Question 11

ThePerfectHacker · #1 $Old$ December 18th, 2006, 08:14 AM

Here are two questions on group theory. Do not worry this is the most elementary group theory, so nothing complicated. Look below, I made a tutorial that explains what a group is in the most simple terms!

1)Let $G$ be a finite group. The order of the group is not divisible by two. And for all $a,b\in G$ we have,
$(ab)^2=(ba)^2$ .
Show that $G$ is abelian.

2)Let $G$ be a finite group. The order of the group is not divisible by three. And for all $a,b\in G$ we have,
$(ab)^3=a^3b^3$ .
Show that $G$ is abelian.
------------
Hacker's Guide to Group Theory.

Here is a really simple intro (which will be mathematically limited) to such a beautiful theory.

I presume you heard the term "set". If you did not then GET OUT OF THE FORUM. (It is a collection of something, called elements. Like the set of all positive integers.)

Definition: A finite set is a set which has a finite number of elements. For example, $S=\{1,2,3\}$ is finite. And $S=\{0,1,2,...\}$ is infinite.

Definition: A binary operation is something that operates two elements into another element. For example, $+$ is a binary operation on $\{0,1,2,...\}$ as you can see $1+2=3, 2+3=5, 6+9=15,...$ . It takes two elements and transfroms them into another element.

Definition: A closed non-empty set under some binary operation. Is a set such that the result after the binary operation is still in the set. For example $\{0,1,2,3,...\}$ is closed under $+$ because the sum of any two of these is again an element in the set. Mathematicians like to use the notation " $\in$ " which means "element of". Thus, symbolically
$a*b\in S$ where $a,b$ any two elements in $S$ and $*$ is the binary operation. For example, $T=\{1,2,3,...\}$ is not closed under $/$ (division).

It should be made clear here that $a*b$ is not necesarry the same as $b*a$ as in our division example just above.

Definition: An "identity element" in a non-empty set with a binary operation such that has an element $e$ which has the following property,
$a*e=e*a=a$
For any $a\in S$ .
For example $N=\{0,1,2,3,...\}$ has an identity element $0$ because we have,
$x+0=0+x=x$ for any $x\in N$ .

Definition: An "inverse of an element" in a set with an identity element is an element that when operated with this element both ways returns back the identity element.
For example, $N=\{0,1,2,3,...\}$ it has no inverse for $1\in N$ because we need that,
$1+x=x+1=0$ . BUT. If we introduce the negative, $N'=\{-3,-2,-1,0,1,2,...\}$ then we do. Here is another example $Q=\{\mbox{positive rationals}\}$ under multiplication. First we show that $Q$ has an identity element, because $1*x=x*1=x$ for any $x\in Q$ . And it has an inverse for an element, all you do is flip the fraction. Thus, if you want the inverse of $1/2$ you flip it to get $2/1$ which is the inverse. Because $(2/1)*(1/2)=(1/2)*(2/1)=1$ . The important thing I said was positive. Because If I did not say that then zero has no invese

.

Definition: A binary operation on a non-empty set is said to be "associative" when we have $a*(b*c)=(a*b)*c$ . The standard sets we look upon before are all associative. If we definie $*$ on $N=\{0,1,2,...\}$ , to be $a*b=a$ then it is associative. Because $1*(2*3)=(1*2)*3$ . It is unusual to find an non-associative example, because they are one of the most important properties in a binary operation.

Definition: A non-empty set with some binary operation $*$ is a "group" when it is: closed, associative, has identiy element, for each element has inverse.

Just to mention some notation $a^{-1}$ means the "inverse of $a$ " and as you guessed it means the inverse of $a$ . It happens to be that the inverse is unique, that there is no ambiguity in writing this, mathematicians say, "well-defined".

I would also like to mention. That mathematicians are lazy and they do not want to waste time writing $a*b$ . Rather they juxtapose the two elements $ab$

Here are some important theorems, that you might use to prove the above problem.

Definition: The "order of an element" (if it exists) in a group is the smallest positive integer such that $a^n=e$ . That means $\underbrace{aaa....a}_n=e$ (identity element). The "order of a group" (finite) is the number of elements in it.

Theorem: In a finite group the order of any element divides the order of the group!

Theorem: In a group $(ab)^{-1}=b^{-1}a^{-1}$ .

Funny. I just realized I explained everything except the most important term, "abelian". When we have $a*b=b*a$ this is called commutative. Thus, a commutative group is called "abelian". After a great mathematician Neils Henrik Abel, who made fundamental work on commutative groups.
Now you know enough to show this.

There is another important rule, I forgot to mention. Sorry.
Theorem: In a group we have "left-right cancellation laws". Meaning that if $ax=ay$ we can cancel the left elements to get $x=y$ . We cannot conclude that $ax=ya$ and cancel. Also, if we have $xa=ya$ again cancelation laws say $x=y$ . (Note: There is no such thing as division by zero in group theory. So nothing to worry about).

galactus · #2 $Old$ December 18th, 2006, 05:20 PM

I just recently started studying a little group theory, PH, so please, be gentle.

Since no one else has taken a stab. What the heck.

Assume G is a finite group and its order isn't divisible by 2, and $(ab)^{2}=(ba)^{2}$

$g:G\rightarrow{G}$
$g(x)=x^{2}$

This is a homomorphism. If $x^{2}=1$ for $x\neq{1}$ , then the cycle group created by x is a subgroup of G of order 2.

Lagranges theorem says no way, since the order of G is not divisible by 2.

(I think.). And given y in G, there is one y in G where $x=y^{2}$

Our hyp. says:

$(ab)^{2}=(ba)^{2}$
$abab=baba$

Make the appropriate cancellations:

$(ba)^{2}=b^{2}a^{2}$
$baba=bbaa$

We can write $ab=ba$

Sorry, if I went off on a tangent or babble.

ThePerfectHacker · #3 $Old$ December 18th, 2006, 05:50 PM

Quote:

Originally Posted by galactus $View Post$

This is a homomorphism. If $x^{2}=1$ for [math]x\neq{1}

Again, I did not select problems that require anything that is even slightly advanced, like homomorphisms. But I do not see how, $\phi: G\to G$ defined as squaring the element is a homomorphism. Because we require that $\phi(xy)=\phi(x)\phi(y)$ not $\phi(xy)=\phi(yx)$ . If the problem was $(ab)^2=a^2b^2$ then yes that would be a homomorphism. But then the solution is simple expand $abab=aabb$ cancelation laws state that $ab=ba$ . Q.E.D.

Which textbook, you use?

Funny, I was with my advisor today registering for math classes. He was speaking on the phone with somebody telling that he will teach an algebra class. He mentioned he is going to use "my book". I wonder if he can to that conclusion before I mentioned it

.

galactus · #4 $Old$ December 18th, 2006, 06:18 PM

Thanks for the correction on the homomorphism. You are so right. I knew that. Just boo-booed. I was mixing it up with the other problem. That, I believe is the homomorphism. I was on the right track with the cancellations then?. Good. Anyway, I have been using Fraleigh's 5th edition. It's what I have on hand and it only cost a few dollars.

ThePerfectHacker · #5 $Old$ December 18th, 2006, 06:37 PM

Quote:

Originally Posted by galactus $View Post$

Thanks for the correction on the homomorphism. You are so right. I knew that. Just boo-booed. I was mixing it up with the other problem. That, I believe is the homomorphism. I was on the right track with the cancellations then?. Good. Anyway, I have been using Fraleigh's 5th edition. It's what I have on hand and it only cost a few dollars.

I would like to make a recommendation. Earlier versions of his book start out the same, the first three sections (do not know how to call them). It changes when you reach the fourth chapter. In the fourth chapter, in earlier versions, I think in your version, he immediately goes into more advanced group theory. In the newer versions, like mine, he avoids that chapter until the need arises. Meaning he goes into Rings and Fields after section 3 and then into Prime and Maximal Ideals. And then into some Field theory (the chapter with Algebraic extensions). And only after all of that he starts doing group theory on a more advanced level. I agree with that approach, I think that chapter is more difficult than those and there is no need to learn it. But maybe, you do not care about field theory and want to jump into more group theory.

ThePerfectHacker · #6 $Old$ December 27th, 2006, 12:18 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

1)Let $G$ be a finite group. The order of the group is not divisible by two. And for all $a,b\in G$ we have,
$(ab)^2=(ba)^2$ .
Show that $G$ is abelian.

I got this problem somewhere off the internet.
The problem and solution are not mine.

We have for all elements,
$(ab)^2=(ba)^2$
Left multiply,
$(ab)^{-1}(ab)^2=(ab)^{-1}(ba)^2$
$(ab)^{-1}(ab)(ab)=(ab)^{-1}(ba)^2$
$(ab)=(ab)^{-1}(ba)(ba)$
Right multiply,
$(ab)(ba)^{-1}=(ab)^{-1}(ba)(ba)(ba)^{-1}$
$(ab)(ba)^{-1}=(ab)^{-1}(ba)$
Square,
$[(ab)(ba)^{-1}]^2=[(ab)^{-1}(ba)]^2$
Using the propery of this group,
$[(ab)(ba)^{-1}]^2=[(ba)(ab)^{-1}]^2$
$[(ab)(ba)^{-1}]^2[(ba)(ab)^{-1}]^{-2}=e$
$[(ab)(ba)^{-1}]^2[(ab)(ba)^{-1}]^2=e$
$\{[(ab)(ba)^{-1}]^2\}^2=e$
Since, the order is not divisible by 2 by Lagrange's theorem it must by the identity,
$[(ab)(ba)^{-1}]^2=e$
Thus,
$(ab)(ba)^{-1}=e$
$ab=ba$ .
Q.E.D.
Thus the group is abelian.

Quote:

2)Let $G$ be a finite group. The order of the group is not divisible by three. And for all $a,b\in G$ we have,
$(ab)^3=a^3b^3$ .
Show that $G$ is abelian.

This problem appears in Hernstein "Topics in Algebra".
Somewhere near the begining of the book. It has a star next to it indicating a difficult problem.

The solution was given to me by a professor I discuss math with. I like his solution very much but it is a bit too advanced. I will present it and then present my revised elementary version of what he did which can be followed by the tutorial on group theory above.
~~~
The proof comes down to considering the product,*)
$aba^{-1}$ .
If we cube it we have,
$(aba^{-1})^3=ab^3a^{-1}$
But by the property by the group we have,
$a^3b^3a^{-3}$
Thus,
$ab^3a^{-1}=a^3b^3a^{-3}$
Cancellation laws and moving elements around,
$a^2b^3=b^3a^2$ .
~~~
Now we reach the important point in the proof.
Namely, each element can be expressed as a cube of some other element.
~~~
This is what the professor did,
$\phi: G\to G$
Defined as $\phi(x)=x^3$
Is a homomorphism.
Because the order is not divisible by 3.
$\ker \phi = \{ e\}$
Thus, $\phi$ is injective on itself.
And any injective map on the finite set itself is surjective.
Thus, any element is expressable as a cube of some other element.
---
What he really did was employed Dirichlet's Pigeonhole Principle only in math language.
---
He is my elementary version.
Consider all the elements of the group,
$\{x_1,x_2,...,x_n\}$
Now cube them,
$\{x_1^3,x_2^3,...,x_n^3\}$
Note none are equal to each other, because if,
$x_i^3=x_j^3$
$x_i^3x_j^{-3}=(x_ix_j^{-1})^3=e$
Order is not divisible by three implies,
$x_ix_j^{-1}=e \to x_i=x_j$ .
A contradiction.
Thus, $\{x_1,x_2,...,x_n\}$ und $\{x_1^3,...,x_n^3\}$ are the same (not necessarily in the same order) (by the Pigeonhole principle).
Thus, any element is expressable as a cube.
~~~
Returning back to the problem we found that,
$a^2b^3=b^3a^2$
But, $b^3$ can represent any element for it is a cube.
Thus,
$a^2c=ca^2$ (for any $c\in G$ ).**)

By the conditions,
$(ab)^3=a^3b^3$
$ababab=aaabbb$
$baba=aabb$
$(ba)^2=a^2b^2$
But we said squares commute,
$(ba)^2=b^2a^2$
$baba=bbaa$
$ab=ba$
Q.E.D.
The group is abelian.

Bruno J. · #7 $Old$ June 18th, 2009, 06:13 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

I got this problem somewhere off the internet.
The problem and solution are not mine.

That is not a very pretty solution.

We know $(ab)^2=(ba)^2$ . We know there is no element of order 2.

Suppose $x^2=y^2$ ; then $(xy^{-1})^2=1$ . Since there is no element of order 2, we must have $xy^{-1}=1$ ; i.e. $x=y$ .
Hence $ab=ba$ for all $a,b$ .

halbard · #8 $Old$ July 9th, 2009, 08:09 PM

Sorry, Bruno J., but how does $x^2=y^2$ imply that $(xy^{-1})^2=1$ without assuming that $x^2y^{-2}=xy^{-1}xy^{-1}$ ? That is, without assuming that $xy^{-1}=y^{-1}x$ , i.e. that $xy=yx$ ?

TheAbstractionist · #9 $Old$ July 10th, 2009, 05:48 PM

Quote:

Originally Posted by halbard $View Post$

Sorry, Bruno J., but how does $x^2=y^2$ imply that $(xy^{-1})^2=1$ without assuming that $x^2y^{-2}=xy^{-1}xy^{-1}$ ? That is, without assuming that $xy^{-1}=y^{-1}x$ , i.e. that $xy=yx$ ?

Good point. As a counterexample, we can take $G=S_3$ and $x=(1\,2),\ y=(1,3).$ Then $x^2=y^2$ but $xy^{-1}$ has order 3. (Of course $S_3$ is not a group of odd order, but Bruno iddn’t make any assumption about group orders in going from $x^2=y^2$ to $(xy^{-1})^2=1.)$