Proof. The ‘if’ part. Take any $e,f\in B$ that satisfy the antecedent of the implication in Equation (1-1), that is, $ef=e^2=e$ and $fe=f^2=f$ . Multiplying $ef=e$ by e on the right and using $e^2=e$ , we get $efe=e$ , and similarly, multiplying $fe=f$ by f on the right and using $f^2=f$ , we get $fef=f$ . Hence, f is an inverse of e. However, we have $e=e^2=e^3$ so that e is an inverse of itself. Since e has at most one inverse, we conclude that $e=f$ so that Equation (1-1) holds. By symmetry, the implication in Equation (1-2) holds as well.

The ‘only if’ part. Take any $a\in B$ and suppose that both b and c are inverses of a. Letting $e:=ba$ and $f:=ca$ , we have $ef=baca=ba=e$ , $e^2=baba=e$ , and similarly, $fe=caba=ca=f$ , $f^2=caca=f$ . Since $(B,\cdot )$ satisfies the implication in Equation (1-1), we get $e=f$ , that is, $ba=ca$ . By symmetry, the implication in Equation (1-2) ensures $ab=ac$ . Now we have $b=bab=cab=cac=c$ .

Proof. Let $a\in \langle E(S)\rangle $ be a regular element and b its inverse. The FitzGerald trick [Reference FitzGerald9, Lemma 1] shows that $b\in \langle E(S)\rangle $ as well. Since $a=aba$ and $b=bab$ generate the same ideal in $(\langle E(S)\rangle ,\cdot )$ , Proposition 2.3 implies $a=b$ . Thus, $a=a^3$ , whence a and $a^2$ also generate the same ideal in $(\langle E(S)\rangle ,\cdot )$ . Now Proposition 2.3 implies $a=a^2\in E(S)$ .

Proof. If the semigroup $(B,\cdot )$ is represented as $\mathcal {B}_{G,I}$ for some group $(G,\cdot )$ and some nonempty set I, then each nonzero idempotent is of the form $(i,e,i)$ , where $i\in I$ and e is the identity element of the group $\mathcal {G}$ . Take an arbitrary element $b=(\ell ,g,r)\in I\times G\times I$ . Then $b=(\ell ,e,\ell )b$ , whence $fb=f(\ell ,e,\ell )b$ for each $f\in E(S)$ . The product $f(\ell ,e,\ell )$ lies in B and so it is regular. By Lemma 2.4, $f(\ell ,e,\ell )$ is an idempotent. If ${f(\ell ,e,\ell )\ne 0}$ , then $f(\ell ,e,\ell )=(\,j,e,j)$ for some $j\in I$ . Multiplying the equality through by $(\,j,e,j)$ on the right yields $f(\ell ,e,\ell )(\,j,e,j)=(\,j,e,j)$ , whence $j=\ell $ . We conclude that either $f(\ell ,e,\ell )=(\ell ,e,\ell )$ or $f(\ell ,e,\ell )=0$ , whence either $fb=b$ or $fb=0$ .

Proof. Let Equation (2-1) be a Brandt series in a semigroup $(S,\cdot )$ . Take any regular element $a\in ~S$ and let $j\in \{0,1,\ldots ,h\}$ be the least number with $a\in S_j$ . Recall that each inverse of a generates the same ideal of $(S,\cdot )$ as does a. Therefore, if $j=0$ , all inverses of a lie in the least ideal of $(S,\cdot )$ which is a group. Then all inverses of a are equal since they coincide with the group inverse of a in the group $(S_0,\cdot )$ . If $j>0$ , we have $a\in S_j\setminus S_{j-1}$ , and all inverses of a also lie in $S_j\setminus S_{j-1}$ . Their images in the Rees quotient $(S_j/S_{j-1},\cdot )$ are inverses of the image of a, and hence, they coincide as the quotient must be an inverse semigroup. Since nonzero elements of $S_j/S_{j-1}$ are in a 1–1 correspondence with the elements of $S_j\setminus S_{j-1}$ , all inverses of a are equal also in this case. Now Lemma 2.2 shows that $(S,\cdot )$ is a block-group.

Proof. Lemma 2.1 implies that for any $a\in B_{G,I}$ , the power $a^{2m}$ lies in a subgroup of $\mathcal {B}_{G,I}$ . Since $\mathbf u_{0,k,m}=(x_1x_2\cdots x_k)^{2m}$ and $\mathbf u_{1,k,m}=(x_1x_2\cdots x_{k+1})^{2m}$ , the claim holds for $n\in \{0,1\}$ .

For the rest of the proof, assume $n\ge 2$ . If $\tau (\mathbf u_{n,k,m})=0$ , there is nothing to prove. Let $\tau (\mathbf u_{n,k,m})\ne 0$ . Then all elements $\tau (x_i)$ , $i=1,2,\ldots ,n+k$ , must lie in the set $I \times G \times I$ of nonzero elements of $\mathcal {B}_{G,I}$ . Hence, $\tau (x_i)=(\ell _i,g_i,r_i)$ for some $\ell _i,r_i\in I$ and $g_i\in G$ .

First, consider the case $k=0$ . Since $\mathbf u_{n,0,m}=x_1x_2\cdots x_n\,(x_nx_{n-1}\cdots x_1)^{2m-1}$ , we have $\tau (\mathbf u_{n,0,m})=(\ell _1,g,r_1)$ for some $g\in G$ . By Lemma 2.1, we have to show that $\ell _1=r_1$ . We verify that $\ell _j=r_j$ for all $j=n,n-1,\ldots ,1$ by backward induction. Since $x_n^2$ occurs as a factor in the word $\mathbf u_{n,0,m}$ , we must have $\tau (x_n^2)\ne 0$ , whence $\ell _n=r_n$ . If $j>1$ , both $x_{j-1}x_j$ and $x_jx_{j-1}$ occur as factors in $\mathbf u_{n,0,m}$ . We then have

$$ \begin{align*} r_{j-1}&=\ell_j\quad\kern8.5pt\text{since } \tau(x_{j-1}x_j)\ne 0,\\&=r_j\quad\kern8.8pt\text{by the induction assumption,}\\&=\ell_{j-1}\quad \text{since } \tau(x_jx_{j-1})\ne 0. \end{align*} $$

Now, let $k>0$ . Since

$$ \begin{align*} \mathbf u_{n,k,m}= x_1x_2\cdots x_{n+k}\,(x_nx_{n-1}\cdots x_1\cdot x_{n+1}x_{n+2}\cdots x_{n+k})^{2m-1}, \end{align*} $$

we have $\tau (\mathbf u_{n,k,m})=(\ell _{1},g,r_{n+k})$ for some $g\in G$ . Here we have to show that $\ell _1=r_{n+k}$ . Since $\tau (x_1x_2\cdots x_nx_{n+1})\ne 0$ and $\tau (x_{n+k}x_n)\ne 0$ , we have

(3-4) $$ \begin{align} r_{1}=\ell_{2},\,r_{2}=\ell_{3},\ldots,r_{n-1}=\ell_{n},\,r_{n}=\ell_{n+1}\quad \text{and}\quad r_{n+k}=\ell_{n}. \end{align} $$

Since $\tau (x_nx_{n-1}\cdots x_1x_{n+1})\ne 0$ , we also have

(3-5) $$ \begin{align} r_{n}=\ell_{n-1},\,r_{n-1}=\ell_{n-2},\ldots,r_{2}=\ell_{1},\,r_{1}=\ell_{n+1}. \end{align} $$

Therefore, we obtain

$$ \begin{align*} r_{n+k}\stackrel{({3-4})}{=}\ell_{n}\stackrel{({3-4})}{=}r_{n-1}\stackrel{({3-5})}{=}\ell_{n-2}\stackrel{({3-4})}{=}r_{n-3}=\cdots=\begin{cases} \ell_{1}&\text{if } n \text{ is odd},\\ r_{1}&\text{if } n \text{ is even}. \end{cases} \end{align*} $$

In addition, if n is even, then

$$ \begin{align*} r_{1}\stackrel{({3-5})}{=}\ell_{n+1}\stackrel{({3-4})}{=}r_{n}\stackrel{({3-5})}{=}\ell_{n-1}\stackrel{({3-4})}{=}r_{n-2}\stackrel{({3-5})}{=}\ell_{n-3}=\cdots=\ell_{1}. \end{align*} $$

We see that $r_{n+k}=\ell _{1}$ in either case.

Proof. Let $(S,\cdot )$ be an $(h,m)$ -semigroup and Equation (2-1) its Brandt series. We induct on h. If $h=0$ , then $(S,\cdot )$ is an m-group, and any m-group satisfies $x^m=x^{2m}$ .

Let $h>0$ and let a be an arbitrary element in S; we have to verify that

(3-6) $$ \begin{align} a^{2^hm}= a^{2^{h+1}m}. \end{align} $$

If a lies in a subgroup of $(S,\cdot )$ , then $a^m=a^{2m}$ since all subgroups of $(S,\cdot )$ are m-groups. Squaring this h times gives Equation (3-6). If a lies in no subgroup of $(S,\cdot )$ , then $a^2\in S_{h-1}$ — this is clear if $a\in S_{h-1}$ or the factor $(S_h/S_{h-1},\cdot )$ is a zero semigroup and follows from Lemma 2.1 if $(S_h/S_{h-1},\cdot )$ is a Brandt semigroup. By the induction assumption, $(a^2)^{2^{h-1}m}= (a^2)^{2^hm}$ , that is, Equation (3-6) holds again.

Proof. By Lemma 3.2, the semigroup $(S,\cdot )$ satisfies $x^{2^hm} = x^{2^{h+1}m}$ , whence for each $p\ge 2^hm$ and each $a\in \langle E(S)\rangle $ , the elements $a^p$ and $a^{p+1}$ generate the same ideal in the subsemigroup $(\langle E(S)\rangle ,\cdot )$ . The periodic semigroup $(S,\cdot )$ is a block-group by Lemma 2.8, whence the subsemigroup $(\langle E(S)\rangle ,\cdot )$ is $\mathscr {J}$ -trivial by Proposition 2.3. Therefore, we have $a^p=a^{p+1}$ , that is, $(\langle E(S)\rangle ,\cdot )$ satisfies the identity $x^p=x^{p+1}$ .

Letting

$$ \begin{align*}a_j:=\begin{cases}\tau(x_j)&\text{if } k=1,\\ \tau(\mathbf v_{n,2^hm,j}^{(k-1)})&\text{if } k>1, \end{cases} \end{align*} $$

for $j=1,2,\ldots ,2n$ , we have from Equations (3-1) and (3-3)

$$ \begin{align*} \tau(\mathbf v_{n,2^hm}^{(k)})&=a_1a_2\cdots a_{2n}\,(a_na_{n-1}\cdots a_1\cdot a_{n+1}a_{n+2}\cdots a_{2n})^{2^{h+1}m-1}&&\\ &=a_1a_2\cdots a_{2n}\,(a_1\cdots a_{n-1}a_n\cdot a_{n+1}a_{n+2}\cdots a_{2n})^{2^{h+1}m-1}&&\text{by Lemma}\ 2.9\\ &=(a_1\cdots a_{n-1}a_n\cdot a_{n+1}a_{n+2}\cdots a_{2n})^{2^{h+1}m}&&\\ &=(a_1\cdots a_{n-1}a_n\cdot a_{n+1}a_{n+2}\cdots a_{2n})^{2^{h}m}&&\text{as } x^{2^hm} = x^{2^{h+1}m} \\ &=((a_1\cdots a_{n-1}a_n\cdot a_{n+1}a_{n+2}\cdots a_{2n})^{2^{h}m})^2= (\tau(\mathbf v_{n,2^hm}^{(k)}))^2.&& \end{align*} $$

Hence, $\tau (\mathbf v_{n,2^hm}^{(k)})\in E(S)$ as required.

Proof. If $\tau (x_k)\in \langle E(S)\rangle $ for all $k\in \{1,2,\ldots ,2n\}$ , then $\tau (\mathbf v_{n,2^hm}^{(1)})\in E(S)$ by Lemma 3.3. Otherwise, let $\{k_1,k_2,\ldots ,k_{p+q}\}$ with

$$ \begin{align*} 1\le k_1<k_2<\cdots<k_p\le n< k_{p+1}<k_{p+2}<\cdots<k_{p+q}\le 2n \end{align*} $$

be the set of all indices k such that $\tau (x_k)\notin \langle E(S)\rangle $ . (Here, $p=0$ or $q=0$ is possible but $p+q>0$ .) Since $S=\langle E(S)\rangle \cup S_1$ , we have $\tau (x_{k_1}),\tau (x_{k_2}),\ldots ,\tau (x_{k_{p+q}})\in S_1$ . By Lemma 2.5 and its dual, either $\tau (\mathbf v_{n,2^hm}^{(1)})=0$ or removing all $\tau (x_k)$ such that $\tau (x_k)\in \langle E(S)\rangle $ does not change the value of $\tau (\mathbf v_{n,2^hm}^{(1)})$ . If $\tau (\mathbf v_{n,2^hm}^{(1)})=0$ , the claim holds. Otherwise, consider the substitution $\tau '\colon X_{p+q}^{(1)}\to S_1$ given by $\tau '(x_s):=\tau (x_{k_s})$ for all $s\in \{1,2,\ldots ,p+q\}$ . Then $\tau (\mathbf v_{n,2^hm}^{(1)})=\tau '(\mathbf u_{p,q,2^hm})$ . By Lemma 3.1, the element $\tau '(\mathbf u_{p,q,2^hm})$ lies in a subgroup of the Brandt semigroup $(S_1,\cdot )$ . Therefore, $\tau (\mathbf v_{n,2^hm}^{(1)})$ belongs to a subgroup of $(S,\cdot )$ .

Proof. Let $\mathcal {G}=(G,\cdot )$ be an $[m,k]$ -group and $a \in G$ . Since $a^m$ is the identity element of $\mathcal {G}$ , we have $a^{2m-1}=a^{-1}$ . Consider any substitution $\tau \colon X_{2n}^{(1)}\to G$ and let $a_i:=\tau (x_i)$ . Then

(3-7) $$ \begin{align} \tau(\mathbf v_{n,m}^{(1)})=a_1a_2\cdots a_{2n}\,(a_na_{n-1}\cdots a_1\cdot a_{n+1}a_{n+2}\cdots a_{2n})^{-1}=a_1a_2\cdots a_n\cdot a_1^{-1}a_2^{-1}\cdots a_n^{-1}. \end{align} $$

Recall that the commutator of elements a and b of a group is defined as ${[a,b]:=a^{-1}b^{-1}ab}$ . It is easy to verify that the right-hand side of Equation (3-7) can be written as a product of commutators, namely,

(3-8) $$ \begin{align} a_1a_2\cdots a_n\cdot a_1^{-1}a_2^{-1}\cdots a_n^{-1}=[a_1^{-1},a_2^{-1}][(a_2a_1)^{-1},a_3^{-1}]\cdots [(a_{n-1}a_{n-2}\cdots a_1)^{-1},a_n^{-1}]. \end{align} $$

Therefore, $\tau (\mathbf v_{n,m}^{(1)})$ belongs to the derived subgroup $(G^{(1)},\cdot )$ of $\mathcal {G}$ . We use this as the induction basis and show by induction on h that any substitution $\theta \colon X_{2n}^{(h)}\to G$ sends the word $\mathbf v_{n,m}^{(h)}$ to the hth derived subgroup $(G^{(h)},\cdot )$ of $\mathcal {G}$ .

Indeed, let $h>1$ . Since for each $j\in \{1,2,\ldots ,2n\}$ , the word $\mathbf v_{n,m,j}^{(h-1)}$ is obtained from the word $\mathbf v_{n,m}^{(h-1)}$ by renaming its variables, $\theta (\mathbf v_{n,m,j}^{(h-1)})\in G^{(h-1)}$ by the induction assumption. Combining Equations (3-3) and (3-7), we see that

$$ \begin{align*} \theta(\mathbf v_{n,m}^{(h)})=\theta(\mathbf v_{n,m,1}^{(h-1)})\theta(\mathbf v_{n,m,2}^{(h-1)})\cdots\theta(\mathbf v_{n,m,n}^{(h-1)})\cdot(\theta(\mathbf v_{n,m,1}^{(h-1)}))^{-1}(\theta(\mathbf v_{n,m,2}^{(h-1)}))^{-1}\cdots(\theta(\mathbf v_{n,m,n}^{(h-1)}))^{-1}. \end{align*} $$

Now rewriting the last expression as in Equation (3-8), we see that $\theta (\mathbf v_{n,m}^{(h)})$ is a product of commutators of elements from $G^{(h-1)}$ . Hence, $\theta (\mathbf v_{n,m}^{(h)})\in G^{(h)}$ . Since the group $\mathcal {G}$ is solvable of derived length at most k, its kth derived subgroup is trivial, whence the word $\mathbf v_{n,m}^{(k)}$ is sent to the identity element of $\mathcal {G}$ by any substitution $X_{2n}^{(k)}\to G$ . This means that $\mathcal {G}$ satisfies the identity $\mathbf v_{n,m}^{(k)}= 1$ .

Proof. We have to verify that $\tau (\mathbf v_{n,2^km}^{(k+1)})\in E(S)$ for every substitution $\tau \colon X_{2n}^{(k+1)}\to S$ . Recall from Remark 1 that the word $\mathbf v_{n,2^hm}^{(k+1)}$ is the image of the word $\mathbf v_{n,2^hm}^{(k)}$ under the substitution that sends every variable $x_{i_1\ldots i_k}\in X^{(k)}_{2n}$ to the word $\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)}$ obtained from $\mathbf v_{n,2^hm}^{(1)}$ by renaming its variables. Hence, if all elements of the form $\tau (\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)})$ lie in $\langle E(S)\rangle $ , then $\tau (\mathbf v_{n,2^hm}^{(k+1)})\in E(S)$ by Lemma 3.3. So, we may assume that at least one element of the form $\tau (\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)})$ lies in $S_1\setminus \langle E(S)\rangle $ .

By the definition of an $(h,m,k)$ -semigroup, $(S_1,\cdot )$ is either a zero semigroup or a Brandt semigroup over an $[m,k]$ -group. If $(S_1,\cdot )$ is a zero semigroup, we have $\tau (\mathbf v_{n,2^km}^{(k+1)})=0$ since $S_1$ is an ideal of $(S,\cdot )$ and each factor of the form $\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)}$ occurs in the word $\mathbf v_{n,2^km}^{(k+1)}$ at least twice. If $(S_1,\cdot )$ is a Brandt semigroup, Lemma 2.5 and its dual imply that either $\tau (\mathbf v_{n,2^hm}^{(k+1)})=0$ or removing all factors $\tau (\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)})$ that lie in $\langle E(S)\rangle $ does not change the value of $\tau (\mathbf v_{n,2^hm}^{(k+1)})$ . In the former case, the claim holds, and in the latter case, the remaining elements belong to subgroups of $(S_1,\cdot )$ by Lemma 3.4. If some of these elements belong to different maximal subgroups of $(S_1,\cdot )$ , then $\tau (\mathbf v_{n,2^hm}^{(k+1)})=0$ by the definition of multiplication in Brandt semigroups. Thus, it remains to consider only the situation when, for all $i_1,i_2,\ldots ,i_k\in \{1,2,\ldots ,2n\}$ , all elements $\tau (\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)})\in S_1\setminus \langle E(S)\rangle $ belong to the same maximal subgroup $(G,\cdot )$ of $(S_1,\cdot )$ .

Denote by e the identity element of $(G,\cdot )$ and define the substitution $\overline {\tau }\,{\colon} X_{2n}^{(k)}\to G$ by

$$ \begin{align*} \overline{\tau}(x_{i_1\ldots i_k}):=\begin{cases} \tau(\mathbf v_{n,2^hm,i_1,\ldots, i_k}^{(1)})&\text{if }\tau(\mathbf v_{n,2^hm,i_1,\ldots, i_k}^{(1)})\text{ belongs to } G,\\ e&\text{otherwise}, \end{cases} \end{align*} $$

for each $i_1,i_2,\ldots , i_k\in \{1,2,\ldots ,2n\}$ . As we know that $\tau (\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(1)})\in \langle E(S)\rangle \cup G$ for all $i_1,i_2\ldots , i_k\in \{1,2,\ldots ,2n\}$ and $fg=gf=g$ for all $g\in G$ , $f\in E(S)$ , we conclude that $\tau (\mathbf v_{n,2^hm}^{(k+1)})=\overline {\tau }(\mathbf v_{n,2^hm}^{(k)})$ . Since $(G,\cdot )$ is an $[m,k]$ -group (and so a $[2^hm,k]$ -group), it satisfies the identity $\mathbf v_{n,2^hm}^{(k)} = 1$ by Lemma 3.5. Hence, $\tau (\mathbf v_{n,2^km}^{(k+1)})=\overline {\tau }(\mathbf v_{n,2^hm}^{(k)})=e$ . Since the substitution $\tau $ is arbitrary, $(S,\cdot )$ satisfies the identity $\mathbf v_{n,2^hm}^{(k+1)} = (\mathbf v_{n,2^hm}^{(k+1)})^2$ .

Proof. We induct on h. If $h=1$ , then the claim follows from Lemma 3.6. Let $h>1$ . The Rees quotient $(S/S_1,\cdot )$ is an $(h-1,m,k)$ -semigroup whose Brandt series starts with the zero term. By the induction assumption, $(S/S_1,\cdot )$ satisfies the identity $\mathbf v_{n,2^{h-1}m}^{(kh-k+h-1)} = (\mathbf v_{n,2^{h-1}m}^{(kh-k+h-1)})^2$ for any $n\ge 2$ . By Lemma 3.2, $(S/S_1,\cdot )$ satisfies the identity $\mathbf v_{n,2^hm}^{(kh-k+h-1)} = (\mathbf v_{n,2^hm}^{(kh-k+h-1)})^2$ as well. This readily implies that any substitution $X_{2n}^{(kh-k+h-1)}\to S$ sends the word $\mathbf v_{n,2^hm}^{(kh-k+h-1)}$ to either an idempotent in $S\setminus S_1$ or an element in $S_1$ . Recall from Remark 1 that the word $\mathbf v_{n,2^hm}^{(kh+h)}$ is the image of the word $\mathbf v_{n,2^hm}^{(k+1)}$ under the substitution that sends every variable $x_{i_1\ldots i_{k+1}}\in X^{(k+1)}_{2n}$ to the word $\mathbf v_{n,2^hm,i_1,\ldots ,i_{k+1}}^{(kh-k+h-1)}$ obtained from $\mathbf v_{n,2^hm}^{(kh-k+h-1)}$ by renaming its variables. Therefore, for every substitution $\tau \colon X_{2n}^{(kh+h)}\to S$ , all elements of the form $\tau (\mathbf v_{n,2^hm,i_1,\ldots ,i_{k+1}}^{(kh-k+h-1)})$ lie in either $E(S)$ or $S_1$ .

Now consider the substitution $\overline {\tau }\colon X_{2n}^{(k+1)}\to E(S) \cup S_1$ defined by

$$ \begin{align*} \overline{\tau}(x_{i_1\ldots i_{k+1}}):=\tau(\mathbf v_{n,2^hm,i_1,\ldots,i_{k+1}}^{(kh-k+h-1)})\quad \text{for each } i_1,i_2,\ldots, i_{k+1}\in\{1,2,\ldots,2n\}. \end{align*} $$

Clearly, $\tau (\mathbf v_{n,2^hm}^{(kh+h)})=\overline {\tau }(\mathbf v_{n,2^hm}^{(k+1)})$ . However, $(\langle E(S) \rangle \cup S_1,\cdot )$ is an $(h,m,k)$ -semigroup satisfying the assumption of Lemma 3.6. By this lemma, the semigroup $(\langle E(S) \rangle \cup S_1,\cdot )$ satisfies the identity $\mathbf v_{n,2^hm}^{(k+1)}= (\mathbf v_{n,2^hm}^{(k+1)})^2$ . This ensures that the element $\tau (\mathbf v_{n,2^hm}^{(kh+h)})=\overline {\tau }(\mathbf v_{n,2^hm}^{(k+1)})$ is an idempotent. Since the substitution $\tau $ is arbitrary, $(S,\cdot )$ satisfies the identity in Equation (3-9).

Proof. Let $(S,\cdot )$ be an $(h,m,k)$ -semigroup with Brandt series in Equation (2-1). Consider the Rees quotient $(S/S_0,\cdot )$ . This is an $(h,m,k)$ -semigroup whose Brandt series starts with the zero term. By Proposition 3.7, the semigroup $(S/S_0,\cdot )$ satisfies Equation (3-9). Recall from Remark 1 that the word $\mathbf v_{n,2^hm}^{(kh+h+k)}$ is the image of the word $\mathbf v_{n,2^hm}^{(k)}$ under the substitution that sends every variable $x_{i_1\ldots i_k}\in X^{(k)}_{2n}$ to the word $\mathbf v_{n,2^hm,i_1,i_2,\ldots ,i_k}^{(hk+h)}$ obtained from $\mathbf v_{n,2^hm}^{(hk+h)}$ by renaming its variables. Then, for every substitution $\tau \colon X_{2n}^{(kh+h+k)}\to S$ , the element $\tau (\mathbf v_{n,2^hm,i_1,\ldots ,i_k}^{(kh+h)})$ is an idempotent of $(S/S_0,\cdot )$ . Therefore, $\tau (\mathbf v_{n,2^hm,i_1,\ldots ,i_k}^{(kh+h)})$ lies in either $E(S)\setminus S_0$ or $S_0$ . Recall that by the definition of an $(h,m,k)$ -semigroup, $(S_0,\cdot )$ is an $[m,k]$ -group. Denote by e the identity element of this group. For each $f\in E(S)$ , the product $fe$ lies in $S_0$ because $S_0$ is an ideal in S. Then $fe$ is an idempotent in $S_0$ by Lemma 2.4 which applies since $(S,\cdot )$ is a periodic block-group by Lemmas 3.2 and 2.8. Hence, $fe=e$ since a group has no idempotents except its identity element. Consequently, for every $g\in S_0$ , we have ${fg=feg=eg=g}$ . Dually, $gf=g$ for all $g\in S_0$ , $f\in E(S)$ . Now consider the substitution $\overline {\tau }\colon X_{2n}^{(k)}\to S_0$ defined by

$$ \begin{align*} \overline{\tau}(x_{i_1\ldots i_k}):=\begin{cases} \tau(\mathbf v_{n,2^hm,i_1,\ldots, i_k}^{(kh+h)})&\text{if }\tau(\mathbf v_{n,m,i_1,\ldots, i_k}^{(kh+h)})\text{ belongs to } S_0,\\ e&\text{otherwise}, \end{cases} \end{align*} $$

for each $i_1,i_2,\ldots , i_k\in \{1,2,\ldots ,2n\}$ . As we know that $\tau (\mathbf v_{n,2^hm,i_1,\ldots , i_k}^{(kh+h)})\in E(S)\cup S_0$ for all $i_1,i_2\ldots , i_k\in \{1,2,\ldots ,2n\}$ and $fg=gf=g$ for all $g\in S_0$ , $f\in E(S)$ , we conclude that $\tau (\mathbf v_{n,2^hm}^{(kh+h+k)})=\overline {\tau }(\mathbf v_{n,2^hm}^{(k)})$ . Since $(S_0,\cdot )$ is an $[m,k]$ -group (and so a $[2^hm,k]$ -group), it satisfies the identity $\mathbf v_{n,2^hm}^{(k)} = 1$ by Lemma 3.5. Hence, $\tau (\mathbf v_{n,2^hm}^{(kh+h+k)})=\overline {\tau }(\mathbf v_{n,2^hm}^{(k)})=e$ . Since the substitution $\tau $ is arbitrary, $(S,\cdot )$ satisfies the identity $\mathbf v_{n,2^hm}^{(kh+h+k)} = (\mathbf v_{n,2^hm}^{(kh+h+k)})^2$ .

Proof. Clearly, each finite semigroup possesses a principal series. By Corollary 2.7, every principal series of a block-group is a Brandt series. Let h be the height of a Brandt series of $(S,\cdot )$ , let m be the least common multiple of the exponents of subgroups of $(S,\cdot )$ , and let k be the maximum of the derived lengths of these subgroups. Then $(S,\cdot )$ is an $(h,m,k)$ -semigroup. Now Proposition 3.8 applies, showing that $(S,\cdot )$ satisfies all identities $\mathbf v_{n,2^hm}^{(kh+h+k)} = (\mathbf v_{n,2^hm}^{(kh+h+k)})^2$ with $n\ge 2$ . Hence, one can choose ${q=2^hm}$ and $r=kh+k+h$ .

Proof. Proposition 3.9 implies that the semigroup $(S,\cdot )$ satisfies the identities $\mathbf v_{n,q}^{(r)} = (\mathbf v_{n,q}^{(r)})^2$ for all $n\ge 2$ and some $q,r\ge 1$ , and therefore, Theorem 4.1 applies.

Proof of Theorem 1.1. We aim to prove that for any finite nonabelian solvable group $\mathcal {G}=(G,\cdot )$ , the identities of the power group $(\mathcal {P}(G),\cup ,\cdot )$ admit no finite basis. It is shown in [Reference Pin24, Proposition 2.4], see also [Reference Almeida1, Proposition 11.1.2], that the semigroup $(\mathcal {P}(G),\cdot )$ is a block-group and all maximal subgroups of $(\mathcal {P}(G),\cdot )$ are of the form $(N_G(H)/H,\cdot )$ , where $(H,\cdot )$ is a subgroup of $\mathcal {G}$ and $N_G(H):=\{g\in G\mid gH=Hg\}$ is the normalizer of $(H,\cdot )$ in $\mathcal {G}$ . Hence, each subgroup of $(\mathcal {P}(G),\cdot )$ embeds into a quotient of a subgroup of $\mathcal {G}$ and so it is solvable. We see that the multiplicative reduct of $(\mathcal {P}(G),\cup ,\cdot )$ is a block-group with solvable subgroups.

First, suppose that $\mathcal {G}$ is not a Dedekind group, that is, some subgroup $(H,\cdot )$ is not normal in $\mathcal {G}$ . Then, of course, $1<|H|<|G|$ and one can find an element $g\in G$ with $g^{-1}Hg\ne H$ . Denote by E the singleton subgroup of $\mathcal {G}$ , let $J:=\{A\subseteq G\mid |A|>|H|\}$ , and consider the following subset of $\mathcal {P}(G)$ :

(4-2) $$ \begin{align} B:=\{E,\,H,\,g^{-1}H,\,Hg,\,g^{-1}Hg\}\cup J. \end{align} $$

Observe that $g^{-1}H\ne Hg$ : indeed $g^{-1}H\cdot Hg=g^{-1}Hg$ while $Hg\cdot g^{-1}H=H$ , so that the assumption $g^{-1}H=Hg$ forces the equality $g^{-1}Hg=H$ which contradicts the choice of g. It is easy to verify that B is closed under union and element-wise multiplication of subsets so that $(B,\cup ,\cdot )$ is a subsemiring in $(\mathcal {P}(G),\cup ,\cdot )$ . Define a map $B\to B_2^1$ via the following rule:

(4-3) $$ \begin{align} \begin{array}{cccccc} \displaystyle A\in J &\displaystyle E &\displaystyle Hg &\displaystyle g^{-1}H &\displaystyle H &\displaystyle g^{-1}Hg \\ \displaystyle\downarrow & \displaystyle\downarrow & \displaystyle\downarrow & \displaystyle\downarrow & \displaystyle\downarrow & \displaystyle\downarrow \\ \displaystyle\left(\begin{matrix} 0&0\\0&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 1&0\\0&1\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&1\\0&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&0\\1&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 1&0\\0&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&0\\0&1\end{matrix}\right) \end{array}. \end{align} $$

It can be routinely verified that this map is a homomorphism from the subsemiring $(B,\cup ,\cdot )$ onto the ai-semiring $\mathcal {B}_2^1$ . Since $\mathcal {B}_2^1$ is the homomorphic image of a subsemiring in $(\mathcal {P}(G),\cup ,\cdot )$ , all identities of $(\mathcal {P}(G),\cup ,\cdot )$ hold in $\mathcal {B}_2^1$ . Thus, Theorem 4.2 applies to $(\mathcal {P}(G),\cup ,\cdot )$ , showing that its identities admit no finite basis.

If $\mathcal {G}$ is a Dedekind group, then since $\mathcal {G}$ is nonabelian, it has a subgroup isomorphic to the 8-element quaternion group; see [Reference Hall12, Theorem 12.5.4]. This subgroup is isomorphic to a subgroup in $(\mathcal {P}(G),\cdot )$ , and by [Reference Jackson, Ren and Zhao14, Theorem 6.1], the identities of every ai-semiring whose multiplicative reduct possesses a nonabelian nilpotent subgroup admit no finite basis. We see that $(\mathcal {P}(G),\cup ,\cdot )$ has no finite identity basis in this case as well.

Proof of Theorem 1.2. The theorem claims that the identities of the semiring $(H(X),\cup ,\cdot )$ of all Hall relations on a finite set X admit a finite basis if and only if $|X|=1$ . The ‘if’ part is obvious, since if $|X|=1$ , then $|H(X)|=1$ , and the identities of the trivial semiring have a basis consisting of the single identity $x=y$ .

For the ‘only if’ part, assume that $X=X_n:=\{1,2,\ldots ,n\}$ and $n\ge 2$ . The semigroup $(H(X_n),\cdot )$ is a block-group; see, for example, [Reference Gaysin, Volkov, Romeo, Volkov and Rajan10, Proposition 2]. It can be easily verified that the ai-semiring $\mathcal {B}_2^1$ embeds into $(H(X_2),\cup ,\cdot )$ via the following map:

$$ \begin{align*} \begin{array}{cccccc} \displaystyle\left(\begin{matrix} 0&0\\0&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 1&0\\0&1\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&1\\0&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&0\\1&0\end{matrix}\right) & \displaystyle\left(\begin{matrix} 1&0\\0&0\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&0\\0&1\end{matrix}\right)\\ \displaystyle\downarrow & \displaystyle\downarrow & \displaystyle\downarrow &\displaystyle \downarrow & \displaystyle\downarrow & \displaystyle\downarrow \\ \displaystyle\left(\begin{matrix} 1&1\\1&1\end{matrix}\right) &\displaystyle\left(\begin{matrix} 1&0\\0&1\end{matrix}\right) &\displaystyle\left(\begin{matrix} 0&1\\1&1\end{matrix}\right) &\displaystyle\left(\begin{matrix} 1&1\\1&0\end{matrix}\right)&\displaystyle\left(\begin{matrix} 1&0\\1&1\end{matrix}\right)&\displaystyle \left(\begin{matrix} 1&1\\0&1\end{matrix}\right) \end{array}. \end{align*} $$

The matrices in the bottom row are matrices over the Boolean semiring $\mathbb {B}\kern1.6pt{:=}\kern1.6pt(\{0,1\},+,\cdot )$ with ${0\cdot 0\kern1.6pt{=}\kern1.6pt0\cdot 1\kern1.6pt{=}\kern1.6pt1\cdot 0\kern1.6pt{=}\kern1.6pt0\kern1.6pt{+}\kern1.6pt0\kern1.6pt{=}\kern1.6pt0}$ , ${1\cdot 1\kern1.6pt{=}\kern1.6pt1\kern1.6pt{+}\kern1.6pt0\kern1.6pt{=}\kern1.6pt0\kern1.6pt{+}\kern1.6pt1\kern1.6pt{=}\kern1.6pt1\kern1.6pt{+}\kern1.6pt1\kern1.6pt{=}\kern1.6pt1}$ , and we mean the standard representation of binary relations as matrices over the Boolean semiring (in which Hall relations correspond to matrices with permanent 1). Obviously, the ai-semiring $(H(X_2),\cup ,\cdot )$ embeds into $(H(X_n),\cup ,\cdot )$ for each $n\ge 2$ , whence so does $\mathcal {B}_2^1$ . Therefore, all identities of $(H(X_n),\cup ,\cdot )$ hold in $\mathcal {B}_2^1$ .

Let $\mathrm {Sym}_n$ stand for the group of all permutations of the set $X_n$ . It is known that all subgroups of the semigroup of all binary relations on $X_n$ embed into $\mathrm {Sym}_n$ ; see, for example, [Reference Montague and Plemmons23, Theorem 3.5]. In particular, all subgroups of the semigroup $(H(X_n),\cdot )$ with $n\le 4$ embed into $\mathrm {Sym}_4$ , and hence, are solvable. We see that Theorem 4.2 applies to the ai-semiring $(H(X_n),\cup ,\cdot )$ with $n=2,3,4$ , whence the identities of the ai-semiring admit no finite basis.

To cover the case $n>4$ , observe that the semigroup $(H(X_n),\cdot )$ has $\mathrm {Sym}_n$ as its group of units, and for $n>4$ (even for $n\ge 4$ ), this group possesses nonabelian nilpotent subgroups, for instance, the 8-element dihedral group. By [Reference Jackson, Ren and Zhao14, Theorem 6.1], the ai-semiring $(H(X_n),\cup ,\cdot )$ has no finite identity basis in this case as well.

Proof. As in the proof of [Reference Gusev and Volkov11, Theorem 4.2], we employ a family of inverse semigroups constructed by Kad’ourek in [Reference Kad’ourek18, Section 2]. First, for all $n,h\ge 1$ , define terms $\mathbf w_n^{(h)}$ of the signature $(\cdot ,{}^{-1})$ over the alphabet $X_n^{(h)}=\{x_{i_1i_2\cdots i_h}\mid i_1,i_2,\ldots ,i_h\in \{1,2,\ldots ,n\}\}$ by induction on h. Put

$$ \begin{align*} \mathbf w_n^{(1)}:=x_1x_2\cdots x_n\,x_1^{-1}x_2^{-1}\cdots x_n^{-1}. \end{align*} $$

Then, assuming that, for any $h>1$ , the unary term $\mathbf w_n^{(h-1)}$ over the alphabet $X_n^{(h-1)}$ has already been defined, we create n copies of this term over the alphabet $X_n^{(h)}$ as follows. For every $j\in \{1,2,\ldots ,n\}$ , we put

$$ \begin{align*} \mathbf w_{n,j}^{(h-1)}:=\sigma_{n,j}^{(h)}(\mathbf w_n^{(h-1)}), \end{align*} $$

where the substitution $\sigma _{n,j}^{(h)}\,{\colon} X_{n}^{(h-1)}\to X_{n}^{(h)}$ appends j to the indices of its arguments, that is,

$$ \begin{align*} \sigma_{n,j}^{(h)}(x_{i_1i_2\ldots i_{h-1}}):= x_{i_1i_2\ldots i_{h-1}j} \quad\text{for all } i_1,i_2,\ldots,i_{h-1}\in\{1,2,\ldots,n\}. \end{align*} $$

Then we put

$$ \begin{align*} \mathbf w_n^{(h)}:=\mathbf w_{n,1}^{(h-1)}\mathbf w_{n,2}^{(h-1)}\cdots \mathbf w_{n,n}^{(h-1)}(\mathbf w_{n,1}^{(h-1)})^{-1}(\mathbf w_{n,2}^{(h-1)})^{-1}\cdots (\mathbf w_{n,n}^{(h-1)})^{-1}. \end{align*} $$

Now, for any $n\ge 2$ and $h\ge 1$ , let $(S_n^{(h)},\cdot ,{}^{-1})$ stand for the inverse semigroup of partial one-to-one transformations on the set $\{0,1,\ldots ,2^hn^h\}$ generated by $n^h$ transformations $\chi _{i_1i_2\ldots i_h}$ with arbitrary indices $i_1,i_2,\ldots , i_h\in \{1,2,\ldots ,n\}$ defined as follows:

• • $\chi _{i_1i_2\ldots i_h}(p-1)=p$ if and only if in the term $\mathbf w_n^{(h)}$ , the element in the pth position from the left is $x_{i_1i_2\ldots i_h}$ ;

• • $\chi _{i_1i_2\ldots i_h}(p)=p-1$ if and only if in the term $\mathbf w_n^{(h)}$ , the element in the pth position from the left is $x_{i_1i_2\ldots i_h}^{-1}$ .

We borrow the following illustrative example from [Reference Gusev and Volkov11]. Let $n=2$ , $h=2$ ; then

$$ \begin{align*} \mathbf w_2^{(2)}=\underbrace{x_{11}x_{21}x_{11}^{-1}x_{21}^{-1}}_{\mathbf w_{2,1}^{(1)}}\ \underbrace{x_{12}x_{22}x_{12}^{-1}x_{22}^{-1}}_{\mathbf w_{2,2}^{(1)}}\ \underbrace{x_{21}x_{11}x_{21}^{-1}x_{11}^{-1}}_{(\mathbf w_{2,1}^{(1)})^{-1}}\ \underbrace{x_{22}x_{12}x_{22}^{-1}x_{12}^{-1}}_{(\mathbf w_{2,2}^{(1)})^{-1}}, \end{align*} $$

and the generators of the inverse semigroup $(S_2^{(2)},\cdot ,{}^{-1})$ act as shown in Figure 1.

Figure 1 The generators of the inverse semigroup $(S_2^{(2)},\cdot ,{}^{-1})$ .

We need two properties of the inverse semigroups $(S_n^{(h)},\cdot ,{}^{-1})$ .

1. (A) [Reference Kad’ourek18, Corollary 3.2] For each $n\ge 2$ and $h\ge 1$ , any inverse subsemigroup of $(S_n^{(h)},\cdot ,{}^{-1})$ generated by less than n elements satisfies all identities of the involution semigroup $(B_2^1,\cdot ,{}^{-1})$ .

2. (B) [Reference Gusev and Volkov11, Proposition 3.3] For each $n\ge 2$ and $h\ge 1$ and any $m\ge 1$ , the semigroup $(S_n^{(h)},\cdot )$ violates the identity $\mathbf v_{n,m}^{(h)}=(\mathbf v_{n,m}^{(h)})^2$ .

Now, arguing by contradiction, assume that for some k, the involution semigroup $\mathcal {S}$ has an identity basis $\Sigma $ such that each identity in $\Sigma $ involves less than k variables. Fix an arbitrary identity $\mathbf u=\mathbf v$ in $\Sigma $ and let $x_1,x_2,\ldots ,x_\ell $ be all variables that occur in $\mathbf u$ or $\mathbf v$ . Consider an arbitrary substitution $\tau \colon \{x_1,x_2,\ldots ,x_\ell \}\to S_k^{(r)}$ and let $(T,\cdot ,{}^{-1})$ be the inverse subsemigroup of $(S_k^{(r)},\cdot ,{}^{-1})$ generated by the elements $\tau (x_1),\tau (x_2),\ldots ,\tau (x_\ell )$ . Since $\ell <k$ , property (A) implies that $(T,\cdot ,{}^{-1})$ satisfies all identities of the involution semigroup $(B_2^1,\cdot ,{}^{-1})$ . Since by the condition of the theorem, $(B_2^1,\cdot ,{}^{-1})$ satisfies all identities of $\mathcal {S}$ , the identity $\mathbf u=\mathbf v$ holds in $(B_2^1,\cdot ,{}^{-1})$ . Hence, the identity $\mathbf u=\mathbf v$ holds also in the involution semigroup $(T,\cdot ,{}^{-1})$ , and so $\mathbf u$ and $\mathbf v$ take the same value under every substitution of elements of T for the variables $x_1,\ldots ,x_\ell $ . In particular, $\tau (\mathbf u)=\tau (\mathbf v)$ . Since the substitution $\tau $ is arbitrary, the identity $\mathbf u=\mathbf v$ holds in the involution semigroup $(S_k^{(r)},\cdot ,{}^{-1})$ . Since $\mathbf u=\mathbf v$ is an arbitrary identity from the identity basis $\Sigma $ of $\mathcal {S}$ , we see that $(S_k^{(r)},\cdot ,{}^{-1})$ satisfies all identities of $\mathcal {S}$ . In particular, the multiplicative reduct $(S_k^{(r)},\cdot )$ satisfies all identities of the multiplicative reduct $(S,\cdot )$ of $\mathcal {S}$ . By the condition of the theorem, $(S,\cdot )$ satisfies the identity $\mathbf v_{k,q}^{(r)} = (\mathbf v_{k,q}^{(r)})^2$ . However, property (B) shows that this identity fails in $(S_k^{(r)},\cdot )$ , which is a contradiction.

Proof of Theorem 1.3. We aim to prove that for any finite non-Dedekind solvable group $\mathcal {G}=(G,\cdot )$ , the identities of the involution semigroup $(\mathcal {P}(G),\cdot ,{}^{-1})$ admit no finite basis. As observed in the first paragraph of the proof of Theorem 1.1, the multiplicative reduct $(\mathcal {P}(G),\cdot )$ is a block-group with solvable subgroups. By Proposition 3.9, the reduct satisfies the identities $\mathbf v_{n,q}^{(r)} = (\mathbf v_{n,q}^{(r)})^2$ for all $n\ge 2$ and some $q,r\ge 1$ .

As the group $(G,\cdot )$ is not Dedekind, one can choose a subgroup $(H,\cdot )$ of $\mathcal {G}$ and an element $g\in G$ such that $g^{-1}Hg\ne H$ . Consider the subset B of $\mathcal {P}(G)$ defined by Equation (4-2). Applying the element-wise inversion to the elements of B, we see that

$$ \begin{align*} E^{-1}=E,\ H^{-1}=H,\ (g^{-1}H)^{-1}=Hg,\ (Hg)^{-1}=g^{-1}H,\ (g^{-1}Hg)^{-1}=g^{-1}Hg, \end{align*} $$

and $A^{-1}\in J$ for all $A\in J$ . Hence, $(B,\cdot ,{}^{-1})$ is an involution subsemigroup of $(\mathcal {P}(G),\cdot ,{}^{-1})$ and the map in Equation (4-3) is a homomorphism of this involution subsemigroup onto $(B_2^1,\cdot ,{}^{-1})$ . Hence, all identities of $(\mathcal {P}(G),\cdot ,{}^{-1})$ hold in the involution semigroup $(B_2^1,\cdot ,{}^{-1})$ . Now Theorem 5.1 applies to $(\mathcal {P}(G),\cdot ,{}^{-1})$ and shows that its identities have no finite basis.