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Let $T=\mathbb Z_7 \times \mathbb Z_{10}$ and let $\odot$ be the operation defined as follows:

$$\begin{aligned} (a,b)\odot(c,d) = (2+a+c, 3bd)\end{aligned}$$

Find the identity element, the inverse elements and characterize all the invertible elements for $(T, \odot)$.

Identity element

In order to find the identity element for $T$:

$$\begin{aligned} (a,b)\odot(e,\varepsilon) = (2+a+e, 3b\varepsilon)\end{aligned}$$

which leads to

$$\begin{aligned} 2+a+e=a \Rightarrow2+e=0\Rightarrow e=-2 = 5 \text{ (mod 7) }\end{aligned}$$

$$\begin{aligned} 3b\varepsilon=b \Rightarrow3\varepsilon=1 \Rightarrow \varepsilon=7 \text{ (mod 10) }\end{aligned}$$

hence the identity element is $(5,7)$.

Inverse and invertible elements

It's time to find the inverse elements, therefore the following needs to happen:

$$\begin{aligned} (a,b)\odot(\alpha,\beta) = (5,7)\end{aligned}$$


$$\begin{aligned} 2+a+\alpha = 5 \Rightarrow \alpha = 5 -2 -a = 5+5-a=3-a \end{aligned}$$

$$\begin{aligned} 3b\beta = \frac{7}{3}b^{-1} = 9b^{-1} \end{aligned}$$

does it suffice stating that as $10$ is a non-prime number, $\mathbb Z_{10}$ has zero-divisors, so not all the elements are invertible, in particular:

$$\begin{aligned} (\forall (a,b) \in \mathbb Z_7 \times \mathbb Z_{10}) \text{ } \exists b^{-1} \in (\mathbb Z_7 \times \mathbb Z_{10}, \odot) \Leftrightarrow gcd(b,10)=1 \end{aligned}$$

Is my conclusion right? If so, is there any other way (perhaps more elegant) to say just that?

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When finding the identity element, the cancellation $3bε=b \implies 3ε = 1$ won't hold unconditionally because $\Bbb Z_{10}$ is not an integral domain. –  user2468 Jul 15 '12 at 16:37
So how do I get to $\varepsilon$ in this case? –  haunted85 Jul 15 '12 at 16:40
$3ε=1$ if $\text{gcd}(b, 10) = 1,$ i.e. when $b^{-1} \pmod{10}$ exists. –  user2468 Jul 15 '12 at 16:45
@J.D. right, so also in that case $gcd(b,10)=1$ should be enough to make myself sure the cancellation is legally done. –  haunted85 Jul 15 '12 at 16:51
To expand on J.D.'s comment, your problems with proving $(5,7)$ is the identity is one of the many problems that arise when you try to use your scratch work as a proof. Once you have decided $(5,7)$ is a good candidate for being the identity, it is nearly trivial to actually prove that it is. But when you tried to do things the way you did, you run into all sorts of technical problems, not to mention making the problem harder than it needs to be. –  Hurkyl Jul 15 '12 at 19:36
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1 Answer

up vote 1 down vote accepted

The computation of the identity is incorrectly justified, as has been noted. From $$3b\varepsilon \equiv b \pmod{10}$$ we can conclude that $(3\varepsilon-1)b\equiv 0\pmod{10}$. This must hold for all $b$. If $\gcd(b,10)=1$, then we know $b$ is invertible modulo $10$, so then we can conclude that $3\varepsilon\equiv 1\pmod{10}$ is necessary. This forces $\varepsilon\equiv 7\pmod{10}$. So the only possible identity will be $(5,7)$. Then you need to check that this works. But note that we cannot go directly from $3b\varepsilon b\equiv b\pmod{10}$ to $3\varepsilon\equiv 0\pmod{10}$ when we make no assumptions about $b$.

It's bad form to start using symbols like $b^{-1}$ before you have established that such an object actually exists. Rather, you get to $3b\beta\equiv 7\pmod{10}$. At this point, you make the observation that this requires that $\gcd(b,10)=1$ (since $1=\gcd(7,10)=\gcd(3b\beta,10)$, and $\gcd(b,10)$ divides $\gcd(3b\beta,10)$). So you must have $\gcd(b,10)=1$, which is now a necessary condition. To show it is sufficient, you note that setting $\beta=9r$, where $br\equiv 1\pmod{10}$, will do.

The final formula is incorrectly written, since you are not looking for $b^{-1}\in\mathbb{Z}_7\times\mathbb{Z}_{10}$. Rather, what you want to write is:

$$\forall (a,b)\in\mathbb{Z}_7\times\mathbb{Z}_{10}\Bigl(\exists x\in\mathbb{Z}_7\times\mathbb{Z}_{10}((a,b)\odot x=x\odot(a,b)=(2,5))\iff \gcd(b,10)=1\Bigr).$$ Note that the biconditional goes with the existence, and that what you are looking for is not $b^{-1}$, but rather

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