Tell if $(\mathbb Z_6, \odot)$ is a semigroup and if the identity element belongs to it 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?
 A: I suppose addition and multiplication is interpreted modulo 6. (Otherwise it would not be a binary operation on $\mathbb Z_6$.)
I guess you have to find out whether the given semigroup has identity.
This means: Is there an element $e$ such that $a\odot e=e\odot a=e$ for all elements.
$a\odot e=a$ means
$$a+4e+2=a\\4e+2=0.$$
We can easily check that this is fulfilled by $e\in\{1,4\}$. So this semigroup has two right identities.
Since there can be only one identity element, there cannot be left identity. But we can check this anyway. 
Existence of left identity $e$ would mean that for each $a$ we have $e\odot a=a$, i.e.
$$e+4a+2=a\\e=4+3a.$$
The expression $4+3a$ has various values for various $a$'s (namely the values $1$ and $4$), so there is no element $e$ fulfilling this for each $a\in\mathbb Z_6$.
A: Recall that $\rm\:e\:$ is a right identity (or neutral) element for $\bigodot$ if $\rm\,x\bigodot e = x,\,$ for all $\rm\,x,\,$ and similar for a left identity element. You need to check if these equations have solutions for $\rm\,e.\,$ Note that the identity element for this operation need have no relationship to identity elements for other operations (such as the two-sided identity $1$ for multiplication in $\mathbb Z_6$).
