# 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?

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$x + 4 \cdot 1_{\mathbb{Z}_6} + 2 = x$ holds, since $4 + 2 = 0$. – Cocopuffs Jul 5 '12 at 16:22
@Cocopuffs I am afraid I don't understand what you mean. – haunted85 Jul 5 '12 at 16:24
@haunted85 We're working modulo $6$, so $4 \cdot 1 + 2 = 0$. – Cocopuffs Jul 5 '12 at 16:27

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 $a\in\mathbb Z_6$.

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how have you come up with $e \in \{1,4\}$? Have you just solved the equation treating $e$ as an unknown quantity or is it a deduction you made by thinking? Is there a way to know for sure if $1$ and $4$ are all and only right identities? – haunted85 Jul 5 '12 at 16:41
@haunted85 Since we're working modulo $6$, you can easily check all possibilities as there are only six of them ($0,1,2,3,4,5$). However, with enough experience working with modular arithmetic the solutions to an equation modulo some small $n$ are usually obvious. – Alex Becker Jul 5 '12 at 16:50

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 $\mathbb Z_6$).

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