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

Quote:

Originally Posted by ThePerfectHacker

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.