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)?

  • $\begingroup$ Note that $1 \notin E$ alone does not imply that $E$ doesn't have an identity (consider $\{0\}$...). $\endgroup$ – fkraiem May 14 '14 at 9:50
  • $\begingroup$ @fkraiem Doesn't identity refer to: $I*A=A=A*I$, so $A \ne 0$, could only refer to identity $I=1$ for multiplication? $\endgroup$ – Display Name May 14 '14 at 9:54
  • $\begingroup$ @fkraiem was that just a comment about the general case and not mine? $\endgroup$ – Display Name May 14 '14 at 10:06
  • $\begingroup$ $1$ being the identity for multiplication in $\mathbf{Z}$ only means that $1\cdot a = a\cdot 1 = a$ for all $a$ in $\mathbf{Z}$. Given a proper, non-trivial, subset $E$ of $\mathbf{Z}$, it is not immediately clear why there can't be an element $e \in E$ such that $e\cdot a = a\cdot e = a$ for all $a$ in $E$. And it is certainly not the case in arbitrary rings, in order to show that it is true in $\mathbf{Z}$ you need to use additional properties of $\mathbf{Z}$. $\endgroup$ – fkraiem May 14 '14 at 10:06
  • $\begingroup$ So by just saying that $1\notin E$, you have not proved anything, there are tons of counterexamples, $\{0\}$ being the simplest one : $1$ is certainly not in it, and it has an identity (which is $0$). $\endgroup$ – fkraiem May 14 '14 at 10:12


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.

  • $\begingroup$ Can you please further that hint. I did easily show that $e\ne0$, but I am not sure how to prove the second equality. $\endgroup$ – Display Name May 14 '14 at 9:59
  • $\begingroup$ @DisplayName I thought of a cleaner way to do it. Does my updated answer help? You will have to use facts about the integers to prove this. $\endgroup$ – 6005 May 14 '14 at 10:02
  • $\begingroup$ Is it because $e=2k, k\in\mathbb{Z}$ and $e^2 = 2l,l\in\mathbb{Z}$, so $e=e^2$ but $e^2 \gt e$? $\endgroup$ – Display Name May 14 '14 at 10:03
  • $\begingroup$ @DisplayName That works, but the easier way is to use that if the product of two integers is $0$, one of those integers must be $0$. $\endgroup$ – 6005 May 14 '14 at 10:04
  • $\begingroup$ I see what you have done now, very clever! Thank you for that! To finish your statement: Either $e$ is odd($e-1=0$), or $e=0$, contradiction, as $e$ is even, and $e\ne0$ $\endgroup$ – Display Name May 14 '14 at 10:07

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