Talk:Identity of Algebraic Structure is Preserved in Substructure

Consider the semigroup of $2 \times 2$ square matrices with integral elements. It has an identity element $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Now consider the subsemigroup of $2 \times 2$ square matrices of the form $\begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix}$ for $a \in \Z$. It has an identity element $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ which is not the same as the identity element $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ of its supersemigroup. --prime mover (talk) 23:00, 13 March 2013 (UTC)

Said subsemigroup does not contain the identity of the supersemigroup, which is a premise of this theorem and mentioned in the title. --Dfeuer (talk) 23:03, 13 March 2013 (UTC)

Yes yes I know I know I admit that the statement as given here does not actually say that (it specifically states that the identity of the subsemigroup has to be that of the semigroup, but you have to be careful here. Where are you getting these results from anyway?

And please put your braces in until we have resolved the argument about them, you are NOT EXCUSED, it won't break your neck. --prime mover (talk) 23:05, 13 March 2013 (UTC)

This result is probably too trivial to appear in a text. --Dfeuer (talk) 23:09, 13 March 2013 (UTC)