Group theory question

[Mano]

Going through my old textbook, and having difficulty with this proof. Figured someone here might have a solution:

Show that if (xy)^3 = (x^3)(y^3) for all x,y in a group G then G is abelian (i.e. xy = yx for all x,y in G)

[bigpooch]

Are you sure? Seems that I can envision a group H all of
whose elements b satisfy bbb=e where e is the identity but
is not abelian. Then the first condition holds trivially,
but maybe my "picture" is wrong.

I am just considering two sets of rotations of a globe
through a multiple of 120 degrees to generate all of the
elements of the group H. Does H then provide a simple
counterexample? Just off the top of my head, so may be way
off the mark!

Algebraically, can't really seem to get anywhere beyond
yxyx = xxyy for all x,y.

[thylacine]

I don't believe your counterexample exists.

[Mano]

Sorry, I apparently missed the first sentence of the problem when I copied it down - G must be a finite group whose order is not divisible by 3. That would rule out your example (since order of an element must divide the order of the group). Still having a hard time with it, so I restate it here in it's entirety:

Let G be a finite group whose order is not divisible by 3. Suppose that (ab)^3 = (a^3)(b^3) for all a,b elements of G. Prove that G must be abelian.

This is problem #24 on pg. 48 of I.N. Hersteins Topics in Algebra second edition.

[Mike Burchett]

For some reason it seems that if someone throws a probability problem, a partial differential equation, an analysis problem, etc. I am intrigued and try to solve it.

But my only reaction to the group theory problem was a case of the shakes.

Mike

BS Math Cal Poly 2001
Who somehow got a B- in Modern Algebra

[thylacine]

The map sending x to x^3 for all x is an automorphism. Dunno if that helps.

[bigpooch]

Well, it wasn't easy, but I believe I found a solution and
almost certainly it isn't the most elegant! First, it's
not hard to show that

(1) xxyy = yxyx for all x,y or similarly, yyxx = xyxy

Denote the identity of the group G by e.

Now, since G is finite, all of its elements have some finite
order, i.e., for all g, there exists an integer k such that
g^k = e. You may know that the lowest such k is called the
order of g. Now since G has an order not divisible by 3,
all of its elements have an order not divisible by 3.

Now, xy must have some order, so let's look at some cases.

If xy=e (xy has order 1), then x^-1 = y so trivially,
yx=e=xy as inverses are unique in a group so xy=yx.

If xy has order 2, xyxy=e and by (1) xyxy = yyxx = e
so yyxx(x^-1) = x^-1 so that (*):x^-1 = yyx. Now xyxy=e so
that (xy)(xy)=e and thus, xy = (xy)^-1 but the RHS of this
is just (y^-1)(x^-1)=(y^-1)(yyx) from (*) above. Hence,
xy = (y^-1)(yyx)= yx and so x and y commute in this case.

Now, suppose xy has order k>3 (it can't have an order that
is divisible by 3 because of the restriction on the order of
G) so it must have an order that is congruent to +/-1 mod 3.

Thus, either k = 3m+1 or k = 3m-1 for some integer m.
Assume k=3m+1 (a similar argument will follow for x^-1 and
y^-1 in the other case and you just have to show that if x
and y commute, so do x^-1 and y^-1.).

Thus, (xy)^3m+1 = e or (xy)^-1 = (xy)^3m = (xyxyxy)^m =
(xxxyyy)^m. This is the same as saying

xy = (y^-1.y^-1.y^-1.x^-1.x^-1.x^-1)^m

or xy = ((xxxyyy)^-1)^m

and similarly, by symmetry,

yx = (x^-1.x^-1.x^-1.y^-1.y^-1.y^-1)^m

or yx = ((yyyxxx)^-1)^m

Premultiplying by yyy and postmultiplying by xxx the first
of these two equations yields after simplification

yyyxyxxx = (x^-1.x^-1.x^-1.y^-1.y^-1.y^-1)^(m-1)
=((yyyxxx)^-1)^(m-1) = (((yyyxxx)^-1)^m).yyyxxx
=(yx).yyyxxx

Thus, yyyxyxxx = yxyyyxxx and simplifying yields

(2) yyx = xyy; again, by symmetry, xxy = yxx
This means a squared term of one of the factors can be moved
to the other side of the other factor without changing the
result.

Putting it all together: now

xxxyyy = x(xxy)yy
=x(yxx)yy
=xy(xxyy)
=xy(yxyx) by (1)
=(xyy)(xyx)
=(yyx)(xyx)
=yyxxyx
=yy(xxy)x
=yy(yxx)x
=yyyxxx.

(There are other ways to show xxxyyy=yyyxxx but I just show
one way, which again, may not be the "nicest"!)

Thus, xy((xxxyyy)^m)=e but of course, xxxyyy=yyyxxx
so xy((yyyxxx)^m)=e so xy is the inverse of (yyyxxx)^m
but also, yx((yyyxxx)^m)=e so yx is also the inverse of
(yyyxxx)^m, but inverses are unique and hence, xy=yx.

Again, the argument for k=3m-1 will be similar.

Can anyone suggest a "cleaner" or more elegant solution?
I would be seriously interested!

[bigpooch]

After a few minutes thought, the obvious general question
now is:

Suppose p is a prime and G is a group of finite order not
divible by p, ...

For what values of p do you need stricter conditions, or
how do you generalize the conditions for G to be abelian?

[bigpooch]

NOTE:

I forgot to put this in to clarify a part in the ANSWER:

If (xy)^k = e for some k, then
e = ye(y^-1) = y((xy)^k)(y^-1) = (yx)^k.
Hence, if xy has order k iff yx has order k.

I can't assume that the typical reader would know this!

[Robk]

Bigpooch wrote...

[ QUOTE ]
Algebraically, can't really seem to get anywhere beyond yxyx = xxyy for all x,y.

[/ QUOTE ]

From there we get that for all x and y,

x(yxyx) = (x^3)(y^2).

Reversing the roles of x and y gives

xyxy = yyxx which similarly implies

(xyxy)x = (y^2)(x^3).

Since x(yxyx) = (xyxy)x we have that for all x and y,

(x^3)(y^2) = (x^2)(y^3).

Now I claim this shows that if w is the third power of some element, w commutes with all of G. Let y be an arbitrary element of G.

w(y^2) = (y^2)w <=> wyy = yyw <=> w(wyy) = w(yyw).

Similarly,

w(y^2) = (y^2)w <=> wyy = yyw <=> (wyy)w = (yyw)w.

Since w(yyw) = (wyy)w, this gives

(w^2)(y^2) = (y^2)(w^2).

As bigpooch claimed, ywyw = (w^2)(y^2) = (y^2)(w^2) <=>

wyw = yww <=> wy = yw.

Now it remains to show only that every element of G is a third power of some other element. Given an arbitrary element w, we must have that its order is not divisble by 3. If the order of w is of the form 3k + 1, then (w^3k)w = e <=> w = w^(3(-k)). If the order of w is of the form 3k + 2, then (w^3k)(w^2) = (w^3k)(w^3)(w^-1) = e <=> (w^3k+3)(w^-1) = e <=> w = w^(3(k+1)), which completes the proof.

[BruceZ]

(x^3)(y^2) = (x^2)(y^3)

Now *this* is a breakthrough result! /images/graemlins/grin.gif

I think you want (x^3)(y^2) = (y^2)(x^3).

[bigpooch]

Good work! And of course, BruceZ just mentioned the one
line where you made a typo, but everything seems correct!

[Mano]

Thanks to all who posted - this helps alot. Although I got decent grades in college, abstract algebra was the one class that I never felt I really grasped well, yet it really intrests me. I am trying to go through my old course work to see if I can finally get a handle on it. On to quotient groups. /images/graemlins/tongue.gif

[bigpooch]

And to rings and fields and all kinds of isomorphisms? /images/graemlins/smile.gif

[Mano]

My goal is to eventually get to Galios theory. Back in college (many, many years ago) my proffessor touched on the subject (splitting fields and such) and I thought something like "gee, that's neat. Wish I had been paying more attention earlier in the year so I knew what the hell is going on here". Funny how much more motivated I am when I am doing math for the pleasure of doing it instead of because I have to.

[theBruiser500]

[ QUOTE ]
Funny how much more motivated I am when I am doing math for the pleasure of doing it instead of because I have to.

[/ QUOTE ]

Heh, I can relate to that. I failed math last semester and did poorly in physics. Just last week I was reading through my physics textbook just for fun, and I also got a math book out at the library, "When Least is Best." The math book looks good, can anyone here can verify that for me?

danny