Proofs about Groups

[daon]

I am using the standard Group axioms of existance of the inverse, existance of identity elt, closure and associativity.

1st: I think this is correct. If there's an error, please let me know.

If a²b² = (ab)² in a group G then ab=ba.

Pf: a²b² = (aa)(bb) and (ab)² = (ab)(ab). Thus, by our assumption that a²b² = (ab)², we can conclude (aa)(bb) = (ab)(ab). We can multiply both sides by a^-1 and we have a^-1[(aa)(bb)] = a^-1[(ab)(ab)]. By Associativity, this is the same as [a^-1(aa)](bb) = [a^-1(ab)](ab) and once again, [(a^-1a)a)](bb) = [(a^-1a)b](ab).
We know that a^-1a = e (the identity elt), so substituting this in, we have: (ea)(bb) = (eb)(ab) or a(bb) = b(ab). By associativity, this is equivilant to (ab)b = (ba)b, and by RHS cancellation of b, ab=ba.


2nd: Not sure where to start with this one..

Let G be a finite group. Show that the number of elements x of G such that x³ = e (the id elt) is odd. Show that the number of elts x of G s.t. x² \(\displaystyle \neq\) e is even.

Ideas-- I was thinking that I need to show that x = 2k+1 for some k, but am not sure. Also, Since x³=e, then xx²=e meaning that x and x² are inverses. I'm not sure if this will help me, but thats as much as I got.

Thanks in advance,
Daon

 

[pka]

Your #1 is well done.
However are you sure that you have copied #2 two correctly?

As stated it is not correct.
The order of an element divides the order of the group.
If x<SUP>3</SUP>=e then the order of the group must be a multiple of 3.
Thus if a group of order five there are no elements of order 3. But 0 is an even number.
Problem #2 says it is odd.

 

[daon]

Now I'm really confused .. We have never discussed what you said in my class, but I just read it all on Wikipedia.

The problem was written exactly as it is written in the book--"Contemporary Abstract Algebra" 6th edition by Joseph A Gillian, page 54 #29.--just in case you happen to have it.

What modification would make that statement true? I don't know why a book, let alone my Professor, would ask us to prove something untrue.

Thanks,
Daon

 

[pka]

Yes I do have a copy of the text. I think I reviewed it for the publisher: not a good review though. Look on page 75, at corollary 2. You will see the business about the order of an element and the order of the group. In my copy that problem is on page 55. But it is just as you have written it. Looking over that chapter, I see nothing to help in knowing how what he expects you to use. At the least, it seems out of place because the definition of order is in the next chapter.

He may mean that if you have an element such that x<SUP>3</SUP>=e [i.e. if such elements happen to exist in the group] then x<SUP>-1</SUP>=x<SUP>2</SUP>. In other words, all such elements occur in pairs. Therefore, collect all the pairs the add e and you have an odd number of elements. However, that is not the way the problem is stated. Likewise, if x<SUP>2</SUP>\(\displaystyle \not=\)e then those also occur in pairs and e is not one of them.

P.S. I have thought more about this problem. In the preliminary edition I have, it is stated that x\(\displaystyle \not=\)e. However e always has that property, right? Hence, if e is the only element with that property then that is an odd number of elements.

 

[daon]

OK well I may not be understanding the problem correctly.

From your revised question, am I trying to show that If for all x in g, xxx=e, then the order of G is odd? If so, does it mean that I assume ALL elements x in G have the property that x³=e, and I have to show that G is of odd order?

 

[pka]

I apologize if I confused you. Here is what I think the problem means.

In any group, \(\displaystyle e^3 = e\) thus \(\displaystyle e\) has the property. If \(\displaystyle e\) is the only element with that property then there are an odd number of them. Now suppose that \(\displaystyle x \not= e\) has the property that \(\displaystyle x^3 = e\). We showed that also \(\displaystyle \(x^{-1}\)^3 = e\), they occur in pairs. Take all the pairs together with \(\displaystyle e\) then we have an odd number of elements with that property.

 

[daon]

If this is not what you were saying, then I apologize. This is what I have gotten from what you said:

We know e has this property and if some element x has this property, then since x and x² are inverses, x² also has this property. Hence, for any x there is another element x² with this property.

So, if G has k elements having this property, then G also has k inverses of these elements, plus the element e makes an odd number of elements?

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(i made an edit here, but nevermid if you happened to have read it)

Thanks

 

[pka]

EXACTLY!

But I still think that the question is poorly worded and out of sequence in the text.

 

[daon]

Thanks for helping me understand this question. For the second one, is this correct:

The number of elements x of G s.t. x² \(\displaystyle \neq\) e is even:

Since e itself does not statisfy this property, and for every element x that does satisfy x² \(\displaystyle \neq\) e, then x^-1 = a for some other a in G such that a² \(\displaystyle \neq\) e. Therefore for every x, there is a second element a that satisfies this property.

So if G has n elements satsfying this property then n is made up of k elements and k inverses of these elements that also satisfy the propety. So n=2k.

Again, I appreciate you taking the time to help me.