Suppose $G$ is a semigroup in which every equation of the form $ax=b$ or $ya=b$ has a solution. Does this solution have to be unique?
123815
I have some difficulties with a task in algebra. I guess it's trivial and really easy but I can't figure out how to solve it.  I have a set $G$ and a binary operation on it, let it be $\circ$. I have that the operation is associative and that the equations $a\circ x = b$ and $x\circ a = b$ have unique solutions. I have to prove that $(G, \circ)$ is a group.  I already have that the operation is binary and associative, so I have to prove that there is unique identity element and unique inverse element and it will come from the equations, but how exactly?
31703
If $A$ and $B$ are $n\times n$ matrices, and $AB$ is invertible then $A$ and $B$ are invertible.    I started out by writing that since $AB$ is invertible, then for the equation $ABx=b$ has a unique solution for any $b$, $x=(AB)^{-1}b$.  Then I don't know how to break up the composition of AB to prove they are each individually similar.
52166
I'm looking to solve some matrix equations. One of the equations involves a commutator, so my question is as follows: let $A$ be a skew-self-adjoint, traceless matrix, does the equation $[X,Y] = A$ always have a self-adjoint solution? For every size of matrices.  I hope that this is a well-known fact. Perhaps it is related to the fact that the traceless skew-adjoint matrices are $\mathfrak{s}\mathfrak{u}_n$.
57005
Let $f:[0,\alpha]\to\mathbb{R}$ be a solution of the Cauchy problem:  $\begin{cases} f'(t)=(f(t))^2+t \\ f(0)=0 \end{cases} $  The question is: prove that $\alpha&lt;3$.  It is clear that the problem admits a unique solution locally on some $[-\epsilon,\epsilon]$, since $(f)^2+t$ is $C^1$ in $(f,t)$, and we have $f(t)&gt;0$ for all $t&gt;0$, looking at the sign of $f'$. But now I don't know how to proceed. Thank you all!
87658
Prove there exists an isomorphism from the semigroup of natural numbers under multiplication to the semigroup of odd natural numbers under multiplication.  An example would be helpful.
9389
This came up in proving non-regularity of a certain language (powers of 2 over the ternary alphabet). Any clue to the above equation could help me move forward.  Edit: Of course, $x = 1, y = 1$ is a solution. I am looking for non-trivial solutions.
66535
Suppose $G$ is a semigroup and it holds both left and right cancellation. Also for each $a,b\in G$, $xa=b$ has solution in $G$. Prove G is a group.   I know this question looks very "old" style. First I think it's easy for me too since I already practice probelems like "semigroup, $xa=b$ and $ax=b$ have solution , then group" Or "finite semigroup, both left and right cancellation hold, then group ". But I still get stuck... My try:  First of course, for $xa=a$ , there exists $e_a$ such that $e_aa=a$. Now we need to proof $e_a$ is left identity for all elements. Then left inverse exists hence $G$ is a group. But the key is to proof $e_ab=b$ for all $b\in G$.
18882
I've just begun to work learn Linear Algebra on my own through Hoffman and Kunze's book and the first problem set already has a question that I can't solve:  Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.  I can't seem to figure out how to prove this without resorting to case work where you account for the cases where one of the coefficients are zero and when both are.  Is there an elegant way to prove in general that when two systems of linear equations have the same solutions, they are equivalent? The converse is obvious enough though.  Definition of equivalence from the text :     Let us say that two systems of linear equations are equivalent if each equation in each system is a linear combination of the equations   in the other system.
64977
Theorem: Given a system of linear equations $Ax = b$ where $A \in M_{m \times n}\left(\mathbb{R}\right)$, $ x \in \mathbb{R}^{n}_\text{col} $, $ b \in \mathbb{R}^{m}_\text{col}$ Deduce that a solution $x$ exists if and only if $\operatorname{rank}\left(A|b\right) = \operatorname{rank}\left(A\right)$ where $A|b$ is the augmented coefficient matrix of this system  I am having trouble proving the above theorem from my Linear Algebra course, I understand that A|b must reduce under elementary row operations to a form which is consistent but I don't understand exactly why the matrix A|b need have the same rank as A for this to happen. Please correct me if I am mistaken
73477
I was wonderwing about a matrix equation (in some way similar to a very well known theorem :P).  Find all positive integers $n,m$ such that there exists $X$, $Y$, $Z$ matrices $n\times n$, non-singular, such that: $$X^m+Y^m=Z^m$$  Disclaimer: $X$, $Y$, $Z$ must have integer entries. Also, they may have an inverse which is not necessarily in $\mathbb{Z}^{n\times n}$.  I made some approachs (I don't know how to spoil, so if you want to think it yourself stop reading):  As my intuition says to me that there will always be solutions for $n&gt;1$, then, it suffices to show there will be solutions for $n=2$ and $n=3$ (because then you can form a matrix composed by matrices $2\times 2$ and $3\times 3$ in its diagonal).  I showed that for all $m$ odd, and $n=2$ there is a solution.   Can someone give me a hand with the other cases? I would be very grateful.
109664
This is Exercise 1.9.17 of Howie's "Fundamentals of Semigroup Theory".  The Details:  Let $S$ be a commutative semigroup.     Definition 1: Let $\theta_n^S$ for $n\ge 1$ be given by   $$a\theta_n^Sb \iff (\forall x\in S^n)\, xa=xb.$$      Lemma 1: The relation $\theta_n^S$ is a congruence on $S$ and $$\theta_1^S\subseteq\theta_2^S\subseteq \dots$$   Proof: For all $a\in S$, $xa=xa\forall x\in S^n$. Hence $a\theta_n^Sa$, so $\theta_n^S$ is reflexive.  Suppose $a\theta_n^S b$. Then, for $x\in S^n$, $xa=xb$, i.e., $xb=xa$, so $b\theta_n^Sa$. Hence $\theta_n^S$ is symmetric.  Suppose $a\theta_n^S b$ and $b\theta_n^S c$. Fix $x\in S^n$. Then $xa=xb=xc$, so $a\theta_n^S c$. Hence $\theta_n^S $ is transitive.  Suppose $a\theta_n^S b$ and $c\theta_n^S d$. Then for $x=x_1\dots x_n\in S^n$, $$\begin{align} x(ac)&amp;=(xb)c \\ &amp;=(x_1\dots x_{n-1})(x_nb)c \\ &amp;=(x_1\dots x_{n-1})(x_nb)d \\ &amp;=x(bd), \end{align}$$ so that $ac\theta_n^S bd$. Hence $\theta_n^S $ is compatible.    Let $i\in \mathbb{N}$ and $a\theta_i^S b$. Then for any $x=x_1\dots x_i\in S^i$, $xa=xb$, so for any $c\in S$ we have $(cx_1\dots x_i)a=(cx_1\dots x_i)b$; thus $a\theta_{i+1}^S b$. Hence $\theta_1^S\subseteq\theta_2^S\subseteq \dots$. $\square$     Lemma 2: Let $S$ be a monoid. Then for all $n\in\Bbb N$, $\theta_n^S$ is $1_S$, the identity relation on $S$.   Proof: Let $n\in \Bbb N$. If $a1_Sb$, then by definition $a=b$ and so for all $x$ in $S^n$, $xa=xb$, so that $a\theta_n^S b$ and $1_S\subseteq\theta_S^n$. If $a\theta_n^Sb$, then since $1\in S^n$ we have $a=1a=1b=b$, i.e., $a1_Sb$, so $\theta_n^S\subseteq 1_S$. Hence $\theta_n^S= 1_S$. $\square$     Lemma 3: For $n\in \Bbb N$, denote $S/\theta_n^S$ by $S_n$. Then for $n\ge 2$, $$S_n\simeq S_{n-1}/\theta_1^{S_{n-1}}.$$    Proof: User amrsa gives a proof of this (and a proof alone) in an answer below. $\square$  The Question:     Let $S=\langle a\rangle= M(m, r)$ be the monogenic semigroup generated by $a$ with index $m$ and period $r$, where $m&gt;1$. Show that   $$S/\theta_1^S\simeq M(m-1, r)$$ and deduce that   $$S/\theta_n^S\simeq M(m-n, r)$$ for $n&lt;m$.      Show also that $S/\theta_n^S$ is isomorphic to the cyclic group of order $r$ for all $n\ge m$.   My Attempt:  (The above proofs are my answers to the subproblems of Exercise 1.9.17, so I'm not approaching you empty-handed.)  I suppose $S/\theta_1^S\simeq M(m-1, r)$ because $\theta_1^S$ simply removes an $a$ from $\langle a\rangle$. I can see it but I can't prove it yet. That $S/\theta_n^S\simeq M(m-n, r)$ comes from Lemma 3 by the same token.  Any help would be appreciated.
91529
Characterize all finite dimensional algebras (may not be commutative) over a field $K$ without nilpotent elements.   My condition: Let $A$ be any algebra (may not be finite dimensional), then it's easy to prove that $A$ has no nilpotent elements iff the equation $x^2=0$ has a unique solution (the trivial solution).  But is there a more explicit characterization of these finite dimensional algebras?
19189
Let $S$ be a set with a binary operation $xy$ defined on it, with a neutral element, but not satisfying associativity. I want to prove that the inverse isn't necessarily unique.  My attempt to answer this is in the answer below.
90870
In a book I am reading the author states without proof that in an $n$-dimensional vector space $X$, the representation of any $x$ as a linear combination of a given basis $e_{1},e_{2},...,e_{n}$ is unique. How to proof that?
107609
Let $(S, \cdot)$ be a semigroup. There are natural numbers $p,q \geq 2$ such that $a^pb^q=ba$ for all $a,b \in S$. Prove that $S$ is commutative.   I wrote   $$\begin{align} a^{p+1}b^{q+1} &amp;=b^{(q+1)p}a^{(p+1)q} \\  &amp;=b^{p}\cdot(b^q)^p \cdot (a^p)^q\cdot a^q \\  &amp;=b^p\cdot a^p \cdot b^q \cdot a^q \\  &amp;= b^p\cdot b \cdot a \cdot a^q \\  &amp;=b^{p+1}a^{q+1}. \end{align}$$  From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.  Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
55606
Consider three $n \times n$ real valued matrices, $P$, $X$, and $Y$ and the following equation:  $PX=YP$,   where $P$ is the unknown term.  Under what conditions is this equation solvable? Is it there a closed-form solution?  Your help is greatly appreciated.
99525
So my suggestion was:  Assume: Let $S = \{u_1, u_2, ....u_r\}$ be a set of vectors in $\mathbb{R}^n$ . If $r &gt; n$ then the vectors $u_1, u_2 \ldots u_r$ must be linearly dependent.  If one writes the linear system corresponding to $c_1u_1 + c_2u_2 + ... + c_ru_r$ , one will have a homogeneous system of $n$ equations in $r$ unknowns. We know that such a system has infinitely many solutions. Thus in $\mathbb{R}^n$ , a set which is linearly independent cannot contain more than $n$ vectors.