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.
Best Answer
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 $a$.