Let ⋆ be a binary operation on a nonempty set $M$. That is, every pair $(a,b) \in M$ is assigned an element $a$ ⋆$ b$ in $M$. Suppose that ⋆ has the additional property that $(a $ ⋆ $b) $ ⋆$ b= a$ and $a$ ⋆ $(a$ ⋆$ b)= b$ for all $a,b \in M$. (a) Show that $a$ ⋆ $b = b$ ⋆ $a$ for all $a,b \in M$. (b) On which finite sets $M$ does such a binary operation exist?
Problem
Source: Czech and Slovak Match 1996 P2
Tags: Binary operation, algebra
Math_Is_Fun_101
06.11.2021 03:04
(a) \[ a\star b = ((a\star b)\star a)\star a = ((a\star b)\star((a\star b)\star b))\star a = b\star a. \]
ineq1192
06.11.2021 17:20
Math_Is_Fun_101 wrote:
(a)
\[ a\star b = ((a\star b)\star a)\star a = ((a\star b)\star((a\star b)\star b))\star a = b\star a. \]
I think this solusion is incompleted. parmenides51 wrote: Let ⋆ be a binary operation on a nonempty set $M$. That is, every pair $(a,b) \in M$ is assigned an element $a$ ⋆$ b$ in $M$. Suppose that ⋆ has the additional property that $(a $ ⋆ $b) $ ⋆$ b= a$ and $a$ ⋆ $(a$ ⋆$ b)= b$ for all $a,b \in M$. (a) Show that $a$ ⋆ $b = b$ ⋆ $a$ for all $a,b \in M$. (b) On which finite sets $M$ does such a binary operation exist? Let $P(a, b)$ be assertion of $(a*b)*b=a$ and $Q(a, b)$ be assertion of $a*(a*b)=b$. (a) i) $(c*b)*b=b*(c*b)$ for $\forall b, c\in M$. $\textsl{Proof.}$ $P(c, b), P(c, c*b)$ gives $(c*b)*b=c=(c*(c*b))*(c*b)=b*(c*b)$ ii) For given $b\in M$, $\forall a\in M, \exists c\in M$ such that $c*b=a$ $\textsl{Proof.}$ $P(a, b)$ implies $c=a*b$. From i), ii), we are done.