Thread: Question 11
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Old December 18th, 2006, 05:20 PM
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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.