Suppose $S$ is a finite set with a binary operation $*$ given by a Cayley table. While the commutativity of $*$ can be determined on the basis of the symmetry of the table across the upper-left to lower-right diagonal, is there any way we can, by inspecting the table alone, decide if $*$ is associative or not?
21013
Most properties of a single binary operation can be easily read of from the operation's table. For example, given $$\begin{array}{c|ccccc} \cdot &amp; a &amp; b &amp; c &amp; d &amp; e\\\hline a     &amp; e &amp; d &amp; b &amp; a &amp; c\\ b     &amp; d &amp; c &amp; e &amp; b &amp; a\\ c     &amp; b &amp; e &amp; a &amp; c &amp; d\\ d     &amp; a &amp; b &amp; c &amp; d &amp; e\\ e     &amp; c &amp; a &amp; d &amp; e &amp; b \end{array}$$ it is easy to check that it is closed (no elements occur in the table which don't occur as row or column index), commutative (the table is symmetric), has a neutral element (the row and column of $d$ are copies of the index row/column), and has an inverse element for each element (there's a $d$ in each row and column). In other words, almost all important properties can immediately be seen. The only part missing is associativity.  Therefore my question: Is there a simple way to see directly from the operation's table (i.e. without doing explicitly all the calculations) if an operation is associative?
104635
In a set $X$ we define a binary operation $*$ such that  $$\forall x, y \in X,\  (x*y)*y=y*(y*x)=x.$$  Is $*$ commutative and why?
122407
I want to show that $\forall x$,$y\in G$, $(xy)^2=x^2y^2 \iff G $ is an abelian group.  Proof of the backward direction seems easy enough:  Assume $G$ is an abelian group. Then consider $(xy)^2=(xy)(xy)$. By associativity, $(xy)(xy)=xyxy$. Since $G$ is abelian, $xyxy=xxyy=x^2y^2$.   Proof of the forward direction I attempt by contrapositive:  That is, assume $G$ is not an abelian group. Then, consider $(xy)^2=xyxy$ as above. Yet, $xyxy\neq xxyy=x^2y^2$ since we no longer have commutativity.  Is this proof by contrapositive completely correct?
36949
Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification.
89132
Is it (and if yes how?) possible to define an an binary operation $*$ for an arbitrary set $M$ such that $(M,*)$ is a group?  If $M$ is finite or countable infinite this is trivial, but is it also possible for any other sets? And if yes how can you construct $*$? If no can you provide a counterexample?
71556
I don't know how to evaluate this expression:  $A-B\cap C$  Is it true that:  $A-(B\cap C)=(A-B)\cap C$  or, in other words, that the associative property applies to these operations?
77239
Let $a,b$ be two group elements of finite order that commute.   What can be said about the order of $ab$?   I thought that $|ab| = \text{lcm}(|a|,|b|)$. My proof was that $(ab)^n = a^n b^n =e$ if and only if $a^n = b^n = e$ if and only if $|a|,|b|$ both divide $n$. The smallest $n$ such that both orders divide it is the least common multiple of $|a|$ and $|b|$.  By chance I came across . It has an upvote so it's clearly correct. (?)  But it contradicts what I think: It's clear that   $$ (ab)^{\text{lcm}(|a|,|b|)} = e$$  hence $|ab| \mid \text{lcm}(|a|,|b|)$.  It seems to me that this is saying more than $|ab| \mid |a| |b|$. Is it not?     Now my question is: what is the precisest statement that can be made about $|ab|$?  Is there anything more than $|ab| \mid \text{lcm}(|a|,|b|)$ that can be said about $|ab|$?
44051
I have attempted a proof of this, but upon looking at my notes, I feel it might be incorrect: it is noticeably simpler than the one in my notes.  Proposition: any permutation in $S_n$ can be written as a product of disjoint cycles.  Proof (attempt): $s\in S_n$. Let $\langle s\rangle$ act on $X=\{1,\cdots,n\}$. Define $x\sim x'\iff x'\in\text{Orb}(x)$ for $x,x'\in X:$ this is clearly an equivalent relation, hence the orbits partition $X$. Each orbit corresponds a cycle in $s$, so we are done.  Does this hold, or have I missed something vital?
114311
Assume that my definition of a Boolean algebras is the following one: I have a set $B$ with two binary operations $\vee$ and $\wedge$ which both satisfy the commutative, associative and distributive laws. Both the operations have an identity element ($0$ and $1$) and for each element $b\in B$ there exists a complement (i.e. an element $b{'}\in B$ such that $b\vee b{'}=1$ and $b\wedge b{'}=0$).  Now I found the following definition of Boolean Algebra. Let $B$ be a set on which we define a binary operation $\cap$ and an unary operation ${'}$. We also require that those operations satisfy the following rules:  1) $x\cap y=y\cap x$ for each $x,\, y\in B$  2) $(x\cap y)\cap z=x\cap (y\cap z)$  3) $x\cap y{'}=z\cap z{'}$ iff $x\cap y=x$  Are those equivalent definitions? I was trying to work the thing out in the following way.  a) First we may put $0:=z\cap z'$ by property 3).  b) We may define $x\cup y:=(x{'}\cap y{'}){'}$ and put $0{'}=:1$ and we may prove the commutative law for $\cup$.  Anyway I cannot prove the associativity of $\cup$ or either the distributivity or even the fact the $0$ and $1$ are identity elements.
64148
I'm reading Beardon's Algebra and Geometry:     Show that $\mathbb{Z}$ with the operation $m*n=m+(-1)^m n$, is a group.   I've been thinking on how I should show associativity, I thought about doing this:  $$(m*n)*p$$  Then assuming:  $$m*n=m+(-1)^m n$$  I have:  $$(m+(-1)^m n)*p$$  Which results in:  $$[m+(-1)^m n]+(-1)^{[m+(-1)^m n]}p$$  Then:  $$m*(n*p)$$  Again, assuming that:  $$n*p=n+(-1)^n p$$  I have:  $$m*(n+(-1)^n p)$$  Then:  $$m+(-1)^m [n+(-1)^n p]$$  And show that:  $$[m+(-1)^m n]+(-1)^{[m+(-1)^m n]}p=m+(-1)^m [n+(-1)^n p]\tag{A} $$  But I've tested it with some values on Mathematica and found that when:  $$m = 1; n = 0; p = -1;$$  The equality does not hold. I'm stuck again. Can you please provide me some hints on how to do it? I'm not looking for a full answer.    When I was testing the equality on Mathematica, I felt that permutations of $m,n,p$ on the set $\{-1,0,1\}$ could be enough to prove it, for example:  $$\begin {array} { c | c | c } m,n,p &amp; m,n,p &amp; m,n,p \\ \hline   -1, -1, -1 &amp; - 1, -1, 0 &amp; - 1, -1, 1  \\   -1, 0, -1 &amp; - 1, 0, 0 &amp; - 1, 0, 1  \\   -1, 1, -1 &amp; - 1, 1, 0 &amp; - 1, 1, 1  \\   0, -1, -1 &amp; 0, -1, 0 &amp; 0, -1, 1  \\   0, 0, -1 &amp; 0, 0, 0 &amp; 0, 0, 1  \\   0, 1, -1 &amp; 0, 1, 0 &amp; 0, 1, 1  \\   1, -1, -1 &amp; 1, -1, 0 &amp; 1, -1, 1  \\   1, 0, -1 &amp; 1, 0, 0 &amp; 1, 0, 1  \\   1, 1, -1 &amp; 1, 1, 0 &amp; 1, 1, 1  \\ \end {array} $$  But I'm not sure if this is true. What do you think?
88756
I am struggling with the end of my proof.   So far I have used the fact that $S_n=\langle T\rangle$ where $T=\{(i\text{ }j)|1\leq i&lt;j\leq n\}$ to show that $S_n$ is generated by the set $\{(1\text{ }2),(1\text{ }3),\cdots,(1\text{ }n)\}$ of $n-1$ transpositions.   I am struggling with proving that $S_n$ is also generated by the set $\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$ of $n-1$ transpositions. Essentially I want to show that each $(1\text{ }i)$ is a product of these transpositions to reach this conclusion.   Here is what I have so far...    Consider any $(1\text{ }i)$ transposition in $S_n$. Notice by the composition of a permutation that \begin{eqnarray*} (1\text{ }i-1)(i-1\text{ }i)(1\text{ }i-1)&amp;=&amp;\begin{bmatrix}     1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i    \end{bmatrix} \begin{bmatrix}     1       &amp; i-1 &amp; i   \\    1       &amp; i &amp; i-1  \end{bmatrix} \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i \end{bmatrix}\\ &amp;=&amp; \begin{bmatrix}     1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i    \end{bmatrix} \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i       &amp; 1 &amp; i-1 \end{bmatrix}\\ &amp;=&amp; \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i       &amp; i-1 &amp; 1 \end{bmatrix}\\ &amp;=&amp; (1\text{ }i) \end{eqnarray*}  Now we want to use this property, call it STAR, to show that $S_n$ is also generated by the set  $$Y=\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$$ of $n-1$ transpositions. Notice for each $(i\text{ }j)$ in $Y$ that $j-i=1$. Suppose $n=2$. Then $S_2=\{(1),(1\text{ }2)\}$, and we done because $2-1=1$. Suppose $n=3$. Then $S_3$ is generated by the set $\{(1\text{ }2),(1\text{ }3)\}$. By STAR,  $$ (1\text{ }3)=(1\text{ }2)(2\text{ }3)(1\text{ }2) $$ $$ (1\text{ }2)=(1\text{ }1)(1\text{ }2)(1\text{ }1) = (1\text{ }2) $$ Therefore $S_3$ is generated by $\{(1\text{ }2),(2\text{ }3)\}$ and for each $(i\text{ }j)$ we have $j-i=1$.    And here is where I am struggling. I do not know how to apply induction to this. I want to show that each side part will continue breaking down using the property I stated until all of $S_n$ is created. Here is why I think this: $$ (1\text{ }1)(1\text{ }2)(1\text{ }1)=(1\text{ }2) $$ $$ (1\text{ }2)(2\text{ }3)(1\text{ }2)=(1\text{ }3) $$ $$ \cdots $$ $$ (1\text{ }n-1)(n-1\text{ } n)(1\text{ } n-1)=(1\text{ }n) $$ Any input would be much appreciated. Best.
117842
As the title says. Is the matrices $A^*A$ and $AA^*$ hermitian (symmetric if $A$ is real)?
29321
Show that the operation $*$ is commutative is a structural property.   Give a careful proof that the indicated property of a binary structure $\langle S,* \rangle$ is indeed a structural property.  I've started this problem as:  Let $\langle S,* \rangle$ be isomorphic to $\langle T,\Box\rangle$. Also let $f:S \to T$. This means that for $a,b \in S$, then $f(a*b)=f(a)\Box f(b)$ and $f(a*b)=f(b*a)=f(b)\Box f(a)$ and therefore $f(a)\Box f(b)=f(b) \Box f(a)$. This means that an operation $*$ is commutative is a structural property. Does this work?
10058
I know that a set G with a binary operation $*$ is a group, if:   $a*b\in G$, for all $a, b \in G$. $*$ is associative:   $$(a*b)*c=a*(b*c) \\ \text{for all }a, b, c\in G.$$   An identity element $e \in G$ exists, such that   $$a*e = e*a = a\\ \text{for all }a\in G.$$   For all elements $a \in G$, there exists an $a^{-1} \in G$, such that:   $$a*a^{-1} = a^{-1}*a=e.$$  Can I use that to show that the empty set is a group?
12307
I've tried deriving a proof for this using the handshaking lemma, but that was unsuccessful. Could someone please explain to me how they would do this proof?
32240
How can we prove that (the set of all polynomials in infinitely non commutative indeterminants over $k$) is a ring, where $k$ is a ring with unity?   Adding two polynomials is a polynomial, but for commutativity  How can we show the following: $ax_1+bx_2=bx_2+ax_1$ (commutativity of addition) If we have finite number of variable it is easy to prove it, (we take the $n-1$ variables...), but what about the infinite number of variables?  Can someone give me a hint?
84062
Considering Peano axioms we'll define addition, multiplication and exponentiation operations.  We can then prove that addition and multiplication operations are commutative and associative. The proof for addition operation is pretty straightforward but for for multiplication it a bit more difficult. For example, to prove associativity, we have to prove distributivity first.  Moreover, for the definition itself it is not obvious that commutativity should take place, because the definition is not symmetric.  $a \cdot 0 := 0$  $a \cdot S(b) := a \cdot b + a$  Here $S(b)$ is the "next" function from Peano axioms  This definition is not symmetric with respect to $a$ and $b$, so commutivity is not obvious.  For exponentiation operation neither commutivity nor associativity take place. And I am very curious about the logical explanation why that happens  NOTE: The explanation I am looking for is somehow similar to a Abel–Ruffini theorem which explains WHY there is no an algebraic solution for a general polynomial equation