# Thread: help me clean up a grp thry proof

1. ## help me clean up a grp thry proof

My proof of this problem looks clunky...suggestions for cleanup or different approaches are welcome. Sorry, I really need to get laTeX, this looks ridiculous without it.

From 2nd ed. of I.N. Herstein's Topics in Algebra:
Let G be a finite group whose order is not divisible by 3. Suppose that (ab)^3=a^3b^3 for all a,b in G. Prove that G must be abelian.

----
Denote the order of G by o(G) and G's identity element by e.

We first show that there can be no a in G other than e for which a^3=e. Since if G is a finite group and a is in G, then a^o(G)=e (Corollary 2, page 43, a consequence of the definition of the order of a group element). Then a^o(G)=e=a^3n for any integer n>0. But if 3 does not divide o(G) then o(G)=3m+r, where 3 does not divide m>0 and 0<r<3, m,r integers. Thus a^o(G)=a^(3m+r)=a^(3m)a^r=a^r, one of the two distinct powers of a which do not equal a^0=a^3=e, contradicting a^o(G)=e.

Now for all a,b in G, (ab)^3=ababab=a^3b^3. Cancellation shows baba=a^2b^2. The equation must be true when a and b are not each other's inverse and when a does not equal b does not equal e (that is, a, b and e are distinct from each other). But any association on baba (namely, (b)(aba), (ba)(ba), and (bab)(a)) forces a=b. The equation can only be true for all a,b in G if all a,b in G can be commuted. Thus G is abelian.

2. Beautiful problem! (It was posted here before).

Define the homomorphism:
$\displaystyle \phi : G\mapsto G$ as $\displaystyle \phi (x) = x^3$.
Notice that,
$\displaystyle \ker \phi = \{ 1 \}$.
So, $\displaystyle \phi$ is 1-to-1 (it is not divisible by 3).
By Pigeonhole Principle $\displaystyle \phi$ is surjective.

Now consider,
$\displaystyle a^3b^3a^{-3} = (aba^{-1})^3= ab^3a^{-1}$
So,
$\displaystyle a^2b^3 = b^3a^2$
Since $\displaystyle \phi$ is surjective $\displaystyle b^3$ can represent any element in $\displaystyle G$ so,
$\displaystyle a^2x=xa^2 \mbox{ for all }x\in G$.
Hence,
$\displaystyle a^2 \in Z(G)$ the center of $\displaystyle G$ (because it commutes with everything).

Now,
$\displaystyle (ab)^3=a^3b^3$
$\displaystyle ababab=a^3b^3$
$\displaystyle baba=a^2b^2$
$\displaystyle baba=b^2a^2$ ---> since $\displaystyle a^2 \in Z(G)$.
$\displaystyle ab=ba$
Q.E.D.

----------
Try doing the problem after that in Herstein's book

3. Sweet! That's really quite a nice proof. lol there's like, 2 lines that show what I took the whole first paragraph of my proof to show. There's just one part I don't follow, which is how phi is one-to-one. Elaborate a bit? I tried to figure it out but I feel I'm wasting time :P

4. Originally Posted by TheSicklyElf
Sweet! That's really quite a nice proof. lol there's like, 2 lines that show what I took the whole first paragraph of my proof to show. There's just one part I don't follow, which is how phi is one-to-one. Elaborate a bit? I tried to figure it out but I feel I'm wasting time :P
Are you familar with the rule that if the kernel is only the identity element then the map is one-to-one?

Say,
$\displaystyle \phi(x) = 1$
Then,
$\displaystyle x^3 = 1$
That means,
$\displaystyle x=1$ because otherwise if $\displaystyle x\not = 1$ then the order of $\displaystyle x$ is three and hence $\displaystyle \left< x \right>$ is a cyclic subgroup of order 3. But by Lagrange's theorem then 3 is a divisor of the order of this group ---> a contradiction!

5. OK I finally get it; if phi's kernel is 1 and phi(x)=phi(y) you have phi(x)[phi(y)]^(-1)=1 and following that logic, x=y.
Yeah Herstein has a lot of good problems. Thanks for sharing your proof, I definitely learned from it.