43
votes
Is each squared finite group trivial?
It seems that every squared finite group is indeed trivial.
Let $G$ be a squared finite group with the subset $A$ showing the squared-ness of $G$.
For any irreducible representation $\pi$ of $G$, ...
- 2,053
27
votes
Accepted
What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$?
No such space exists. We actually get the stronger statement that every isomorphism $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(\mathbf R)$ is induced by an isomorphism $X \stackrel\sim\...
- 27.3k
27
votes
Accepted
Is each squared finite group trivial?
I think, as implicitly suggested by Yemon Choi, it is possible to explain the proof of the answer of user49822 by making more use of idempotents. Suppose that the finite group $G$ is squared via the ...
23
votes
When $X \times Y \cong X \times Z$ implies $Y \cong Z$ (in the category of finite topological spaces)
The monoid $\mathcal M_{\rm fin}$ is in fact cancellative.
To prove this, we start with a Lemma. Let us write $h(T,Y)$ for the cardinality of the set of continuous maps $T\to Y$ and $i(T,Y)$ for the ...
- 30.7k
22
votes
Accepted
Are there any "simple" monoids with intermediate growth?
Yes. Jan Okninski showed that $$\begin{bmatrix} 1 & 1 \\ 0 &1\end{bmatrix}\ \text{and}\ \begin{bmatrix} 1 & 0\\ 1 & 0\end{bmatrix}$$ generate a semigroup of intermediate growth. ...
- 37.4k
22
votes
Accepted
The number of polynomials on a finite group
$\DeclareMathOperator\Poly{Poly}$Proposition. If $G$ is a simple non-abelian finite group, then $\Poly(G)=G^G$.
(Edit: this observation appears as the main therorem in this paper by Maurer and Rhodes, ...
- 60.4k
21
votes
Accepted
Is the Golomb countable connected space topologically rigid?
[Edit, Dec 6, 2019] I have a pleasure to inform that this problem was finally resolved in affirmative by T.Banakh, D.Spirito and S.Turek who proved the following
Theorem. The Golomb space is ...
- 40.9k
19
votes
Accepted
Semi group of polynomials which all roots lie on the unit circle
A complex polynomial is uniquely determined by its set of roots together with multiplicities. This means that the semigroup of your polynomials is freely generated by the set of point on the unit ...
- 4,730
18
votes
Accepted
Extending $\Bbb N$ to a semiring with isomorphic additive and multiplicative structure
There is an extension $R$: take the closure of $\mathbb N$ by the operations $\text{L}$ (or $\varphi$ in the OP) and its inverse $\text{E}$, which are the logarithm and exponential in base $1.2$. ...
- 5,055
18
votes
Accepted
Monoids of endomorphisms of nonisomorphic groups
For any prime $p$, the endomorphism monoid of $\mathbb{Z}[\frac{1}{p}]$ is a commutative monoid with zero whose submonoid of nonzero elements is the direct sum of a cyclic group of order two (...
- 33.9k
17
votes
Are all free monoids residually finite?
The standard meaning of residually finite here is that for every pair of elements $u, v$ in the free monoid $A^\ast$, if $u \neq v$ then there is a homomorphism $\phi \colon A^\ast \to F$ with $F$ a ...
15
votes
Statements about groups proved using semigroups
At page 4 of J. Meakin's paper Groups and semigroups: connections and contrasts it is explained that
many properties of braid groups, Artin groups of finite type,
Garside groups and the more ...
15
votes
Monoids of endomorphisms of nonisomorphic groups
It is proved in $[$1$]$ that the tetrahedral group $A_4$ of order $12$ and the binary tetrahedral group of order $24$ have isomorphic endomorphism monoids. So this gives a finite example. It is also ...
- 37.4k
15
votes
Accepted
Associativity may fail by little?
The result you quoted appears in this reference: G. Szasz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28.
The Szasz theorem requires that the ...
- 1,879
15
votes
Accepted
Generalized cancelation properties ensuring a monoid embeds into a group
The answer is no. What you call a generalized cancellation rule is called a quasi-identity in universal algebra. Malcev proved in 1939 that there is no finite basis of quasi-identities defining group ...
- 37.4k
15
votes
Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$
This is false as shown by the following digraph.
From $x$ there is an edge to $v_p$, from $v_p$ there is a cycle of length $p$ to itself, and from $v_p$ there are $p-1$ different paths to $y$, of ...
- 18.4k
14
votes
What are the main structure theorems on finitely generated commutative monoids?
Have a look at Grillet's Commutative Semigroups. Let $C$ be a commutative semigroup. The outline of the structure theory is as follows (I'll include references to Grillet; see also V.5.7 for an ...
- 61.6k
14
votes
Accepted
Nielsen-Schreier theorem for monoids
For the question of finite generation, here's a a counterexample that appears in D.B. McAlister and L. O'Carroll,
Finitely generated commutative semigroups. Glasgow Math. J. 11 1970 134–151. MR0269765 ...
- 7,012
14
votes
Accepted
Who invented Monoid?
The name "monoid" was first used in mathematics by Arthur Cayley [*] for a surface of order $n$ which has a multiple point of order $n-1$.
In the context of semigroups the name is due to Bourbaki [...
- 178k
14
votes
Accepted
Is the class of power-associative binars finitely axiomatizable?
No.
Indeed, let $\mathcal{V}_n$ be the variety of magmas generated by the relating identities with variable $y$ saying that for every $k\le n$, all products of $k$ copies of $y$ are equal. Since the ...
- 60.4k
14
votes
Is each squared finite group trivial?
Here is a proof for the abelian case, that perhaps has some chance to generalize.
Suppose $G$ is a squared finite group as witnessed by the subset $A$. Consider the element $$\alpha = \frac{1}{|A|} \...
- 1,432
14
votes
Accepted
Grothendieck group of the Fibonacci monoid
This operation is not mysterious at all! The monoid $(\mathbf N,\circ)$ is isomorphic to a multiplicative submonoid $T$ of the commutative ring $\mathbf Z[\varphi] = \mathbf Z[t]/(t^2-t-1)$, where $\...
- 27.3k
13
votes
Accepted
Can a Shelah semigroup be commutative?
An infinite Shelah semigroup must be a Jonsson semigroup (meaning that it is an infinite semigroup whose proper subsemigroups have lesser power). Therefore the following paper answers the question ...
- 12.1k
13
votes
Accepted
Spaces with unique endomorphism monoids
$\DeclareMathOperator\End{End}$As shown in Todd Trimble’s comment, the set of constant maps $X\to X$ is definable in $\End(X,\tau)$, as it consists of exactly the left-absorbing endomorphisms (i.e., $\...
- 44.9k
12
votes
Ternary associative multiplication
Perhaps this has already been said in some form, but the identity
$$[[a,b,c],d,e] = [a, [b,d,e], [c,d,e]]$$
is exactly the $(2,2,2)$-associative law of clone theory. I think that this is the most ...
- 12.1k
12
votes
What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$?
By @YCor's comment, $X$ has the same number of elements as $\mathbb{R}$ and the actions of $End(X)$ on $X$ is the same as the action of $End(\mathbb{R})$ on $\mathbb{R}$. Now consider the automorphism ...
- 9,549
12
votes
Accepted
For what sets $X$ do there exist a pair of functions from $X$ to $X$ with the identity being the only function that commutes with both?
The answer is no: for every set $X$ there exists a pair in the monoid $X^X$ of self-maps of $X$, with centralizer reduced to $\{\mathrm{id}\}$.
(I first left my original "groupwise" answer ...
- 60.4k
12
votes
Is a solvable group satisfying a semigroup law?
The generators of a free metabelian group of rank two generate a free subsemigroup so there is no semigroup law defining solvable of derived length 2. Much smaller metabelian groups, like the ...
- 37.4k
11
votes
Accepted
Identifying a group without 2-torsion
Using a mixture of computation and thought I believe that I have established that this group is indeed torsion-free. I don't know of any general approach to solving that particular problem. Even if ...
- 36.4k
11
votes
Accepted
Homotopy type of a specific discrete monoid
This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
\begin{align*}
A(x) &=
\begin{cases} 2x &\text{if }x \leq 1/2\\1 &\text{if }x \...
- 51.5k
Only top scored, non community-wiki answers of a minimum length are eligible