Proof of no identity element in a multiplication semi group

Question

I have a question from my workbook:

Let $\mathrm{E}$ denote the set of even integers. This forms a semi-group under multiplication. Show that there is no identity in this semi-group.

Now this is obvious, as $1 \not\subset \mathrm{E}$

How do I prove this problem without using anything but the given multiplication binary operator(without using division or other)?

Answer 1

Hint

Towards contradiction, let $e$ be the identity of this semigroup. Then show that $e \ne 0$. Finally, factor $e = e^2$ as $e(e - 1) = 0$ to derive a contradiction.