# Invertible elements and identity element in $\mathbb Z_6$ under the circle product

Let's define the operation $\odot$ on $\mathbb Z_{6}$ as follows:

\begin{aligned}x\odot y=x+y+xy\end{aligned}

Show:

• $(\mathbb Z_6, \odot)$ is a monoid;
• if $3 \in \mathbb Z_6$ is an invertible element, if so use an appropriate congruential equation to determine its invertible element;
• define all the invertible elements in $\mathbb Z_6$.

I've started proving the last two items first. The element $3$ is not an invertible element in $\mathbb Z_6$ because $gcd(3,6) \neq 1$.

In order to define all the invertible elements in $\mathbb Z_6$ I thought it would be enough to enumerate all the elements $[a] \in \mathbb Z_6 : gcd(a,6)=1$. So I concluded all invertible elements in $\mathbb Z_6$ are:

\begin{aligned}\{1+6k:k\in\mathbb Z\} = 1+6 \mathbb Z\end{aligned} and \begin{aligned}\{5+6k:k\in\mathbb Z\} = 5+6 \mathbb Z\end{aligned}

In order to prove $(\mathbb Z_6, \odot)$ to be a monoid, we require $\odot$ to be associative and $\exists \mathbb 1_{\mathbb Z_{6}} : x \odot \mathbb 1_{\mathbb Z_{6}} = \mathbb 1_{\mathbb Z_{6}} \odot x = x$

Associativity

\begin{aligned}(a \odot b) \odot c = a \odot (b \odot c)\end{aligned} \begin{aligned}(a+b+ab) \odot c = a \odot (b+c+bc)\end{aligned} \begin{aligned}a+b+ab+c+ac+bc+abc = a+b+c+bc+ab+ac+abc\end{aligned}

And the point is proved.

Identity element

\begin{aligned}a \odot \mathbb 1_{\mathbb Z_{6}} = a\end{aligned} \begin{aligned}1_{\mathbb Z_{6}} + a1_{\mathbb Z_{6}} = 0 \end{aligned}

... and now I am stuck because I don't know how to proceed in order to evaluate $1_{\mathbb Z_{6}}$.

Is it correct stating $1_{\mathbb Z_{6}} + a1_{\mathbb Z_{6}} = 0 \Rightarrow a1_{\mathbb Z_{6}} = 0$? How do I take it from here?

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$\mathbb{Z}_6$ is an abelian group, so it is customary to write its identity element as $0$ - in this case, it is the residue class of $0\mod 6.$ Now there is no guarantee that the identities for $\odot$ and modular addition will coincide, but why don't you try it and see what you get. –  user17794 Jul 9 '12 at 16:44
Further, it's impossible to answer the last two questions without verification of the first. The notion of invertability for this monoid is not the same as for multiplication of residues. –  user17794 Jul 9 '12 at 16:47
I think the OP is (confusingly!) using $1_{\mathbb{Z}_6}$ to refer to a hypothetical identity of the structure under consideration, not the identity of $\mathbb{Z}_6$. –  Hurkyl Jul 9 '12 at 17:03

You should not answer the second question first. Your answer is assuming you are talking about the usual multiplicative structure of $\mathbb{Z}_6$, but that is not the case. Once you prove that $(\mathbb{Z},\odot)$ is a monoid, the question is asking you to determine whether $3$ is invertible relative to the operation $\odot$. So checking whether $\gcd(3,6)=1$ may not do you any good. Likewise, the third question is not really asking you to find the elements of $\mathbb{Z}_6$ that are invertible under the usual multiplication, but rather to find the elements that are invertible relative to the operation $\odot$. So you've correctly answered a different question from what you are being asked, and incorrectly answered the question you were really asked.

So, let's talk about $\odot$.

You've shown that $\odot$ is associative, so that means $(\mathbb{Z}_6,\odot)$ is a semigroup. To prove that it is a monoid, you need to find some element $z$ of $\mathbb{Z}_6$ with the property that $a\odot z=z\odot a = a$ for all $a$. So you need $$a = a\odot z = a+z+az.$$ Therefore, we need $a=a+z+az$, which means $0=z+az = z(1+a)$. This has to hold for every $a\in\mathbb{Z}_6$. So... what must $z$ be?

(Note that since $b\odot a = b+a+ba = a+b+ab=a\odot b$, you in fact can save yourself a bit of work; you don't actually need to check that $z\odot a= a$ also holds).

Once you determine what $z$ is, the second part of the question is asking you to determine whether there exists $a\in\mathbb{Z}_6$ such that $3\odot a = z$.

Finally, the problem asks you to figure out all elements $a\in\mathbb{Z}_6$ with the property that there exists $b\in\mathbb{Z}_6$ such that $a\odot b = z$.

Now, general comments. Do not read this until you attempt the problem as outlined above.

The operation you are talking about is known as the "circle operation" (sometimes it is defined as $x\odot y = x+y-xy$, in which case some of the pluses signs below need to be turned into minuses, and vice versa). In a ring, the elements that are invertible under the circle operation are known as "quasi-units", and the group of quasi-units of the ring is known as the "circle group" of the ring. (Associativity of the circle operation guarantees that if $x$ and $y$ are quasi-units, then so is $x\odot y$).

If $R$ is a ring, the identity of the $\odot$ operation is $0_R$, since $x\odot 0_R = x+0_R + x0_r = x$ and $0_R\odot x = 0_R + x + 0_Rx = x$. If $R$ is commutative, then the circle monoid $(R,\odot)$ is commutative.

When $R$ has a multiplicative identity, $1_R$, then the circle group of the ring is isomorphic to the group of units of $R$ by the map $x\mapsto 1_R+x$.

Indeed, if $x$ is quasi-invertible, then there exists $y$ such that $x\odot y = 0_R$ (and $y\odot x = 0_R$ when we don't assume $R$ is commutative). Then $x+y+xy=0_R$, so $$(1_R+x)(1_R+y) = 1_R + x + y + xy = 1_R.$$

Conversely, if $z$ is a unit in $R$, then there exists $w$ such that $zw=1_R$. Then $$(z-1_R)\odot(w-1_R) = (z-1_R)+(w-1_R) + (z-1_R)(w-1_R) = z+w-1_R-1_R+zw - z-w-1_R = 0_R$$ so $z-1_R$ is a quasi-unit. Thus, the map is a bijection between quasi-units and units.

Finally, note that $x\odot y + 1_R = x+y+xy+1_R$, and $(x+1_R)(y+1_R) = xy +x+y+1_R$, so the map is a group homomorphism.

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first of all thank you for your making well clear that the operations are relative to the newly defined operation, but now I feel a bit lost because 'til now I've only managed to manipulate in $\mathbb Z_n$ and the vanilla-flavoured operations. So in regard to the identity element, I think that it must be $z = 0$, but now a new problem arises: $3 \odot a = 3+a+3a = 0$: how do I attempt to solve such equation? I guess I can't proceed like I would in $\mathbb Z$ as $$a = -\frac{3}{4}$$ which I think it's not acceptable, right? –  haunted85 Jul 10 '12 at 8:08
@haunted85: You solve it with congruences. Does there exist an $a$ such that $3+a(3+1)\equiv 0\pmod{6}$? you need $4a\equiv -3\equiv 3\pmod{6}$. Can a multiple of $4$ be congruent modulo $6$ to $3$? All you have to do is check, if you can't think of a better way: test $a=0,1,2,3,4,5$ and see if there is a solution. If not, then $a$ is not $\odot$-invertible. –  Arturo Magidin Jul 10 '12 at 14:23
thank you, this has helped a lot... I have to ask, what if I had $mod 38$ instead of $mod 6$, in order to verify if there's a solution to that equation, would it be ok using the Euclidean method? Now that I've made myself sure that none of those values is a good fit as a solution how do I get to find out all the invertible elements in $\mathbb Z_{6}$ –  haunted85 Jul 10 '12 at 14:39
@haunted85: The modular equation $ax\equiv b\pmod{n}$ has a solution if and only if $\gcd(a,n)$ divides $b$. This holds no matter what $b$ is. As to finding the invertible elements, you want to find the $a$ for which there exists $b$ such that $a\odot b = 0$ (since $0$ is the $\odot$-identity). This amounts to finding what values of $a$ will allow you to solve $a+b+ab\equiv 0\pmod{6}$. You can write this as $(1+a)b\equiv -a\pmod{6}$. You know that $a\equiv 0$ is possible (take $b\equiv 0$. You know that $a\equiv 3$ is impossible. Check the others; not that many to try, so just try them. –  Arturo Magidin Jul 10 '12 at 23:24