

Fact 1. An act $X$ over a group $G$ is unitary cyclic if and only if $X$ is isomorphic to an act $G/H$ for some subgroup $H$ of $G$.

Fact 2. An act $X$ over a group $G$ is unitary if and only if $X\cong {\displaystyle {\coprod }_{i\in I}{X}_i}$ where $X_i\cong G/H_i$ for some (not necessarily distinct) subgroups $H_i$ of $G$.

Fact 3. There exists a homomorphism $G/H\to G/H^{\prime }$ of acts over $G$ if and only if $H\subseteq a^{-1}Ha$ for some $a\in G$ This homomorphism is always surjective (if it exists).

Let $X={\displaystyle {\coprod }_{i\in I}{X}_i}$ be a coproduct of acts over a semigroup $S$. Define a binary relation $\preccurlyeq$ on the index set $I$ as follows:

$i\preccurlyeq j\iff \text{there exists a homomorphism}{X}_i\to X_j.$

Let $X$ be an act over a semigroup $S$ and $\text{End}X$ be its endomorphism semigroup. For $\phi \in \text{End}X$ we consider ${\phi }^0={\text{id}}_X$ (identical automorphism). If $X$ is strongly hopfian then

$\forall \phi \in \text{End}X\exists k\ge 0\ker {\phi }^k=\ker {\phi }^{k+1}.$

Therefore we can put $l\left(\phi \right)=\inf \left\{k\ge 0|\ker {\phi }^k=\ker {\phi }^{k+1}\right\}$. Further, we put $\text{lh}\left(X\right)=\text{sup}\left\{l\left(\phi \right)|\phi \in \text{End}X\right\}$ (hopfian length of $X$). Note that the equality $\text{lh}\left(X\right)=\infty$ is possible even if $X$ is strongly hopfian.

Let an act $X$ be strongly hopfian. For every endomorphism $\phi \in \text{End}X$ we define a length $l\left(\phi \right)=\min \left\{k|\ker {\phi }^k=\ker {\phi }^{k+1}\right\}$. We consider that ${\phi }^0={\text{Id}}_X$ is an identical automorphism. And a hopfian length of $X$ we call $\text{lh}X=\text{sup}\left\{l\left(\phi \right)|\phi \in \text{End}X\right\}$. If $\text{sup}\left\{l\left(\phi \right)|\phi \in \text{End}X\right\}$ does not exist then we say that $\text{lh}X=\infty$.

Let $X={\displaystyle {\coprod }_{i\in I}{X}_i}$ and $Y={\displaystyle {\coprod }_{j\in J}{Y}_j}$ be acts over a semigroup $S$ and $X_i,Y_j$ be their connected components. Further, let $\phi :X\to Y$ be a homomorphism. Consider any $X_i$. As $X_i$ is connected, then $\phi \left(X_i\right)$ is also connected, therefore $\phi \left(X_i\right)\subseteq \phi \left(Y_j\right)$ for some $j\in J$. So we have a mapping $\bar{\phi }:I\to J$. Thus,

$\bar{\phi }\left(i\right)=j\iff \phi \left(X_i\right)\subseteq Y\left(j\right).$

For $i,j\in I$ we put $i\underset{\_}{\prec }j$ if there exists a homomorphism of acts $X_i\to X_j$. Clearly, the relation $\underset{\_}{\prec }$ is reflexive and transitive, i.e. it is a quasi-order.

Theorem 1. Let $X={\displaystyle {\coprod }_{i\in I}{X}_i}$ be a coproduct of the connected acts over a semigroup $S$. Then $X$ is strongly hopfian if and only if the following condition hold:

(i) $X_i$ is strongly hopfian for any $i\in I$;

(ii) a set $J=\left\{i|\text{lh}\left(X_i\right)=\infty \right\}$ is finite;

(iii) there exists a natural number $L$ such that $\text{lh}\left(X_i\right)\le L$ for $i\in I\backslash J$;

(iv) there exists a natural number $K$ such that $k\le K$ for any chain $i_1\underset{\_}{\prec }{i}_2\underset{\_}{\prec }\cdots \underset{\_}{\prec }{i}_k$ of the distinct elements of $I$.

Proof. Necessity. Let $X$ be strongly hopfian. Suppose that (i) does not hold. Then $X_i$ is not strongly hopfian for some $i\in I$. Hence there is $\alpha \in \text{End}{X}_i$ such that $\ker {\alpha }^k\ne \ker {\alpha }^{k+1}$ for all $k$. Put

$\phi \left(x\right)=\{\begin{array}{ll}\alpha \left(x\right)\hfill & \text{if}x\in X_i,\hfill \\ x\hfill & \text{if}x\in X_j\text{for some}j\ne i.\hfill \end{array}$

Obviously, $\ker {\phi }^k\ne \ker {\phi }^{k+1}$ for all $k$. As $\phi$ is an endomorphism, then $X$ is not strongly hopfian which is false.

Suppose (ii) is not fulfilled. Then the set $J$ is infinite. Therefore, there exists a sequence $j_1,j_2,\cdots$ of distinct elements of $J$. For any $j_t$ we take ${\phi }_t\in \text{End}{X}_{j_t}$ such that $\ker {\phi }_t^{t+1}\supset \ker {\phi }_t^t$. Define an endomorphism $\phi \in \text{End}X$ by the rule

$\phi \left(x\right)=\{\begin{array}{ll}{\phi }_t\left(x\right)\hfill & \text{if}x\in X_{j_t},\hfill \\ x\hfill & \text{if}x\in X_i\text{where}i\notin \left\{j_1,j_2,\cdots \right\}.\hfill \end{array}$

It is seen that $\ker {\phi }^k\ne \ker {\phi }^{k+1}$ for all $k\in \mathbb{N}$, and it contradicts with the fact that $X$ is strongly hopfian.

Suppose (iii) is not fulfilled. Then there are $i_1,i_2,\cdots \in I$ such that $\text{lh}\left(X_{i_t}\right)\ge t$ for $t=1,2,\cdots$. Therefore, there exist ${\phi }_t\in \text{End}{X}_{i_t}$ such that $\ker {\phi }_t^{t-1}\subset \ker {\phi }_t^t$. Put $Y={\displaystyle {\coprod }_{t=1}^{\infty }{X}_{i_t}}$. Construct a mapping $\phi :X\to X$ as follows:

$\phi \left(x\right)=\{\begin{array}{ll}{\phi }_t\left(x\right)\hfill & \text{if}x\in X_{i_t}\text{for}\text{some}t,\hfill \\ x\hfill & \text{if}x\notin Y.\hfill \end{array}$

Obviously, $\phi$ is an endomorphism of the act $X$ and we have the strong inclusions $\ker {\phi }^{t-1}\subset \ker {\phi }^t$ for all $t$. It contradicts to the assumption that $X$ is strongly hopfian.

Suppose (iv) is not fulfilled. Construct a sequence of chains ${\text{Γ}}_1,{\text{Γ}}_2,\cdots$ in $I$ such that $\left|{\text{Γ}}_t\right|=t$ for all $t$ and ${\text{Γ}}_i\cap {\text{Γ}}_j=\varnothing$ for $i\ne j$. Take ${\text{Γ}}_1=\left\{i\right\}$ where $i\in I$ is arbitrary. Let ${\text{Γ}}_1,\cdots ,{\text{Γ}}_m$ be constructed. Take a chain $\text{Γ}$ of $k$ elements where $k\ge 1+2+\cdots +\left(m+1\right)$. Put $\text{<msup> Γ <mo> ′ </mo> </msup>}=\text{Γ}\backslash \left({\text{Γ}}_1\cup \cdots \cup {\text{Γ}}_m\right)$. Clearly $\left|\text{<msup> Γ <mo> ′ </mo> </msup>}\right|\ge m+1$. Deleting (if it is necessary) from the chain $\text{<msup> Γ <mo> ′ </mo> </msup>}$ some elements we obtain a chain ${\text{Γ}}_{m+1}$ of $m+1$ elements, and ${\text{Γ}}_{m+1}\cap \left({\text{Γ}}_1\cup \cdots \cup {\text{Γ}}_m\right)=\varnothing$.

For every chain ${\text{Γ}}_t=\left\{i_1\underset{\_}{\prec }{i}_2\underset{\_}{\prec }\cdots \underset{\_}{\prec }{i}_t\right\}$ we define a subact $X^{\left(t\right)}=X_{i_1}\bigsqcup X_{i_2}\bigsqcup \cdots \bigsqcup X_{i_t}$. By definition of the relation $\underset{\_}{\prec }$, there exist the homomorphisms ${\phi }_1^{\left(t\right)}:X_{i_1}\to X_{i_2}$, ${\phi }_2^{\left(t\right)}:X_{i_2}\to X_{i_3}$, $\cdots$, ${\phi }_{t-1}^{\left(t\right)}:X_{i_{t-1}}\to X_{i_t}$. Construct a homomorphism ${\phi }^{\left(t\right)}$ putting

${\phi }^{\left(t\right)}\left(x\right)=\{\begin{array}{ll}{\phi }_i^{\left(t\right)}\left(x\right)\hfill & \text{if}x\in X_i\text{and}i<t,\hfill \\ x\hfill & \text{if}x\in X_t.\hfill \end{array}$

Remark that $\ker {\phi }^{\left(t\right)}\cap \left(X_i\times X_t\right)=\varnothing$ for $i<t$, however $\ker {\left({\phi }^{\left(t\right)}\right)}^2\cap \left(X_{t-1}\times X_t\right)\ne \varnothing$ but $\text{ker}{\left({\phi }^{\left(t\right)}\right)}^2\cap \left(X_i\times X_t\right)=\varnothing$ only for $i<t-1$ and so on. Thus, we have $\ker {\phi }^{\left(t\right)}\subset \ker {\left({\phi }^{\left(t\right)}\right)}^2\subset \cdots \subset \ker {\left({\phi }^{\left(t\right)}\right)}^t$. We can present the act $X$ in the view $X={\displaystyle {\coprod }_{t=1}^{\infty }{X}^{\left(t\right)}}\bigsqcup X^{\prime }$ where $X^{\prime }=X\backslash {\displaystyle {\cup }_{t=1}^{\infty }{X}^{\left(t\right)}}$ is a subact of $X$ or the empty set. Define an endomorphism $\phi :X\to X$ as follows:

$\phi \left(x\right)=\{\begin{array}{ll}{\phi }^{\left(t\right)}\left(x\right)\hfill & \text{if}x\in X^{\left(t\right)}\text{for}\text{some}t,\hfill \\ x\hfill & \text{if}x\in X^{\prime }.\hfill \end{array}$

Since the lengths of the chains is not bounded from above that $\ker {\phi }^t$ are distinguish, which contradicts the strongly hopfness of $X$.

Sufficiency. Assume that the conditions (i)-(iv) hold and $\phi \in \text{End}X$. As $X_i$ are connected then $\phi$ induces a mapping $\bar{\phi }:I\to I$ such that $\bar{\phi }\left(i\right)=j\iff \phi \left(X_i\right)\subseteq X_j$. The set $I$ with the unary operation $\bar{\phi }$ is a unar (in another terminology: monounary algebra, see [15] ).

Because of the condition (iv), ${\bar{\phi }}^K\left(i\right)$ lies in a cycle for any $i\in I$. As the lengths of cycles are less or equal to $K$ then for any cycle $C$ and any $i\in C$ we have ${\bar{\phi }}^{K!}\left(i\right)=\left(i\right)$. Because of (ii) there are finitely many members of infinite length, let they be $X_{j_1},\cdots ,X_{j_s}$. Put $\psi ={\phi }^{K!}$. Select from $X_{j_1},\cdots ,X_{j_s}$ the members which are invariant with respect to $\psi$ (i.e. $\psi \left(X_i\right)\subseteq X_i$). Without loss of generality we may consider that $X_{j_1},\cdots ,X_{j_t}$ are invariant but $X_{j_{t+1}},\cdots ,X_{j_s}$ are not (here $0\le t\le s$). Put ${\psi }_i={\psi |}_{X_{j_i}}$ for $i=1,\cdots ,t$. By the condition (i) each endomorphism of $X_i$ has a finite length. Therefore, we may put $L_0=\text{max}\left\{L,l\left({\psi }_1\right),\cdots ,l\left({\psi }_t\right)\right\}$.

Let $\left(x,y\right)\in \ker {\phi }^{K+2L_{0}K!}$. Then $\left({\phi }^K\left(x\right),{\phi }^K\left(y\right)\right)\in \ker {\phi }^{2L_{0}K!}$. Put $x^{\prime }={\phi }^K\left(x\right)$, $y^{\prime }={\phi }^K\left(y\right)$. Obviously, $x^{\prime }\in X_i$, $y^{\prime }\in X_j$ where $i,j$ belong to a cycle of the unar $I$ Denote this cycle by $C$. Thus $i,j\in C$.

We have ${\psi }^{2L_0}\left(x^{\prime }\right)={\psi }^{2L_0}\left(y^{\prime }\right)$. As the lengths of cycles are less or equal to $K$ then $\bar{\psi }={\bar{\phi }}^{K!}$ is identical mapping on $C$, i.e. $\psi \left(X_i\right)\subseteq X_i$ and $\psi \left(X_j\right)\subseteq X_j$. As ${\psi }^{2L_0}\left(x^{\prime }\right)={\psi }^{2L_0}\left(y^{\prime }\right)$ then $i=j$, therefore $x^{\prime },y^{\prime }\in X_i$.

If $\text{lh}\left(X_i\right)<\infty$, then by (iii) $\text{lh}\left(X_i\right)\le L\le L_0$. If $\text{lh}\left(X_i\right)=\infty$, then $X_i=X_{j_u}$ for some $u\le t$, and therefore $l(\psi |X_i)=l\left({\psi }_u\right)\le L_0$. In both cases

$\ker {\left({\psi |}_{X_i}\right)}^{L_0}=\ker {\left({\psi |}_{X_i}\right)}^{2L_0}$. It follows that ${\psi }^{L_0}\left(x^{\prime }\right)={\psi }^{L_0}\left(y^{\prime }\right)$. It means that ${\phi }^{L_{0}K!}\left(x^{\prime }\right)={\phi }^{L_{0}K!}\left(y^{\prime }\right)$. Therefore ${\phi }^{K+L_{0}K!}\left(x\right)={\phi }^{K+L_{0}K!}\left(y\right)$.

Thus $\left(x,y\right)\in \ker {\phi }^{K+L_{0}K!}$. We proved that $\ker {\phi }^{K+L_{0}K!}\subseteq \ker {\phi }^{K+2L_{0}K!}$. Put $M=K+L_{0}K!$. We have $\ker {\phi }^M\subseteq \ker {\phi }^{M+L_{0}K!}\subseteq \ker {\phi }^{M+1}\subseteq \ker {\phi }^M$. It implies $\ker {\phi }^{M+1}=\ker {\phi }^M$. Thus $X$ is strongly hopfian. $\square$

Corollary. A coproduct $X_1\bigsqcup \cdots \bigsqcup X_n$ of a finite number of acts over a semigroup is strongly hopfian if and only if each of $X_i$ is strongly hopfian.

Proof. Decompose each $X_i$ into a coproduct of connected components and apply Theorem 1. $\square$

Remark. The authors do not know whether a similar statement is true for the hopfness.

Now let us move on to the unitary acts over the groups. Let $G$ be a group and $H$ be its subgroup.

Lemma 1 ( [8] , Lemma 1). A unitary cyclic act $G/H$ is hopfian if and only if the following condition holds:

$\forall a\in GH\subseteq a^{-1}Ha\iff H=a^{-1}Ha.$ $\ast$

Lemma 2. A unitary cyclic act $G/H$ over a group $G$ is strongly hophian if and only if it is hopfian.

Proof. We need to proof only that every hopfian act $G/H$ is strongly hopfian. Let $G/H$ is hopfian and $\phi \in \text{End}\left(G/H\right)$. As any endoomorphism $\phi :G/H\to G/H$ is subjective and $G/H$ is hopfian, then $\phi$ is also injective. Then $\ker \phi ={\Delta }_{G/H}$. Also $\ker {\phi }^n={\Delta }_{G/H}$ for all $n$. Therefore $G/H$ is strongly hopfian. $\square$

Remark. If $G/H$ is hopfian then $\text{lh}\left(G/H\right)=0$.

The authors proved in ( [8] , Thm. 1) the following statement.

Fact 4. A unitary act $X={\displaystyle {\coprod }_{i\in I}\left(G/H_i\right)}$ over a group $G$ is hopfian if and only if each of $H_i$ satisfies ( $\ast$) and there is no an infinity chain $i_1\underset{\_}{\prec }{i}_2\underset{\_}{\prec }\cdots$ of distinct elements of $I$.

A similar statement for the strong hopfness is so.

Theorem 2. A unitary act $X={\displaystyle {\coprod }_{i\in I}\left(G/H_i\right)}$ over a group $G$ is strongly hopfian if and only if the following conditions hold:

(v) $\forall i\in I$ $\forall a\in G$ $H_i\subseteq a^{-1}{H}_{i}a\iff H_i=a^{-1}{H}_{i}a$;

(vi) there exists a natural number $K$ such that for any chain $i_1\underset{\_}{\prec }{i}_2\underset{\_}{\prec }\cdots \underset{\_}{\prec }{i}_k$ of distinct elements of $I$ an inequality $k\le K$ is true.

Proof. The necessity follows from Theorem 1 in [8] (Fact 4) and Theorem 1. Prove the sufficiency. Let (v) and (vi) be satisfied. As (v) is fulfilled then by Lemma 2 the condition (i) of Theorem 1 holds. The condition (ii) holds since $J=\varnothing$. The condition (iii) holds since $\text{lh}\left(G/H_i\right)=0$ (it follows from (v)). Finally the condition (iv) folds since it coincides with (vi). $\square$

Here we give two examples: 1) a hopfian but not strongly hopfian act; 2) a strongly hopfian act of infinity hopfian length. Both acts are unitary acts over group.

Example 1. Let $p$ be a prime number and $G={\mathbb{Z}}_{p^{\infty }}$ be a quasi-cyclic group (a union of the ascending sequence of groups $H_n={\mathbb{Z}}_{p^n}$: cyclic groups of order $p^n$). Then a coproduct $X={\displaystyle {\coprod }_{n\in \mathbb{N}}G/H_n}$ is hopfian but not strongly hopfian act over the group $G$.

Proof. Really, here $I=\mathbb{N}$ (set of natural number). We have an ascending sequence $1\underset{\_}{\prec }2\underset{\_}{\prec }3\underset{\_}{\prec }\cdots$, therefore $X$ is not strongly hopfian by Theorem 1. In the same time $X$ is hopfian since ( $\ast$) holds for any abelian group and its subgroup and there are no a descending sequence of distinct elements of $I$.

Remark. Although the quotient groups $G/H_n$ in the Example 1 are isomorphic to one another (and they are isomorphic to the group $G$), $G/H_n$ and $G/H_m$ are not isomorphic for $m\ne n$ as the acts over the group $G$. It is known that $G/H\cong G/H^{\prime }$ if and only if subgroups $H$ and $H^{\prime }$ are conjugated.

Let us finish the article with another example. Note that any semigroup $S$ is an act over itself. Denote this act by $S_S$. The subacts of this act are exactly the right ideals of the semigroup $S$, and the congruences are exactly the right congruences of the semigroup.

Example 2. Let $T=\left(0,1\right)$ be a semigroup with the usual multiplication. Clearly, $I=\left(0,1/2\right]$ is an ideal of $T$. Let $S=T/I$ be a Rees quotient semigroup. We may think that $S=\left\{0\right\}\cup \left(1/2;1\right)$ with a multiplication

$x\ast y=\{\begin{array}{ll}xy\hfill & \text{if}xy>1/2,\hfill \\ 0\hfill & \text{if}xy\le 1/2.\hfill \end{array}$

Then the act $S_S$ is strongly hopfian with infinite hopfian length. We will provide a scheme of the proof:

(a) to prove that the endomorphisms of the act $S_S$ are exactly the mappings of view ${\phi }_a\left(x\right)=ax$ for $a\le 1$;

(b) to note that ${\phi }_1\left(x\right)$ is identical automorphism and hence $\ker {\phi }_1^k={\Delta }_X$ for every $k$;

(c) to note that ${\phi }_a\left(x\right)$ is nilpotent for $a<1$, i.e. ${\phi }_a^k=0$ for some $k$ (namely, ${\phi }_a^k=0\iff a^k\le 1/2$); therefore $\ker {\phi }_a^k=S\times S$ for $k\ge -\ln 2/\ln a$;

(d) it follows from (c) that $l\left({\phi }_a\right)=\left[-\ln 2/\ln a\right]$;

(e) to note that ${\lim }_{a\to 1}l\left({\phi }_a\right)=+\infty$.