Is S={1,2,3,...,10} under the binary operation x∗y=x a group? I'm a bit confused as to how to go about proving if this is a group. First of all the question says it is a binary operation, but is it? I would've thought a binary operation has to make some of use of both x and y, but this question just says xy= always x, regardless of y. How would I go about trying to prove this has an identity, since xe=x for any e? Or would I say that actually y is the identity?
And from there, how would I go about trying to find an inverse? If I say my identity is y, then I would need an inverse B so that xB=e=y. But since xy=x, there's no way I could ever actually have x*y=y. Does this mean there is no inverse? Thanks for the help. 
 A: The answer to this question might be that it's not a group. Indeed, if there were an identity element $e$, it would have to satisfy $e * n = n * e = n$ for each $n \in S$. But by definition, $e * n = e$, so this would mean that $e = n$ for each $n \in S$.
A: The operation is a binary operation. Contrary to what you expect, the term "binary operation" does not contain a requirement that both inputs are "used" by the operation -- anything that takes two inputs from some set and produces a single output in the same set is a binary operation. It doesn't matter how that single output comes about, as long as each pair of inputs will consistently lead to one and only one output.
Your set with this operation is, however, not a group, since there is no element that works as an identity for it.
(On the other hand, the operation is associative, which is at least something, though it doesn't make it a group operation).
A: In fact, you may stop at the identity (don't bother with inverses), since we should have, for any $x$, that
$$
x=xe=ex=e
$$
and there is clearly no such $e$.
A: In a group you have to prove that $$xe=x ,\forall x$$ but also $$ex=x ,\forall x$$ And in this case the second identity is never true (you can try every value of S for e).
So since there is no identity element, you can't define an inverse. You only have a semigroup (https://en.wikipedia.org/wiki/Semigroup).
