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Let's say I have a set $M$ of $n$ elements, with $n \in \mathbb{N}^*$. I have to answer the following question:

How many binary operations with a zero element can be defined on $M$?

I know that $*$ is a binary operation on $M$ only if $\forall x, y \in M$ we have that $x * y \in M$. I also know that the binary operation $*$ has a zero element on $M$ only if there is an element $e \in M$ such that $x * e = e * x = x$, $\forall x \in M$.

I don't know how to use this information to find the answer to the question. How can I find how many binary operations with a zero element can be defined on $M$? I found a similar question here, but it's simply about the number of binary operations, without requiring a zero element. How does the necessity for a zero element change the answer?

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Hint: Can you prove that there can only be one zero element? How many ways are there to pick the zero element? How many results of the operation are determined once you pick the zero element? Then use the question you linked to for the rest.

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  • $\begingroup$ By answering your last $2$ questions I was able to find the answer, so thank you for that. However, I was unable to answer your first question. How can I prove that there can only be one zero element? I never thought about this before. Is that a requirement in a binary operation? $\endgroup$ – user1502 Feb 23 at 12:05
  • $\begingroup$ If you have two different zero elements, $e$ and $f$, you would have to have $e*f=e$ and $e*f=f$, a contradiction. $\endgroup$ – Ross Millikan Feb 23 at 14:23
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The Cayley table will have $n^2$ places to be filled, each with any of the $n$ elements of $M$. But $2n-1$ entries are determined. So $n^{n^2-2n+1}$.

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