Question

Let $G$ be a semigroup. Then $G$ is a group if for all $a,b\in G,ax=b$ and $ya=b$ have solutions in $G$.

score 492 · Answer 1 · 2014-09-04 21:41:35

492

Views

Let $G$ be a semigroup. Then $G$ is a group if for all $a,b\in G,ax=b$ and $ya=b$ have solutions in $G$.

Published on 04 Sep 2014 - 21:41

score 470 · Answer 2 · 2026-03-25 06:07:03

470

Views

How to prove that: If two binary operations are anti-isomorphic and one of them is associative then the second one also will be associative?

Published on 25 Mar 2026 - 6:07

score 1.2k · Answer 3 · 2026-03-25 14:23:55

1.2k

Views

Why is a monoid with right identity and left inverse not necessarily a group?

Published on 25 Mar 2026 - 14:23

score 57 · Answer 4 · 2014-09-20 11:16:57

57

Views

finite semigroup on one generator,cycle, tail,group,zero element

Published on 20 Sep 2014 - 11:16

score 182 · Answer 5 · 2014-09-22 16:58:20

182

Views

Monoids, Semigroups, and a Reflective Subcategory.

Published on 22 Sep 2014 - 16:58

score 87 · Answer 6 · 2014-09-24 19:58:15

87

Views

Complicated proof in Transformation Semigroups.

Published on 24 Sep 2014 - 19:58

score 194 · Answer 7 · 2014-09-28 21:04:43

194

Views

Showing the full transformation semigroup $\mathscr{T}_n=\langle\zeta, \tau, \pi\rangle$.

Published on 28 Sep 2014 - 21:04

score 183 · Answer 8 · 2014-10-02 12:56:52

183

Views

For semigroups, $S\preccurlyeq T$ iff there exists an injective relational morpism $\mu: S\to T$.

Published on 02 Oct 2014 - 12:56

score 83 · Answer 9 · 2014-10-02 19:07:03

83

Views

How to construct a semi-group over a non-numerical set?

Published on 02 Oct 2014 - 19:07

score 223 · Answer 10 · 2014-10-03 09:33:22

223

Views

How would a category theorist describe Green's relations?

Published on 03 Oct 2014 - 9:33

score 196 · Answer 11 · 2014-10-08 09:11:09

196

Views

The congruence $\{(a^m, a^{m+r})\}^\#$ on $a^+$.

Published on 08 Oct 2014 - 9:11

score 64 · Answer 12 · 2026-02-23 02:57:41

64

Views

Existence of right and left identity in minimalistic algebraic structure

Published on 23 Feb 2026 - 2:57

score 840 · Answer 13 · 2014-10-09 22:57:31

840

Views

Semigroup without left cancellation law and right cancellation law

Published on 09 Oct 2014 - 22:57

score 44 · Answer 14 · 2014-10-09 23:24:59

44

Views

Semigroup of a group and identiy

Published on 09 Oct 2014 - 23:24

score 155 · Answer 15 · 2026-03-25 06:00:08

155

Views

If $\lambda\subseteq\mathscr{L}, \rho\subseteq\mathscr{R}$, then $\lambda\circ\rho=\rho\circ\lambda$.

Published on 25 Mar 2026 - 6:00

List Question

Let $G$ be a semigroup. Then $G$ is a group if for all $a,b\in G,ax=b$ and $ya=b$ have solutions in $G$.

How to prove that: If two binary operations are anti-isomorphic and one of them is associative then the second one also will be associative?

Why is a monoid with right identity and left inverse not necessarily a group?

finite semigroup on one generator,cycle, tail,group,zero element

Monoids, Semigroups, and a Reflective Subcategory.

Complicated proof in Transformation Semigroups.

Showing the full transformation semigroup $\mathscr{T}_n=\langle\zeta, \tau, \pi\rangle$.

For semigroups, $S\preccurlyeq T$ iff there exists an injective relational morpism $\mu: S\to T$.

How to construct a semi-group over a non-numerical set?

How would a category theorist describe Green's relations?

The congruence $\{(a^m, a^{m+r})\}^\#$ on $a^+$.

Existence of right and left identity in minimalistic algebraic structure

Semigroup without left cancellation law and right cancellation law

Semigroup of a group and identiy

If $\lambda\subseteq\mathscr{L}, \rho\subseteq\mathscr{R}$, then $\lambda\circ\rho=\rho\circ\lambda$.

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