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Unformatted text preview: Miscellaneous Algebra facts Jonathan L.F. King University of Florida, Gainesville FL 326112082, USA [email protected] 14 April, 2008 (at 06:59 ) Basic. For us, a semigroup is a triple ( S, • , e ), where • is an associative binary operation on set S , and e ∈ S is a twosided identity elt. ♥ 1 Use LI for “left inverse” and RI for “right inverse”. 1 : Prop’n. In a semigroup ( S, • , e ) : i : For each x ∈ S : If x has at least one LI and one RI , then x has a unique LI and RI , and they are equal. ii : Suppose every elt of S has a rightinverse. Then S is a group. ♦ Proof of (i) . Suppose λ is a LI of x , and ρ a RI. Then λ = λ [ xρ ] = [ λx ] ρ = ρ. And if two LIs, then λ 1 = ρ = λ 2 . Proof of (ii) . Given x ∈ S , pick a RI r and a RI to r , call it y . Now x = x • [ ry ] = [ xr ] • y = y . Hence r is both a left and right inverse to x . Etc. 2 : Lemma. Suppose O is an associative binop on S , and e ∈ S is a righthandidentity elt. Suppose that each y ∈ S has a righthand inverse, y . Consequently: Whenever y O y = y , then y = e . 3 : Moreover, each y is also a left inverse to y , and e is also a left handidentity. Thus ( S, O , e ) is a group, ♦ Proof of (3) . Note y = y O [ y O y ] = [ y O y ] O y . By (3 hypothesis ), then, y = [ y ] O y = e . ♥ 1 What I’m calling a semigroup is usually called a monoid . Proof of (2) . First let’s show that every RHInverse, y , is also a LHInverse to y . Let b := [ y O y ]. Cour tesy (3), it is enough to show that b O b = b . And b O b = y O [ y O y ] O y , by assoc. , = y O e O y = y O y note ==== b. We can now show that e is also a left hand identity. After all, e O y = [ y O y ] O y = y O [ y O y ] = y O e , since y is a LHInverse. I.e, e O y = y . Henceforth, groups ♥ 2 are the subject. Cyclic groups I use Cyc N for the order N cyclic group . By de fault, it is written multiplicatively, but I may write (Cyc N , · ) or (Cyc N , +) to indicate specifically. The infinite group Cyc ∞ is iso to ( Z , +). For y ∈ G we use Periods G ( y ) for the set of in tegers k with y k = e . A subgroup H ⊂ G de termines a similar set. Let P H ( y ) = P H,G ( y ) be { k ∈ Z  y k ∈ H } . So Periods( y ) is simply P H ( y ), when H is the trivial subgp { e } . 4 : Periods Lemma. Fix G,H,y as above, and let P H mean P H ( y ) . If P H is not just { } , then P H = N Z , where N is the least positive element of P H . For Gsubgroups H ⊃ K , then, HOrd G ( y ) • KOrd G ( y ) • Ord G ( y ) . ♦ ♥ 2 Here is my generic footnote: Typical group notation: ( G, · , e ) or (Γ , · , ε ) or ( G, · , 1) or ( G, + , 0). The symbol for the neutral ( i.e, identity ) element may change, according to whether the group name is a Greek letter, or whether the group is writ ten multiplicatively or additively. A vectorspace might be writ ten as ( V , + , ). A group of functions , under composition, might be written ( G, ◦ ,Id )....
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This note was uploaded on 01/26/2012 for the course MAC 3472 taught by Professor Jury during the Fall '07 term at University of Florida.
 Fall '07
 JURY
 Calculus, Algebra

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