Let the operation $*$ be defined in $\mathbb Z_{10}$ for every $a,b \in \mathbb Z_{10}$ as follows:
$\begin{aligned} a*b=3ab-a-b+4\end{aligned}$
determine:
- if $(\mathbb Z_{10}, *)$ has an identity element;
- if $0,1,2,6$ are invertible in $(\mathbb Z_{10}, *)$ and, if that is the case, calculate the inverses.
We know that $\varepsilon$ is an identity element $\Leftrightarrow (\forall a\in \mathbb Z_{10})(a*\varepsilon = \varepsilon *a = a)$. In my case (given that $* $ is commutative):
$\begin{aligned} a*\varepsilon =a \Leftrightarrow3a\varepsilon-a-\varepsilon+4 = a\end{aligned}$
so
$\begin{aligned} \varepsilon = (2a-4)(3a-1)^{-1}\end{aligned}$
As the identity element $\varepsilon$ is bound to the value of the $a$ variable, then there isn't an unique identity for every element in $\mathbb Z_{10}$ therefore can I state that the identity element does not belong to $(\mathbb Z_{10}, *)$?
Moreover is it wrong using the everytime different $\varepsilon$ to find the $a^{-1}$ of $0,1,2,6$?