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I am currently working on the following question:

$\mathbb{Z}n$ is the set of equivalence classes determined by division mod $n$ for $n\ge 1$, $n \in \mathbb{N}$. Let $n$ = $11$. Consider the binary option $*$ on $\mathbb{Z}_{11}$ defined by,

$x*y = x\oplus_{11}y\oplus_{11}2$ for all $x, y \in \mathbb{Z_{11}}$.

Recall that by definition $x\oplus_{11}y = x + y$ mod $11$.

  • What is the identity element for $(\mathbb{Z}_{11},*)$?
  • Prove that $(\mathbb{Z}_{11},*)$ is a monoid
  • Which elements of $(\mathbb{Z}_{11},*)$ are invertible?
  • Is $(\mathbb{Z}_{11},*)$ a group?

I can write $x*y$ as $((x + y)mod11+2)mod11$.

To find the identity element, I must find an element $e$ of $\mathbb{Z}_{11}$, such that $e*x=x*e=x$ for all elements $x$ of $\mathbb{Z}_{11}$. Can I just test different values of $e$ until I arrive at the solution?

Once I have the identity element, I need to show that the binary operation is associative in order to show that $(\mathbb{Z}_{11},*)$ is a monoid. I am not sure how to approach this. I have read "Proving ... is a monoid" but am unsure as to how to apply the method in this case.

As for the remaining parts, I'll hopefully get them once I am sure of the method for the first two parts.

Corrections and guidance appreciated. Thank you!

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  • $\begingroup$ Yes. Try values of $e$ in $\mathbb{Z}_{11}$ until you arrive at a solution. For monoid part, use $xmodn + ymodn = (x+y)modn = (xmodn+ymodn)modn$. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 17 '16 at 15:44
  • $\begingroup$ @expiTT--I--1z0 Ok. I've gotten, 9, as the identity element. For the monoid proof, do I expand both sides of: (xy)*z = x*(yz) and show that they are equal? $\endgroup$ – Smaointe Nov 17 '16 at 17:10
  • $\begingroup$ Yes. That's correct. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 17 '16 at 18:07
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Hint: The map $f\colon x\mapsto x\ominus_{11} 1$ has the property that $f(x*y)=f(x)\oplus_{11}f(y)$.

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