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March 3rd, 2009, 09:17 AM  #1 
Newbie Joined: Mar 2009 Posts: 4 Thanks: 0  Help with Written Proofs
I'm currently taking Modern/Abstract Algebra and I have several questions on proofs I have to turn in. Several of them I don't know how to do, even though I have asked my teacher for help. Please help if you can! 1) Prove that, if G is a group with the property that the square of every element is the identity, then G is Abelian. Here is what I have so far: "In order for G, a group, to be an Abelian group, the operation * must be commutative. Commutativity: for all a, b in G, ab = ba. We are given that, for any a and b in G, a*a = a^2 = e and b*b = b^2 = e. Then (a^2)*(b^2) =e*e = e2 = e and (b^2)*(a^2) = e*e = e2 = e. Thus, (a^2)(b^2) = (b^2)(a^2) and so (ab)^2 = (ba)^2 and so (ab)(ab) = (ba)(ba)." I need: ab = ba. 2) Prove that, in any group, an element and its inverse have the same order. I have proved the finite case, but I don't know how to prove the infinite case. 3) Let H = {a + bi  a, b are in R, ab ? 0}. Prove or disprove that H is a subgroup of C under addition. I can't find where I did this problem before, so I don't really have anything. Help? 4) Let a be a group element that has infinite order. Prove that <a^i> = <a^j> if and only if i is equal to (+/)j. Basically, I have the same issue as #2, which is that I don't know how to prove the infinite case. I guess I don't feel like I know anything definite about an element with infinite order. 5) Let G be a group. Prove or disprove that H = {g^n  g is in G} is a subgroup of G. This is supposed to be a proof by counter example. My teacher suggested I use A4, the set of even permutations of 4 symbols (or the alternating group of degree 4), but we just learned about it and I don't understand it fully yet. Do I need to write it out? Do I just follow the steps of a subgroup test until it fails? I think that's it. I'm sorry for putting so many questions in one place. Any help would be much appreciated! 
March 3rd, 2009, 10:46 AM  #2 
Senior Member Joined: Nov 2008 Posts: 199 Thanks: 0  Re: Help with Written Proofs
for 1 start with x.y.x.y = 1. what happens when you multiple each side on the right by y.x? for 2 think what happens when you multiply g'...g' by g...g (where g' is the inverse of g) and we have the same number of gs as g's? Suppose the order of g is greater than the order of g'. Let these orders be N and M respectively (so M<N). Note that this makes M finite (assuming we're not worrying about uncountable cardinals). Then M(g).M(g') = 1, but also M(g).M(g') = M(g).1 = M(g). So M(g) = 1. This is a contradiction. You can do something similar for the other case. for 3 consider (a + bi) and (c + di) in H. What do you get when you mulitply (a  c)(b  d)? Use this to see the values for which this product is negative. Using this show the subgroup test is not satisfied. for 4 it is straight forward to show the equality occurs if i = +/ j. I'm not sure what your problem with infinity is here; I imagine we can assume both i and j are finite (otherwise the question gets dangerously close to not making sense). So, for the next part assume that i<j and show that a^i is not an element of <a^j> (using the fact that a has infinite order). for 5 you've been told you're looking for a counterexample so write out A4 and find an n for which the statement does not hold with G = A4 (subgroup test will probably help as you suggest). I hope this is helpful (and correct). 
March 5th, 2009, 10:54 AM  #3  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Help with Written Proofs
I don't understand problem 4. If g has infinite order and g^7 = g^7, then g^14 = g^7g^7 = g^7g^7 = 1 (or e if you prefer) and so g has order 7, a contradiction. I'd expect the question to be "Let a be a group element that has infinite order. Prove that a^i = a^j if and only if i is equal to j.". Where did I go wrong? Quote:
 
March 5th, 2009, 02:04 PM  #4 
Senior Member Joined: Nov 2008 Posts: 199 Thanks: 0  Re: Help with Written Proofs
The multiplication in 3 is because (a+bi) + (c+di) = (a+b) + ((c)+(d)i), and we require (ac)(bd)>=0 (and ab>=0, cd>=0 from the setup), for subgroup test. In hindsight it's (slightly) easier without the subgroup test: you can just look for values for which (a+c)(b+d)<0 and the conditions on ab and cd are preserved. e.g a=0, b=2, c=1, d=1. Then ab=0, cd=1 and (a+c)(b+d)=1(1) = 1.

March 5th, 2009, 02:08 PM  #5  
Member Joined: Aug 2008 Posts: 84 Thanks: 0  Re: Help with Written Proofs Quote:
 
March 5th, 2009, 03:14 PM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Help with Written Proofs Quote:
 
March 6th, 2009, 07:43 AM  #7  
Newbie Joined: Mar 2009 Posts: 4 Thanks: 0  Re: Help with Written Proofs Quote:
 
March 6th, 2009, 07:46 AM  #8  
Newbie Joined: Mar 2009 Posts: 4 Thanks: 0  Re: Help with Written Proofs Quote:
x*y*x*e*x = y*x x*y*x*x = y*x x*y*e = y*x x*y = y*x That's great! Thanks!  
March 6th, 2009, 08:32 AM  #9 
Senior Member Joined: Nov 2008 Posts: 199 Thanks: 0  Re: Help with Written Proofs
my guess is CRGreathouse misread ab>=0 as a,b>=0. In that case H is closed under addition but not inverses, in the original case H is closed under inverses but not addition.

March 6th, 2009, 09:08 AM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Help with Written Proofs Quote:
 

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