Recent content by rideabike

Sounds like a good "dot-dot-dot" proof, but maybe your professor doesn't like those. I.e. show the process of a couple steps, then "dot-dot-dot", then show the final step!

Homework Statement Prove the collection of all finite order elements in an abelian group, G, is a subgroup of G. The Attempt at a Solution Let H={x\inG : x is finite} with a,b \inH. Then a^{n}=e and b^{m}=e for some n,m. And b^{-1}\inH. (Can I just say this?) Hence...

Yeah that's exactly it, I was being sloppy. My understanding of a cyclic group is that there exists some element of G, say a, such that, for every x in G, x=a^n for some integer n. How does this follow from |x|=p?

So |x|\inG = p? Would this relate to a group being cyclic?

That's good, thanks!

If h\notinS, then h=e, which isn't what we want. I was trying to show that if you take all the elements and their inverses out of S, then you're left with a single element whose inverse must be itself. There's probably a better way to show this.

Homework Statement Let (G,*) be a finite group of even order. Prove that there exists some g in G such that g≠e and g*g=e. [where e is the identity for (G,*)] Homework Equations Group properties The Attempt at a Solution Let S = G - {e}. Then S is of odd order, and let T={g,g^-1...

Okay, I see that now, it's recursive. Then you could replace a with pk for some integer k. Then (pk)^n=pb^n p^(n-1)k^n=b^n Could the same logic as before be used to show p divides b, hence a and b have a common divisor, producing a contradiction?

How do we know p doesn't divide a?

Since gcd(a,b)=1 and a does not divide b^n for any n≥1, by Euclid's Lemma a divides p, which is a contradiction since p is prime. Or does this not work because a is not necessarily greater than 1?

Homework Statement Let p be a prime and let n≥2 be an integer. Prove that p1/n is irrational. Homework Equations We know that for integers a>1 and b such that gcd(a,b)=1, a does not divide b^n for any n≥ 1. The Attempt at a Solution To prove irrationality, assume p^(1/n)=a/b for...

Right. And we want to show there's an injection from B to A. Would it be that since there's an injection from B to C and and injection from C to A, there must be an injection from B to A?

I know, I don't really know where to start. Schroder-Bernstein maybe?

Homework Statement Prove that if A,B, and C are nonempty sets such that A \subseteq B \subseteq C and |A|=|C|, then |A|=|B| The Attempt at a Solution Assume B \subset C and A \subset B (else A=B or B=C), and there must be a bijection f:A\rightarrowC...