Group Theory Midterm Study Guide 2

Proofs today, folks. I'll be proving the following theorems: Commutativity of Disjoint Cycles, Lagrange's Theorem (and the requisite 4th Property of the Lemma of Cosets), and the Corollaries of Lagrange's Theorem. Oh, just so there's no confusion, I'm not copying this out of my text; this is coming from memory (as much as possible) as an exercise for cementing these proofs for the exam (where I will be asked to give at least one of them).

Commutativity of Disjoint Cycles
Claim: If cycles A and B are disjoint, then AB=BA.
Pf: Define a set S such that S={a1,..., an, b1,..., bn, c1,..., cn}, where a's are elements mapped by A, b's are elements mapped by B, and c's are elements mapped by neither A nor B. Then, I need to show that (AB)(x)=(BA)(x) for all x in S. Let x=ai. Then, (AB)(ai)=A(B(ai))=A(ai)=a(i+1) and (BA)(ai)=B(A(ai))=B(a(i+1))=a(i+1). Now, let x=bi. Then, (AB)(bi)=A(B(bi))=A(b(i+1))=b(i+1) (BA)(bi)=B(A(bi))=B(bi)=b(i+1). Now, let x=ci. (AB)(ci)=A(B(ci))=A(ci)=ci and (BA)(ci)=B(A(ci))=B(ci)=ci. Then, (AB)(x)=(BA)(x) for all x in S. QED.

Property 4 of the Lemma of Cosets
Claim: aH=bH or aH&bH={0}.
Pf: Assume that aH and bH are left cosets of H in G. Let c be an element of aH and an element of bH. Property 3 of the Lemma of Cosets states that xH=yH if and only if x is an element of yH. Then, because c is an element of both aH and bH, cH=aH and cH=bH. Therefore, aH=bH. QED.

Lagrange's Theorem
Claim: The order of the group G is divisible by the order of the subgroup H; the number of distinct left cosets of H in G is |G|/|H|.
Pf: Let G be a finite group and let H be subgroup of G. Let a1H,..., anH be the distinct left cosets of H in G. Then, G=a1H&...&anH. However, by Property 4 of the Lemma of Cosets, these are disjoint cosets. Then, |G|=|a1H|+...+|anH|. However, |aiH|=|H|. Then, |G|=|a1H|+...+|anH|=n|H|. So, the order of G is divisible by the order of H. QED.

Corollary 2 of Lagrange's Theorem
Claim: The order of an element a of a group G divides the order of group G.
Pf: Let there be a group G with an element a. Then, 'a' is a subgroup of G. By Lagrange's Theorem, |'a'| divides |G|, but |'a'|=|a|. Then, |a| divides |G|. QED.

Corollary 3 of Lagrange's Theorem
Claim: Groups of prime order are cyclic.
Pf: Let G be a group of prime order with an element a such that a does not equal e. Then, n|'a'|=|G| by Lagrange's Theorem. However, because a does not equal e, |'a'| must not be 1. Because |G| is prime, the only other possibility for |'a'| is |G|. So, |'a'|=|G|. QED.

Corollary 4 of Lagrange's Theorem
Claim: a^|G|=e.
Pf: Let a be an element of a group G. Then, by Corollary 2 of Lagrange's Theorem, |a|k=|G|. So, a^|G|=a^(|a|k)=(a^|a|)^k=e^k=e. QED.

Ok, there are some other properties that MAY be on the exam. But I don't think I want to prove those ones. They're a pain. I almost guarantee that he'll ask for the proof of Lagrange's Theorem (and the requisite Property 4 of the Lemma of Cosets), and then he'll ask for one of the following: the Corollaries of Lagrange's Theorem, Commutativity of Disjoint Cycles, or some isomorphism properties. I've got it in the bag, kids. Ok, time to go review the definitions, then start making a reference sheet of groups -- that's going to be clutch.