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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 $0,1,2,6$?

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    @dado: You're right that $(3a-1)^{-1}$ may not exist, and that this is$a$problem for the OP's approach -- but note that it needs to be an inverse with respect to ordinary multiplication in $\mathbb Z_{10}$, not with respect to $*$.2012-10-28

2 Answers 2

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Let us look at your equation. In order for $b$ to be an identity, we want $(3a-1)b\equiv 2a-4\pmod{10}.$ It is not obvious that there is no solution $b$ that works for all $a$. And in fact there is such a $b$. Hint: Can we manage to have $3b\equiv 2\pmod{10}$?

Remark: The second question strongly hints that there might be an identity element! There are only $10$ objects to worry about. So it is not unreasonable to try thm one at a time. For "small" structures, and even for larger ones, it is often a good idea to dig in and compute.

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    @haunted85: The assertion $(2a-4)\equiv 2a+4$ is very wrong. I can't understand how it could be made. Calculate. If $b=4$, then $(3a-1)b=12a-4\equiv 2a-4\pmod{10}$.2012-10-28
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If it has an identity, then it should work for all elements.

As you calculated, for $a=4$ (whence $3a-1=11\equiv 1$ is invertible $\pmod{10}$), we must have $\varepsilon=4$ If that works for all, then good, if not, then having an inverse is not a well defined notion.