0
$\begingroup$

Maybe this will be a trivial question but I can't find a solution for this. I checked already on a lot of questions here and on google but can't find a solution for my specific problem.

Given an operation $\ast: Z \times Z \xrightarrow[]{} Z$ defined as: $a\ast b = 3a + b$ prove that $\ast$ is 1)commutative, 2)associative and 3)exist the identity element. I proved 1) and 2) without problem, is neither commutative and associative(correct me if I'm wrong) but I have some problem with 3). From definition I know that exist the identity element $\iff$ $ \forall a \in Z \quad \exists u \in Z: \quad a\ast u = u \ast a = a$. So essentially I must solve 2 equation one for left side identity element and another one for right side identity element, in my case: $$ a \ast u = 3a+u $$ I should solve the equation: $3a+u=a$. So the left identity element will be $ u= -2a$

Similarly for the other side: $$ u \ast a = 3u+a $$ I should solve the equation: $3u+a=a$. So the left identity element will be $ u= 0$. If I did everything correct $u_l \neq u_r$.

My question arises from this situation, I should consider that identity element $u$ doesn't exists because $u_l \neq u_r$ or I should write only that $u_l \neq u_r$ without specifing if $u$ exists? Briefly, does the identity element exists in this case?

Therefore, is right to say that when an operation isn't commutative then will never exist the identity element? This conclusion seems almost strange to me because if I consider a simple non-abelian monoid my conclusion isn't correct. Thanks all in advance.

$\endgroup$
1
$\begingroup$

You have the definition of identity element wrong. $u$ is an identity element if for all $a$ one has $a * u = a = u * a$. That is, you have your quantifiers wrong, it is not "for all $a$ there is $u$", but rather "there is $u$ such that for all $a$". In other words, $u$ must be independent of $a$.

In your case you have a left identity, but not a right one, because the equation $3 a + u = a$ does not have a solution $u$ which is independent of $a$.

$\endgroup$
  • $\begingroup$ You're absolutely right! thanks $\endgroup$ – Alfonse Jun 23 '16 at 15:48
0
$\begingroup$

well, in case there is a left identity which is'nt a right identity and a right one that is'nt a left identity , we do'nt usually call any of those an identity.

notice that, if there is an identity $e$, and a left(right) identity $l$($r$), then $$e=el=l(r=re=e)$$, so if there is an left identity that is not a right identity, there is no identity element.

about your specific question, in the first equation you did'nt get a "solution", but some relation, so only because $u$ depends on $a$, you cansay that there is no element that sucj that the right identity equation holds for all elements.

finally, if a monoid is'nt comutative, it does'nt mean that there is no element that act "comutativly", but that not all element act like that.

$\endgroup$
  • $\begingroup$ Thanks a lot!! So is right to generalize and say that when an equation solution is dependent from some parameters, then there's no identity element? $\endgroup$ – Alfonse Jun 23 '16 at 15:48
  • $\begingroup$ yes, but it is not that simple to recognize if that is the case. $\endgroup$ – user160823 Jun 23 '16 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.