Some Proofs on Limits and Continuity

Here are a couple of proofs that I think are nice. By nice, I mean, I like how I proved the claim because the proof is concise, clear, and rigorous.

1. Let f:R->R be a continuous function. For all r in Q, let f(r)=0.
   Claim: f(x)=0 for all x in R.
   Pf: Because f is continuous, for all c in R, for all V=N(f(c);ß), there exists U=N(c;∂) such that f(U) is a subset of V. So, V=(f(c)-ß, f(c)+ß); U=(c-∂, c+∂); and f(U) is contained in V. By the density of the rationals, there is at least one rational number in U, so f(U)=0&f(U\Q). Now, if c is rational, then V=(-ß, ß), which obviously contains 0, but must also contain f(U\Q) for all ß>0. That is, f(U\Q) is between 0 and ß, for all ß, so f(U\Q)=0. If c is irrational, then V=(f(c)-ß, f(c)+ß), which must contain 0. But if f(c)≠0, then, if ß≤f(c), 0 is not in V. So, f(c)=0. qed.
2. This one is pretty facile, but I still liked how I proved it.
   Let f:D->R, with c in D', and lim(x->c)f(x)=L>0.
   Claim: There is V=N*(c, ß): f(x)>0, for all x in R.
   Pf: Pick ß such that ß0. Then, ß>|f(x)-L| implies that f(x)-L>-ß, so f(x)>L-ß. But, by choice, L>ß>0, so L-ß>0, so f(x)>0. That is, let V be the deleted neighborhood of c, N*(c;ß), were L>ß>0. Thus, for all x in V&D, ß>|f(x)-L| implies that f(x)>L-ß<, so f(x)>0. qed.

Proofs and Theorems for Exam

Here are some proofs that I may need iterate on the exam. Following that are some theorems that I'll probably need to know.

• Claim: A finite integral domain is a field.
  Pf: Let D be a finite integral domain with unity 1. Let a be any nonzero element in D. I will show that a is a unit. If a=1, it is its own inverse, so I am done. So, assume that a =! 1. Then, let D contain a, a^2, a^3, a^4.... D is finite by assumption, so there must be two positive integers i and j such that i>j and a^i=a^j. Then, a^(i-j)*a^(j)=1*a^j. By cancellation, a^(i-j)=1. Since a=!1, i-j>1. Then, a*a^(i-j-1)=1. So, a^(i-j-1) is the inverse of a. qed.
  
• Claim: The characteristic of an integral domain is 0 or prime.
  Pf: It suffices to show that if the additive order of 1 is finite, it must be prime, because, in a ring with unity, the additive order of 1 is characteristic of the ring. Suppose that 1 has order n and that n=st, where 1<=s, t<=n. Then, 0=n*1=(st)*1=(s*1)*(t*1). There are no zero-divisors in an integral domain, so either s*1=0 or t*1=0. Since n is the least positive integer with the property that n*1=0, we must have s=n or t=n. Thus, n is prime. qed.
  
• Let R be a commutative ring with unity and let A be an ideal of R.
  Claim: R/A is an integral domain iff A is prime.
  Pf: Suppose that R/A is an integral domain and ab in A. Then, (a+A)(b+A)=ab+A=A. So, either a+A=A or b+A=A; that is, either a in A or b in A. Then, A is prime. To prove the converse, observe that R/A is a commutative ring with unity for any proper ideal A. Suppose that A is prime and that (a+A)(b+A)=0+A=A. Then, ab in A. Then, since A is prime, a in A or b in A. Then, a+A=A or b+A=A. That is, there can be no zero-divisors. qed.
  

There are two more that I am encouraged to know, but I don't think I'll need to know proofs. He generally only asks for proof of a theorem found in the book, but gives us three or four different options and lists five different theorems on the study guide. So, I'll be fine if I know these three.
Here are some theorems that I'll probably need to know:

• Cancellation: Let a, b, c be in an integral domain. If a=!0 and ab=ac, then b=c. NOTE: THIS IS NOT DIVISION.
  
• A finite integral domain is a field.
  
• For every prime p, Z(p) is a field.
  
• Let R be a ring with unity 1. If 1 has infinite additive order, then the characteristic of R is 0. Else, if 1 has order n, then the characteristic of R is n.
  
• The characteristic of an integral domain is 0 or prime.
  
• R/A is an integral domain iff A is prime.
  
• R/A is a field iff A is maximal.
  
• The kernel of any homomorphism is an ideal the ring.
  
• Every ideal A of a ring R is the kernel of a ring homomorphism r-->r+A from R to R/A.
  
• A ring with unity contains Z(n) or Z.
  
• A field contains Z(p) (if char=n) or Q (if char=0).
  

Ok, that's about all I need to know. Not bad, really. Its just the first midterm. Not bad at all.

Nasty little problem,,,

This is a problem that's been stumping me for a few hours now.
Let R be a commutative ring with unity. Let x and y be in R, such that x is a unit and y is nilpotent.
Claim: (x+y) is a unit.
Side Bar: Ok, here's what that means: x is unit implies that there exists b in R such that xb=1, or, that x has a multiplicative inverse. y is nilpotent implies that there is a positive integer n such that y^n=0. I want to show that there exists c in R such that (x+y)c=1, or that (x+y) has a multiplicative inverse. Observe that, if we set c to y^(n-1), then we get x*y^(n-1). If I can show that that has a multiplicative inverse, then I'll be done. But I'm not sure that that is the way to go. How about this instead: (x+y)(a+b)=1. Hmmm, that has potential. Then, xa+xb+ya+yb=1. Now, let a=x and let b=-y. Then, xa+xb+ya+yb=x^2-y^2. Wait, multiply this by x^2+y^2. Then, we have x^4-y^4. Do this enough times, and eventually, you'll get to y^n or y to some number that is greater than n, which amounts to the same thing, namely, y goes to 0. Then, we are left with x^m. Does x^m have a multiplicative inverse? Well, of course, it does. x^m=x*x^(m-1), and x is a unit. So, multiply x^m by x^-m and you get 1. Pretty facile, I know, but it's worth explaining these things. So, now for the actual proof.
Pf: Observe that multiplying x+y by (x^-m)(x-y)*II(x^i+y^i), i=2, m, where m is sufficiently large (n is fine), you have (x^-m)(x^m-y^m). But, y^m=0, so, you have x^-m*x^m, which is 1. So, x+y is a unit; specifically, its multiplicative inverse is (x^-m)(x-y)*II(x^i+y^i), i=2, m, where m is a number such that the product of the nonzero even numbers preceding m is greater than or equal to n. QED.

I hope... I bet there is an easier way to SHOW that there must be an inverse, other than actually FINDING the inverse... I'll keep thinking on this.

Welcom Back to Blogdom

Ok, ok, I'm back. Keep your yappers shut, I know you all missed me.
So I'm taking Ring and Field Theory this term, which is a branch of group theory.
Here is a proof that has been stumping me for a bit now. I think it's clever more than anything...

Suppose that R is a commutative ring and |R|=30, with an ideal I, where |I|=10.
Claim: I is a maximal ideal of R.
Sidebar: The order of R is 30, and, by Lagrangre's Theorem, the order of any subset divides the order of the group. Then, there are supgroups of order 10 and order 15 (call them I and J). But we aren't talking about groups, we're talking about rings. We know that there is a subring of order 10 (I). So, I need to show that I is not a subset of J or that J is not a subring (although it is a subgroup). A third possibility is to show that all groups of order 30 are isomorphic to Z(30), at which point I can fairly easily prove that I is maximal. No, I just Googled it: not every group of order 30 is isomorphic to Z(30). Wait. Wait. Wait. In I, there are elements of order 2 (because |I|=10=5*2), whereas in J, there are no elements of order 2 (because |J|=15=5*3). So, I is not a subset of J. And, by Lagrange's Theorem, there can be no other subgroups. So, I is maximal. Incidentally, J is maximal, too. Now, the formal proof.
Pf: Let R be a commutative ring, with |R|=30, and two ideas, I and J, where |I|=10 and |J|=15. By Lagrange's Theorem, there can be subgroups (and, so, subrings) of orders that divide the order of the group (or, rather, ring). Then, there can be no proper subring of order greater than 15. So, either I is maximal or J is maximal (or both). If I is not maximal, then I is a subset of J. However, I contains elements of order 2 (by Lagrange's Theorem and because 10=2*5), whereas J does not (by Lagrange's Theorem again, and because 15=3*5). So, I cannot be a subset of J. So, I is maximal. Incidentally, if there even exists a subring of order 15, then J is also maximal. QED.

Here's another that's been stumping me:
Question: Is the ring 2Z isomorphic to the ring 3Z?
Answer: Assume that 2Z is isomporphic to 3Z, where P is such an isomorphism. Then, by principal, 3Z is isomorphic to 2Z, where P(inv) is such an isomorphism. The ring 3Z has unity 1, so P(inv)(1)=1*, where 1* represents the unity of 2Z. However, no such unity exists. Then, 1 cannot map to 1*, so there is no such isomorphism P(inv). So there is no such isomorphism P, which contradicts our initial assumption.
Question: Is the ring 2Z isomorphic to the ring 4Z?
Answer: This one is a little more tricky. It feels obvious to me, frankly, that they are. They are both infinitely countable, so any mapping can be one-to-one and onto. Hmmm, let's see. Let P be such an isomorphism. Then, P(2)=4n, n in Z. Then, P(2*2)=P(2+2), implies that P(2)*P(2)=P(2)+P(2), which implies that 4n*4n=4n+4n, which implies that 16n^2=8n, which implies that 2n=1, which is impossible. So, we have a contradiction, so P cannot exist. Well, well, well, my intuition was wrong. The two aren't isomorphic. Incidentally, I can use this same method for the previous problem.

That all took me way too long. Time to get it formally written up, then off to read more ethics.

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.

And now, a stumper

Ok, this one's stumping me. Here goes:
Let G be a nontrivial group with no nontrivial proper subgroups.
Claim: |G| is prime.
Sidebar: Here's the problem. Well, the first one. It's easy to say that, if G is of prime order p, then the possible orders of its subgroups are either 1 or p. However, what's there to stop a group G of non-prime order n from not having any nontrivial proper subgroups? This relates back to what I was saying in my last post about how the converse of Lagrange's Theorem is not true. So, I really need to show that, if a group G is of non-prime order n, then it MUST have a nontrivial proper subgroup. Oh, wait. Let H be a subgroup of G, whose order is n and non-prime, and let a belong to G, where |a|=m and m does not equal n; by a corollary of Lagrange's Theorem, m|n. Then, 'a' is a nontrivial proper subgroup of G. So, G MUST have a nontrivial proper subgroup. However, if G is of prime order p, then all of its elements must be of order 1 or of order p, by the same corollary, so 'a'=G or 'a'=e, for all a in G. Great. Excellent. Half down. Now, the second problem. What if G is infinite? Obviously, I need to prove that G CANNOT BE infinite. Then, if it can't be infinite, it must be finite, and, as I've already showed, if it's finite, it's of prime order. So. How do I prove that it can't be infinite? Same way I showed that if its finite, its of prime order. That is to say, I need to show that any infinite group must have a nontrivial proper subgroup. How do I do this? Let's assume that G is an infinite group and that a belongs to G. Then, 'a' must be infinite as well. No help there, right? How about this; it's a little edgy, but I'll give it a go. Let G be an infinite group, and let a belong to G. Then, a possible subgroup of G is {e, a, a^-1}; this is a subgroup because it includes the identity and the inverse of each element, and it's closed. Note that, if G is of prime order, this subgroup doesn't fly: the subgroup has order 3, and, by Lagrange's Theorem, the order of the group is a multiple of the order of the subgroup, so the order of G cannot be prime, which contradicts our assumption. Ok. This feels a little sketch. But I'll go ahead and write it up formally now:

Claim: If G is a nontrivial group with no nontrivial proper subgroups, |G| is prime. \
Pf: Assume that G is a nontrivial group with no nontrivial proper subgroups. First, I need to show that G cannot be infinite. I will do this by contradiction. Assume that G is infinite, that a belongs to G, and |a}=2. Then, define a subgroup H, where H={e, a}. Then, H is a nontrivial proper subgroup of G, which contradicts our assumption that G has no nontrivial proper subgroups. So, G cannot be infinite.
Now, I need to show that |G| is prime. I will do this by contradiction again. Assume that |G| is non-prime. Let a be a non-identity element of G, let |G|=n, let |a|=m, and let m and n be distinct. Then, 'a' is a subgroup of G (I won't prove this; it's in my text as Thm 3.4, so I feel free to call upon it at will). However, because m and n are distinct, 'a' does not equal G (note that, if the orders of G and a are equal, 'a'=G). Then, 'a' is a nontrivial proper subgroup of G, contradicting our assumption that G has no nontrivial proper subgroups. Therefore, |G| must be prime. QED.
Ok, here's the problem I see with this proof: In a finite group G with non-prime order n, what if there are no elements that are of orders other than 1 and n? That is to say, what's stopping a group of order n having 1 identity and then n-1 elements of order n? That's a problem. I kinda just ran myself in a circle, maybe. The first half is solid, though. I'll keep thinking on it.

A proof

This is a proof of which I'm rather proud.
Claim: |U(n)| is even when n>2.
Pf: It is known that Aut(Z(n)) is isomorphic to U(n). Let a,b be elements if Z(n). Then, a+b=b+a, so Z(n) is Abelian. If a group G is Abelian, then there exists an automorphism P(g)=g^-1. As proof, note that inverses of elements are unique, so P is one-to-one and onto. Further, letting x,y be elements of G, P(ab)=(ab)^-1=(b^-1)(a^-1)=(a^-1)(b^-1)=P(a)P(b), showing that P preserves the operation. This suffices to show that P(g)=g^-1 is an automorphism on a group G. Then, P is an element of Aut(Z(n)). Finally, observe that P(P(g))=g; that is to say, the inverse of the inverse of g is g. This shows that the inverse of P is P, or, more precisely, that P is of order 2. Then, Aut(Z(n)) has at lease one element of order 2. By the properties of isomorphisms acting on elements, if G is a finite group, then G and P(G) have exactly the same number of elements of every order. Then, since U(n) is isomorphic to Aut(Z(n)) and Aut(Z(n)) has an element of order 2, U(n) has an element of order 2. Then, by a corollary of Lagrange's Theorem (specifically, the corollary states |a| divides |G|, or, the order of a group's element a divides the order of the group G), because U(n) contains an element of order 2, the order of U(n) must be a multiple of 2; that is, |U(n)| must be even. QED.
I bet there is an easier way to do this, but the problem specifically required the use of the noted corollary. And, really, this proof is pretty sexy -- it basically just compiles a bunch of other proofs and says, look, the claim's true. That's the definition of a sexy proof right?