#1
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Group theory question
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) |
#2
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Re: Group theory question
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. |
#3
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Re: Group theory question
I don't believe your counterexample exists. |
#4
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I omitted some info from original problem
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. |
#5
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Re: Group theory question
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 |
#6
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Re: I omitted some info from original problem
The map sending x to x^3 for all x is an automorphism. Dunno if that helps. |
#7
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ANSWER
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! |
#8
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GENERALIZATION
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? |
#9
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NOTE on ANSWER
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! |
#10
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Neater Solution
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. |
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