Question on Operations

[bZNyQ7C2]

This question is about abstract algebra, specifically binary structures. I'm trying to work through Charles Pinter's A Book of Abstract Algebra. One of his early exercises asks us to list all 16 operations that are possible on a two-element set $\displaystyle \left\{ {a,b} \right\}$ and to say (among other things) which of them have identity elements.

Here is one of those 16. Only one of the 16 is a group, and this isn't it. I present it the way Dr. Pinter asks his readers to do.

$\displaystyle \begin{array}{l} \left( {a,a} \right) \to a\\ \left( {a,b} \right) \to a\\ \left( {b,a} \right) \to a\\ \left( {b,b} \right) \to b \end{array}$

I am not sure whether this structure has an identity element. To me, it looks like $\displaystyle a$ has two left identity elements and two right identity elements as well, but $\displaystyle b$ seems to have only one two-sided identity, which is itself. I know that in a group we can't have two identities, but this structure isn't a group and isn't represented as a group.

Should I say that this structure has no identity element, in which case we don't have to ask the question about whether there are inverses, or should I say that $\displaystyle a$ has two identities but $\displaystyle b$ has only one, in which case $\displaystyle a$ has no inverse but $\displaystyle b$ does?

Or am I missing the point altogether?

 

[HallsofIvy]

Yes. there are 2*2= 4 ordered pairs of two objects and there are then $\displaystyle 2^4= 16$ values that can be assigned to those so there are 16 binary operations. "e" is an identity if and only if (a, e)= (e, a)= a and (b, e)= (e, b)= b. Obviously if there is an identity it is either a or b. If the identitoy were a then we would have to have (a, a)= a and (a, b)= (b. a)= b. Is that true? If the identity were b then we would have to have (a, b)= (b, a)= a and (b, b)= b. Is that true?

 

[Dr.Peterson]

I think it's easiest to see if there is an identity element by looking at an operation table:

    a   b
  +-------
a | a   a
b | a   b

The row and column for an identity element both have to match the headings. Is that true for either element here?

You said,

To me, it looks like $\displaystyle a$ has two left identity elements and two right identity elements as well, but $\displaystyle b$ seems to have only one two-sided identity, which is itself. I know that in a group we can't have two identities, but this structure isn't a group and isn't represented as a group.

A left identity element must yield e*x = x for every element x, not just for some one element x. It is meaningless to speak of an element x having an identity. If there were two left identities, then there would be two entire rows in the table that read "a b".

 

[bZNyQ7C2]

Thanks to both of you for your help.

I want to be sure I understand here. "If the identity were a then we would have to have (a, a)= a and (a, b)= (b. a)= b. Is that true? " Well, (a,a) = a but that is not enough; we must also have (a,b) = (b,a) = b and they don't, so a is not the identity. On the other hand, "If the identity were b then we would have to have (a, b)= (b, a)= a and (b, b)= b. Is that true?" Yes, both those statements are true. So b is the identity. Am I right?

If we go on to the question of inverses, it looks like b has an inverse which is b, because (b,b) = b, but a does not have any inverse, because neither b nor a, combined with a, will produce b. Since this structure is associative and has an identity and is even commutative, it is only the lack of an inverse for a that prevents it from being a group. Am I right there?

One more thing. Should I understand what you two are saying about identities to be true by definition, or does any part of it need to be proved?

 

[Steven G]

Thanks to both of you for your help.

I want to be sure I understand here. "If the identity were a then we would have to have (a, a)= a and (a, b)= (b. a)= b. Is that true? " Well, (a,a) = a but that is not enough; we must also have (a,b) = (b,a) = b and they don't, so a is not the identity. On the other hand, "If the identity were b then we would have to have (a, b)= (b, a)= a and (b, b)= b. Is that true?" Yes, both those statements are true. So b is the identity. Am I right?

Yes, what you said is 100% correct. Just to be a bit more clear, e is the identity element if eg=ge=g for EVERY element g in the set under the given operation.

If we go on to the question of inverses, it looks like b has an inverse which is b, because (b,b) = b, but a does not have any inverse, because neither b nor a, combined with a, will produce b. Since this structure is associative and has an identity and is even commutative, it is only the lack of an inverse for a that prevents it from being a group. Am I right there?

Yes, this is not a group as not all the elements have an inverse. Did you really check for associativity? The operation may be commutative but that does not prevent it from being a group. If it were a group and had the commutative property we call this an abelian group or a commutative group.

 

[bZNyQ7C2]

@Jomo ... Thanks for your reply, but I'm still a little confused. In my second paragraph, I claimed that b is the identity for this structure. And you said that was 100% correct.

I got an email notice about your reply to this thread, where you said that this structure does NOT have an identity, but I don't see that in your comment above. I'm guessing that you edited your reply to delete that. Am I correct that you changed your mind, and you now agree that b is the identity for this structure?

You also seem to agree that b has an inverse, but a does not, and that is sufficient to show that we don't have a group here.

Yes, I did check it for associativity. It was a little tedious, but I did it. I'm attaching a pdf file to show you what I did, just in case you want to look at it. If you're not interested, I will understand. This structure is number 5. Number 7 is the one that is a group. I'm about halfway through the exercise.

Hmm ... I'm still wondering if what you guys say about the identity is the definition, or whether any of it needs to be proved.

 

[Steven G]

I thought that i remembered that the Cayley table did not have an id elt, so I initially stated that. But then I felt that I should look to be sure and I realized I was wrong (I really do not want to go back to the corner for this!) so I updated my reply.

 

[bZNyQ7C2]

Thanks. If it turns out you are wrong as often as I am, I will be very surprised.

 

[Dr.Peterson]

Hmm ... I'm still wondering if what you guys say about the identity is the definition, or whether any of it needs to be proved.

What definition were you given? What part do you think would need to be proved?

 

[Steven G]

Thanks. If it turns out you are wrong as often as I am, I will be very surprised.

I do make many silly mistakes but I catch most of them before opening my mouth (or my keyboard). Always check what you do to be sure that you believe it. It works. Knowing you definitions also helps!