Identity element

Originally posted by adam warlock
Consider that the set has n elements: a_1, a_2,..., a_n

Take the "product" a_1*a_2*...*a_n.

From this, the result will follow.

So you're saying I should have done what I did for the type II elements to the entire set. Ok. give me a few hours.

If the set consists of only type I and type II elements then if z is all the elements multiplied together, by my earlier argument z² = z (and z is a type I element). Suppose the set is a, b, c, d, ..., k, l

let Ω(a) = b*c*d*...*k*l
a*Ω(a) = z
So Ω( · ) is doing the job of an inversion operator. Which doesn't yet prove anything but I can see how I could get there.

But I'm still worried about type III elements. Suppose the set consists of a, b, c and a² = b; b² = c; c² = c; then
z = a*b*c
z² = a*a*b*b*c*c = b*c*c = b*c != z (as a is not an identity element).
I'll check to see that this doesn't generate a contradiction.

Let S = {a, b, c} and a*a = b; b*b = c; c*c = c

Then consider z = a*b*c

A) a*z = a*a*b*c = b*b*c = c*c = c
B) b*z = b*a*b*c = a*b*b*c = a*c
C) c*z = c*a*b*c = a*b*c*c = z

Now there are three cases, for your axioms to be sufficient I should be able to generate a contradiction for each possibility for z.

1) z = a
From A we have: a*z = a*a = c, but a*a = b from my earlier definition. So we have a contradiction.
2) z = b
From A) a*z = c => a*b = c
From B) b*z = a*c => b*b = a*c, b*b = c => a*c = c
From C) c*z = z => c*b = b
I haven't generated a contradiction and the multiplication table is complete:
*|a b c
----------
a|b c c
b|c c b
c|c b c
Which is commutative and complete. Consider the product:
a*a*c
(a*a)*c = b*c = b
a*(a*c) = a*c = a
So the table isn't associative.

3) z = c
From A) ac = c
From B) bc = ac = c
From C) c*c = c
The multiplication table is incomplete (a*b is not specified) I'll check all multiplication tables for associativity.

Originally posted by DeepThought
So you're saying I should have done what I did for the type II elements to the entire set. Ok. give me a few hours.

If the set consists of only type I and type II elements then if z is all the elements multiplied together, by my earlier argument z² = z (and z is a type I element). Suppose the set is a, b, c, d, ..., k, l

let Ω(a) = b*c*d*...*k*l ...[text shortened]... as a is not an identity element).
I'll check to see that this doesn't generate a contradiction.

I have forgotten a lot of my group theory, but unless I'm mistaken, using the hint we have that

a_1 * ... * a_n = a_k for some 1 <= k <= n (by finiteness)
which implies a_k * a_1 * ... * a_{k-1} * a_{k+1} * ... * a_n = a_k (by commutativity of *)
which implies a_k * (a_1 * ... * a_{k-1} * a_{k+1} * ... * a_n) = a_k (by associativity of *)
which implies (a_1 * ... * a_{k-1} * a_{k+1} * ... * a_n) = 1_k (the identity element for a_k)

(I don't like saying "identity element for a_k", but as for whether it is an identity for all other elements (aka, the identity) I'm still working on it ...)

Carrying on for case (3) in my above post,
we have a*a = b; b*b = c; c*c = c
a*b*c = c;
consequently
a*c = c
b*c = c
Now consider a*a*b
(a*a)*b = b*c = c
a*(a*b) = c by associativity
i) a*b = a
then a*(a*b) = a*a = b != c so a*b != a
ii) a*b = b
then a*(a*b) = a*b = b != c so a*b != b
iii) a*b = c
then a*(a*b) = a*c = c which doesn't contradict our other results, we have as the only candidate for consistency:
a*a = b; b*b = c; c*c = c
a*b = c; a*c = c; b*c = c

This is associative as the only result of a product which is not c is a*a = b, and whatever that is multiplied by will come out as c.

This is a counterexample. The set {a, b, c} with a product with the multiplication table
*|a b c
---------
a|b c c
b|c c c
c|c c c
is commutative, associative and closed under the multiplication operator. There is an axiom missing.

The post that was quoted here has been removed

😴😴😴

To summarise my current thinking:
The case in the above post indicates we need an axiom to rule out Type I elements.

A) S is a finite set.
B) S is closed under the product *
C) for all a, b in S, a*b = b*a
D) for all a, b, c in S, (a*b)*c = a*(b*c)
We need an extra axiom to rule out elements of type III, for example:
E) for all a in S there exists b such that b*b = a;

Suppose there are n subsets which form loops on squaring, i.e. a*a = b; b*b = c; c*c = a; but potentially with more elements. As I showed earlier (see the post on page 1) for each of these subsets there must be one element (I haven't shown it's the same for each one yet) formed from the product of each element in the subset, which is not in the subset and is an identity with respect to elements in that subset.

The product of all the elements in the set then has the property that z*z = z, as there are only type I (i.e. x*x = x) elements or type II elements (i.e. a*b*c * a*b*c = a*a*b*b*c*c = b*c*a = a*b*c). Suppose the set has eight elements a,b,c,d,e,f,g,h
d*d = d and e*e=e
a*b*c = z1
z1*z1 = a*a*b*b*c*c = z1
f*g*h = z2, z2*z2 = z2

Consider the product of elements z.
z = a*b*c*d*e*f*g*h
z*z = z1*z1*d*d*e*e*z2*z2 = z1*d*e*z2 = z

By closure z, z1 and z2 can each be one of d and e. Without loss of generality take z = e
Now there are three essentially different cases. z1 = e and z2 = e or z1 = d, z2 = e or z1 = d, z2 = d

Case 1) z = e = a*b*c*d*e*f*g*h = z1*d*e*z2 = e*d*e*e = d*e
Case 2) z = e = d*d*e*e = d*e
Case 3) z = e = d*d*e*d = d*e

So however we choose them d is an identity element for e.
For case 1)
a*e = a*(a*b*c) = a*a*b*c = b*b*c = c*c = a
since e = a*b*c; so a*e = a; b*e = b c*e = c
since e = f*g*h; we also have f*e = f; g*e = g and h*e = h
so the only problem element is d
For case 2 and 3)
a*e = a*e*d = a*(a*b*c)*e = a*e (I haven't shown it's an inverse)

So even with this extra axiom there's a problem

Originally posted by DeepThought
Carrying on for case (3) in my above post,
we have a*a = b; b*b = c; c*c = c
a*b*c = c;
consequently
a*c = c
b*c = c
Now consider a*a*b
(a*a)*b = b*c = c
a*(a*b) = c by associativity
i) a*b = a
then a*(a*b) = a*a = b != c so a*b != a
ii) a*b = b
then a*(a*b) = a*b = b != c so a*b != b
iii) a*b = c
then a*(a*b) = a*c = c which doesn't contra ...[text shortened]... mmutative, associative and closed under the multiplication operator. There is an axiom missing.

Thinking about it all I've done here is show that a = b. So, consider the two element set {a,b} with a*a = b; b*b = b; a*b = b*a = b
(a*a)*a = b*a = a*b = a*(a*a)
(a*a)*b = b*b = b
a*(a*b) = a*b = b
a*(b*b) = a*b = b
(a*b)*b = b*b = b
a*(b*b) = a*b = b
(b*b)*b = b*b = b
b*(b*b) = b*b = b
So the two element set is commutative, associative, complete but has no identity element. If b were the identity we'd need a*b = a; but we have a*b = b. a can't be the identity since a*a = b.

This is a counter-example, so there is an axiom missing.

The post that was quoted here has been removed

Are you an atomaton? I'm beginning to lose count on the amount of times you have repeated this rhetoric... which leads me to believe a short circuit may exist in your CPU. If you'd like...since you only seem to understand numbers ( and apparently see no need for the understanding of basic human communication principals ), I can enumerate the all the instances of similar ( if not exact ) responses you have made recently in this forum. Then, ( at least I hope ) you may be able to run a self diagnostic program to begin to flush out the obvious glitch in your simulated human personality program. Savvy?

Originally posted by joe shmo
Are you an atomaton? I'm beginning to lose count on the amount of times you have repeated this rhetoric... which leads me to believe a short circuit may exist in your CPU. If you'd like...since you only seem to understand numbers ( and apparently see no need for the understanding of basic human communication principals ), I can enumerate the all the insta ...[text shortened]... ram to begin to flush out the obvious glitch in your simulated human personality program. Savvy?

Just take into account the fact that she never even once addressed the fact that there is a distinction to be made between a group, a set and a set defined with an operation in it (the crucial distinction being between a set with a * operation and a group) and that she failed to grasp those distinctions ending making a correction to my post that was a mathematical non sequitur.

Also I never said that she claimed to be a mathematical genius (she only acts like it). Like she acts like she has a lot of mathematical knowledge but never once made a substantial mathematical remark ever since I came back to this forum (maybe she did one before and I missed it) and till she has the gall to correct me as if she knew what she's talking about.

The post that was quoted here has been removed

Please do that.

And with the free time that you'll end up getting from not having to write such long replies to me take the time to freshen up your abstract algebra, set theory and group theory.

Also I'm not native from Portugal.

Originally posted by Agerg
I have forgotten a lot of my group theory, but unless I'm mistaken, using the hint we have that

a_1 * ... * a_n = a_k for some 1 <= k <= n (by finiteness)
which implies a_k * a_1 * ... * a_{k-1} * a_{k+1} * ... * a_n = a_k (by commutativity of *)
which implies a_k * (a_1 * ... * a_{k-1} * a_{k+1} * ... * a_n) = a_k (by associativity of *)
which implies ...[text shortened]... hether it is an identity for all other elements (aka, the identity) I'm still working on it ...)

Yes that's the easy and elegant proof I was thinking of (DeepThought you also have the right idea but like I said previously your divide and conquer strategy is interesting but is tantamount to be killing flies with a cannon.) My only caveat is that strictly speaking your first step does not only use commutativity but also associativity.

And your parenthetical remark already touches on a point that I want us to tackle. With your proof it is easy to see that every element of the set has an identity element.
But can we prove without further assumptions that such an element is unique?
In order to properly introduce the previous question I think that a few clarifying remarks are needed.

First of all let us talk about what is a group. A group is defined to be a set of elements together with an operation that combines two elements of the said group to produce a third element that is still a member of the group. Furthermore the said operation needs to obey the four group axioms:
1 - Closure
2 - Associativity
3 -Identity element
4 - Inverse element.

Hence for a group the existence of the identity and inverse element have to be postulated. Also take note of the fact that for a group nothing is said about the number of elements of the said group.

In our case we were able to prove the existence of the identity element (does that imply that an inverse element also exists?) but in order to do that we needed our operation to be commutative (a group is a group whether or not the * operation is commutative as long all four axioms are respected) and the proof also used the fact that the set of elements is finite (can we prove that an identity element exists even if the number of elements of the group isn't bounded?).

Hence we have some similarities between our construct and the concept of a group, but we also have some glaring differences. The answer to the previous questions will help us elucidate the gray areas and will also provide the motivation for further questions and ensuing clarifications.

Originally posted by DeepThought
Thinking about it all I've done here is show that a = b. So, consider the two element set {a,b} with a*a = b; b*b = b; a*b = b*a = b
(a*a)*a = b*a = a*b = a*(a*a)
(a*a)*b = b*b = b
a*(a*b) = a*b = b
a*(b*b) = a*b = b
(a*b)*b = b*b = b
a*(b*b) = a*b = b
(b*b)*b = b*b = b
b*(b*b) = b*b = b
So the two element set is commutative, associative, compl ...[text shortened]... can't be the identity since a*a = b.

This is a counter-example, so there is an axiom missing.

Great example @DeepThought! It kinda looks you are reading my mind and are foreshadowing a couple of things I want us to look into in the future (this simple exercise I provided will actually teaches us a lot about groups, sets and operations as we go further down the rabbit hole.)

First, I'll repost the original statement of my problem with an emphasis added because judging by your example and @Agerg's reply, people kinda missed it:

Show that a set with a finite number of elements which is closed under the operation * (* is commutative) contains at least one identity element.

Your construction of such a set is:

{a,b} with
a*a = b
b*b = b
a*b = b*a = b

You end your post with the affirmation "This is a counter-example, so there is an axiom missin". No, this is not a counter example. Your second condition is b*b=b. Hence b is its own identity element. The problem is that you're implicitly assuming one of two things (which are related):
1 - there needs to be more than one identity element
2 - the identity element needs to be unique.

In your construction there is only one identity element(hence it is trivially unique), but in no way this contradicts my problem statement since I've never said that such an element is unique nor that all elements of the set need to have an identity element (and with this you're forcing me to present a spoiler).

Originally posted by adam warlock
Great example @DeepThought! It kinda looks you are reading my mind and are foreshadowing a couple of things I want us to look into in the future (this simple exercise I provided will actually teaches us a lot about groups, sets and operations as we go further down the rabbit hole.)

First, I'll repost the original statement of my problem with an empha ...[text shortened]... the set need to have an identity element (and with this you're forcing me to present a spoiler).

To be an identity element it has to have the property that e*x = x for each element x in the group. Just one isn't enough. That is the problem I was trying to solve and why I was struggling. I should have spotted there was a problem since commutativity implies there is a unique identity element.

Besides my earlier categorization is sufficient, a type I element is a*a=a, if there are type II elements (a*a = b; b*b = c; c*c=a then there has to be an element different from these of type I), type III elements imply the existence of type II elements or type I elements to end the chain on because the group is finite. So I'd shown it, just not in the simple way you expected.

Earlier I suggested eliminating type III elements. That's no good because Z_2 (-1 and 1 with the normal rules of multiplication) consists of an identity and a type III element.

Originally posted by adam warlock
Yes that's the easy and elegant proof I was thinking of (DeepThought you also have the right idea but like I said previously your divide and conquer strategy is interesting but is tantamount to be killing flies with a cannon.) My only caveat is that strictly speaking your first step does not only use commutativity but also associativity.

And your pa ...[text shortened]... ray areas and will also provide the motivation for further questions and ensuing clarifications.

With your proof it is easy to see that every element of the set has an identity element.

I'd dispute that, partly because of my two element example, but also because of the following consideration:

Suppose the set S = {a, b, c, ..., k, l}, and is closed under * which is associative and commutative.
a*b*c*···*k*l is the product of all the elements and is in the set (call it e), then we have:
e = e*(a*b*··*d*f*···*k*l) = e*z
z acts as an identity with respect to e. Pick and arbitrary element, x. Then
e*x = e*x*z
So for the element e*x = y we have that:
y = z*y
This means that z acts as an identity with respect to all elements obtainable by multiplying e by each member of the set; but this is not guaranteed to span the set we need an extra axiom for that.

However suppose z is indeed the identity. Then z = a*b*c*d*f*···*k*l, and so for an arbitrarily selected element c:
z = a*b*c*d*···*k*l
z = c*(a*b*d*···*k*l)
and the product in brackets is the inverse of c. Since c was arbitrarily chosen this shows that each element has an inverse provided there is an identity.

It remains to be shown that if an identity exists then it is unique, and that z = e. Which my divide and rule strategy did for all groups consisting of one cycle of type II elements and a single type I element (the identity).