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Math Help - Two semi group problems

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    Two semi group problems

    1. Suppose that S is a semigroup, and that ax=b and ya=b has a solution for every a and b in S, prove that S is a group.

    My proof:

    Since ax=b and ya=b has a solution for every a and b, then ax=a and ya=a has a solution, therefore they are the left and right identity, implies that the identity exist in S. [Question: Do I need to show that y = x?]

    Now, for inverse I'm not so sure if I can cancel any elements by multiplying y^-1 or x^-1, am I allow to do that?

    2. Suppose that S is a finite semigroup such that the cancellation law holds, prove that it is a group.

    Suppose that x,y,z are elements of S such that zx = zy, then x = y.

    then z^{-1}zx=z^{-1}zy, now, am I allow to cancel any elements here? I guess I can't since I haven't find the inverses yet...

    And why does it have to be finite? What if it is infinite semigroup?

    Thank you very much for the helps!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    1. Suppose that S is a semigroup, and that ax=b and ya=b has a solution for every a and b in S, prove that S is a group.

    My proof:

    Since ax=b and ya=b has a solution for every a and b, then ax=a and ya=a has a solution, therefore they are the left and right identity, implies that the identity exist in S. [Question: Do I need to show that y = x?]
    Yes you do need to show that y=x. Also, you have found an element that acts like an identity when multiplying a particular element a. But you need to find a single identity element for the whole of S.

    So, for the given element a, you know that there is an element e such that ae=a, and there is an element f such that fa=a. Now let b be another element of S. There is an element y such that ya=b. Therefore be=yae=ya=b, and by a similar argument fb=b. This shows that e acts as a right identity for every element of S, and f acts as a left identity for every element of S. Therefore in particular e=fe=f. Hence e is an identity.

    Quote Originally Posted by tttcomrader View Post
    Now, for inverse I'm not so sure if I can cancel any elements by multiplying y^-1 or x^-1, am I allow to do that?
    No, you can't do that, because you don't yet know that inverse elements exist. However, given an element a, you do know that there is an element x such that ax=e, and there is an element y such that ya=e. But then y=ye=yax=ex=x, so y=x is an inverse for a. So each element has an inverse, and now you just have to check that the group axioms hold.

    Quote Originally Posted by tttcomrader View Post
    2. Suppose that S is a finite semigroup such that the cancellation law holds, prove that it is a group.

    Suppose that x,y,z are elements of S such that zx = zy, then x = y.

    then z^{-1}zx=z^{-1}zy, now, am I allow to cancel any elements here? I guess I can't since I haven't find the inverses yet...
    Again, you don't know that inverses exist, so you can't use that argument.

    Suppose that S contains n elements (n is finite). Given an element a in S, think about the set of products ax as x runs through S. There are n of these products and (by the cancellation property) they are all distinct. So they must constitute all the elements of S. Therefore, given an element b of S, one of those products ax must be equal to b. In other words, the equation ax=b has a solution. Similarly, the equation ya=b has a solution. So now we can use the result from problem 1 to conclude that S is a group.

    Quote Originally Posted by tttcomrader View Post
    And why does it have to be finite? What if it is infinite semigroup?
    Suppose S is the set of natural numbers, with ordinary multiplication as the semigroup operation. The cancellation property holds, but S is not a group.
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