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:
$\begin{aligned} (a,b)\odot(\alpha,\beta) = (5,7)\end{aligned}$
so
$\begin{aligned} 2+a+\alpha = 5 \Rightarrow \alpha = 5 -2 -a = 5+5-a=3-a \end{aligned}$
$\begin{aligned} 3b\beta = \frac{7}{3}b^{-1} = 9b^{-1} \end{aligned}$
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?