# How many binary operations with a zero element can be defined on a set $M$ with $n$ elements in it?

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?

• 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? – user1502 Feb 23 at 12:05
• If you have two different zero elements, $e$ and $f$, you would have to have $e*f=e$ and $e*f=f$, a contradiction. – Ross Millikan Feb 23 at 14:23
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 $$n^{n^2-2n+1}$$.