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March 3rd, 2009, 08:17 AM   #1
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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!
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March 3rd, 2009, 09:46 AM   #2
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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).
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March 5th, 2009, 09:54 AM   #3
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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?

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Originally Posted by pseudonym
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.
It's complex addition, not multiplication. H is closed under addition, but doesn't have inverses.
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March 5th, 2009, 01:04 PM   #4
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Re: Help with Written Proofs

The multiplication in 3 is because (a+bi) + -(c+di) = (a+b) + ((-c)+(-d)i), and we require (a-c)(b-d)>=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.
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March 5th, 2009, 01:08 PM   #5
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Re: Help with Written Proofs

Quote:
Originally Posted by CRGreathouse
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?
It's not equality of elements--it's equality of subgroups generated by a^i and a^j.
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March 5th, 2009, 02:14 PM   #6
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Re: Help with Written Proofs

Quote:
Originally Posted by Spartan Math
It's not equality of elements--it's equality of subgroups generated by a^i and a^j.
Got it, thanks. I knew there was something I was missing.
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March 6th, 2009, 06:43 AM   #7
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Re: Help with Written Proofs

Quote:
Originally Posted by CRGreathouse
It's complex addition, not multiplication. H is closed under addition, but doesn't have inverses.
I don't see how H doesn't have inverses. The identity I guess would be 0+0i and then wouldn't the inverse of a+bi just be (-a) - bi?
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March 6th, 2009, 06:46 AM   #8
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Re: Help with Written Proofs

Quote:
Originally Posted by pseudonym
for 1 start with x.y.x.y = 1. what happens when you multiple each side on the right by y.x?
x*y*x*y*y*x = y*x
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!
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March 6th, 2009, 07:32 AM   #9
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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.
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March 6th, 2009, 08:08 AM   #10
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Re: Help with Written Proofs

Quote:
Originally Posted by pseudonym
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.
Right, I saw a comma that wasn't there -- maybe a spot on my monitor. I'm now 0/2 -- the only questions I got right were the ones I didn't post because you beat me to it!
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