I don't understand "[x][/2]y= y= y[x][/2]". What does "[x][/2]" mean?
First, let's show that S is a monoid. We have for any x, and any y:
x^{2}y = y = yx^{2}.
We'd like to show that we have a UNIQUE identity, that it doesn't matter which x we square. In particular, for y = x'^{2}, we have:
x^{2}x'^{2} = x'^{2} and with x = x' and y = x^{2} we have:
x^{2}x'^{2} = x^{2}, so any square is equal to any other. We can thus call this unique element of S, e.
But then: x^{2} = e, for any element x of S, so we see that every element of S has an inverse: itself. Thus S is a group.
Since every element of this group is its own inverse, we have:
(ab)^{2} = abab = e, so it follows that:
(ab)^{2} = e = e^{2} = a^{2}b^{2} so that:
a(abab)b = a(aa)(bb)b
(aa)(ba)(bb) = (aa)(ab)(bb)
e(ba)e = e(ab)e
ba = ab, so S is abelian.
1) x^{2}=I & x^{-1}=x, all x, by OP and def of I and inverse
2) x^{2}=I -> x=I, all x
But I can't prove 2), which implies commutativity.
EDIT: I see Deveno has answered OP, but not shown x=I, if it can be shown. I started to read it but the first thing I came across was "monoid," and unnecessary jargon just fries me.
Monoids are semi-groups with identity. They are sort of a "half-way" stage between semi-groups and groups. The natural numbers (with 0) form a monoid under addition, and the set of all functions f:S-->S form a monoid under composition.
It is simply not true that x^{2} = e implies x = e. Here is a counter-example:
Let S be the set of 4 matrices:
$\left\{\begin{bmatrix}1&0\\0&1 \end{bmatrix}, \begin{bmatrix}-1&0\\0&-1 \end{bmatrix}, \begin{bmatrix}-1&0\\0&1 \end{bmatrix},\begin{bmatrix}1&0\\0&-1 \end{bmatrix}\right\}$
This set forms an abelian group under matrix multiplication, and we have for every matrix A in S, that A^{2} = I, but 3 of the matrices are clearly NOT the identity.
OP: S is a semi group.
If for every x,y Є S x^{2}y = y = yx^{2}, show S is Abelian
(xy)y=x, OP
(xy)(yx)=I, x^{2}=I, and then multiply both sides by yx and note (yx)^{2}=I
xy=yx to give commutativity so
group has a closed associative binary operation, Identity, inverse, and commutativity, ie, it is Abelian.
Found a mistake in my proof x=I so will accept your matrix algebra for now. Spare me the definition of irrelevant jargon, which anybody can look up- like a HS kid who has learned a big word and can't wait to use it even though it is unnecessary.
EDIT:I can't show x=I. Thanks Deveno for matrix example but not for "monoids." I do note that Deveno solved OP in post #5, though a bit wordy for my taste but quite clear and precise after I forced myself past "monoid."
In the reals, can you show x^{2}=1 -> x=+-1 necessary (clearly sufficient) without using < or >, or big words?
Monoids are becoming more and more common: they are the preferred structure for describing many constructions in computer programming (strings together with concatenation and the "empty string" form a very "general" monoid). Square matrices form monoids under matrix multiplication (you may recall that we have an identity matrix, but not all matrices are invertible). Monoids, rather than groups, are the natural object to study PROCESSES with, as not ALL processes are "reversible".
In the reals (actually, in any field except a field of characteristic 2) from x^{2} = 1, we obtain:
x^{2} - 1 = (x + 1)(x - 1) = 0.
Since the product is 0, one of the factors must be 0 (the non-zero elements of a field are CLOSED under multiplication), leading to:
x + 1 = 0 --> x = -1, or:
x - 1 = 0 --> x = 1.
(in a field of characteristic 2, 1 + 1 = 0, so 1 = -1, and we only get one root: namely, 1).
If you "mod out the parenthetical remarks", there are no "big words".
I was trying to make the point that OP could be solved strictly by the definitions. Introducing "monoid" confuses that procedure by implication.
But I do appreciate x^{2}=1 -> x=+-1.
Sorry to take away your last word, which you deserve, and for burying your correct clear precise (wordy) solution of OP in post #5. In spite of "monoid."
I thought about using minus sign without definiton in B=-A in your example but I accept it's ok.