Let the operation $\odot$ be defined in $\mathbb Z_6$ as follows:
$a \odot b = a +4b+2$
check if $(\mathbb Z_6, \odot)$ is a semigroup and if the identity element belongs to it.
This is the way I have solved this exercise:
Let $x,y,z \in \mathbb Z_6$ then in order for $(\mathbb Z_6, \odot)$ to be a semigroup, the following condition must be met:
$(x\odot y)\odot z = x\odot (y\odot z)$
Considering only the first part of the equation:
$\begin{aligned} (x\odot y)\odot z &= (x+4y+2)\odot z \\ &= (x+4y+2)+4z+2 \\ &=x+4y+4z+4 \end{aligned}$
now considering the second part of the equation:
$\begin{aligned} x\odot (y\odot z) &= x \odot (y+4z+2) \\ &= x+4(y+4z+2)+2 \\ &= x+4y+16z+10 \\ &= x+4y+4z+4 \end{aligned}$
So I conclude stating that $(\mathbb Z_6, \odot)$ is a semigroup. When it comes to verifying the presence of the identity element within the semigroup, some confusion arises:
$x \odot 1_{\mathbb Z_6} = x+4\cdot 1_{\mathbb Z_6} + 2 \neq x $
and also
$1_{\mathbb Z_6} \odot x = 1_{\mathbb Z_6} +4x+2 \neq x$
so the identity element does not belong to $(\mathbb Z_6, \odot)$. Is my solution right or am I wrong?