Let the operation $*$ be defined in $\mathbb Z_{10}$ for every $a,b \in \mathbb Z_{10}$ as follows:

$$\begin{aligned} a*b=3ab-a-b+4\end{aligned}$$


  • 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):

$$\begin{aligned} a*\varepsilon =a \Leftrightarrow3a\varepsilon-a-\varepsilon+4 = a\end{aligned}$$


$$\begin{aligned} \varepsilon = (2a-4)(3a-1)^{-1}\end{aligned}$$

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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    $\begingroup$ Are you sure $(3a-1)$ is invertible in $\Bbb Z_{10}$? If it doesn't have identity element (meant: identity for all), then the other question just doesn't make sense. $\endgroup$ – Berci Oct 28 '12 at 16:43
  • $\begingroup$ $(3a-1)^{-1}$ makes sense only if the element $3a-1$ is invertible. This, in turn, implies that there is an identity element. So your argument seems a bit dodgy. $\endgroup$ – user39280 Oct 28 '12 at 16:46
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    $\begingroup$ Since your identity, if it exists, must work for all ten possible values of $a$, you have ten modular equations it must satisfy. Write down the equations for, say, $a=0$ and $a=1$. Do they have any common solution? If so, does that solution work for the other $a$s too? $\endgroup$ – hmakholm left over Monica Oct 28 '12 at 16:47
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    $\begingroup$ @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 $*$. $\endgroup$ – hmakholm left over Monica Oct 28 '12 at 16:50

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.

  • $\begingroup$ At this point I feel a bit confused. I understand that if $b$ isn't the same for every element in $(Z_{10}, *)$ then there's no identity element for that structure. Does this mean that for any $\mathbb Z_{k}$, where $k$ is not prime, there are not invertible elements? $\endgroup$ – haunted85 Oct 28 '12 at 17:08
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    $\begingroup$ If you are talking about the standard multiplication on $\mathbb{Z}_k$, then of course $0$ is never invertible. If $k$ is prime, then $0$ is the only non-invertible element. If $k$ is composite, then yes, there always are non-invertible elements other than $0$. As to "weird multiplications," like the one in this problem, one cannot give a general answer. $\endgroup$ – André Nicolas Oct 28 '12 at 17:32
  • $\begingroup$ So to sum everything up in this case given that $b$ is not the same for every $a$ in $(\mathbb Z_{10}, *)$, then $(\mathbb Z_{10}, *)$ has no idenity at all. If $(3a-1)$ results in being not invertible, then $a$ isn't an invertible element, instead if $(3a-1)$ is invertible then $a$ is invertible and has an inverse element. Am I correct? $\endgroup$ – haunted85 Oct 28 '12 at 17:40
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    $\begingroup$ If there is no $b$ that works universally, then you are right. But there is a $b$ that works universally. Note that if $b=4$, then $3b\equiv 2\pmod{10}$, and calculation will show that $(3a-1)b\equiv 2a+4$ for all $a$. $\endgroup$ – André Nicolas Oct 28 '12 at 17:44
  • $\begingroup$ I am very sorry, I just can't see it. If you set $b = 4$ then $(2a-4) \equiv (2a+4)$. Let $a = 1$ then $(2-4) \equiv (2+4) \Rightarrow -2 \equiv 6 \Rightarrow 8 \equiv 6$, which is not true. And it's not clear to me why that $3b\equiv 2\pmod{10}$ is important. $\endgroup$ – haunted85 Oct 28 '12 at 18:06

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.


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