01 May '08 15:252 edits

A "Semigroup" is a set S under an operation * such that,

- a*(b*c)=(a*b)*c (it is associative) &

- a*b is contained in S (it is closed).

NOTE: I'll usually just write a*b as ab from now on.

Now, let S be commutative. That is, ab=ba for all a and b in S.

Now, if for all a, b in S there exists x, y in S such that xab=y, prove that S is a group. That is, prove that there already exists an element 1 in S such that a1=1a=a for all a in S, and, further, that there exists a' in S such that a'*a=1. (S contains an identity and inverses).

Sorry if the question seems rather confusing - just ask if you need clarification... ðŸ˜›

EDIT: Sorry for all the "that is"-es. Can't think of anything to replace them though...

- a*(b*c)=(a*b)*c (it is associative) &

- a*b is contained in S (it is closed).

NOTE: I'll usually just write a*b as ab from now on.

Now, let S be commutative. That is, ab=ba for all a and b in S.

Now, if for all a, b in S there exists x, y in S such that xab=y, prove that S is a group. That is, prove that there already exists an element 1 in S such that a1=1a=a for all a in S, and, further, that there exists a' in S such that a'*a=1. (S contains an identity and inverses).

Sorry if the question seems rather confusing - just ask if you need clarification... ðŸ˜›

EDIT: Sorry for all the "that is"-es. Can't think of anything to replace them though...