Group Q: If |a| is odd and aba^(-1)=b^(-1) then b^2=e.

[daon]

If |a| is odd and aba^-1=b^-1 then b²=e.

I have tried many routes here, but I usually end up going in a circle. I think the reason why is that I can't find a way to use the fact that |a| is odd. I tried saying that a^2k+1=e and using that but I still wound up where I started.

Thank you.
-daon

 

[pka]

daon, is it possible that there is a misprint in the question.
I have thought about this all day.
I doubt it is true. If the group is Abelian then it is clearly true.
So we are looking for a counter-example it has to be non-Abelian.

 

[daon]

Yes pka, that question is correct as quoted from the book. Page 84 #30 of Contemporary Abstract Algebra.

 

[daon]

I see what you mean about it needing to be Abelian.

I have been at this all afternoon and I may have figured something out. I'm not sure if this shows that it it abelian, but let me know:

We have aba^-1=b^-1. Taking the inverse of both sides we see: ab^-1a^-1=b. Substituting in aba^-1 for b^-1, we get: a(aba^-1)a^-1=b, Or a²ba^-2=b. Which implies a²b=ba².

You can also use the fact to show that b=a²ba^-2 = a⁴ba^-4 by recursive insertion of b. In this manner I have been able to "produce" the conjecture that b=a^(2k)ba^(-2k) for all integers k>=0 (since k=0 gives b=b). It may be all k, but I haven't had time to think about it.

So, from this I gather that a^2kb=ba^2k. Then, this would make the group Abelian for all elements x such that x=a^2k.

I'm not sure if we can say this, but: For some l, a^2l=a^-1 since |a| is odd (2l+1), we have a^-1b=ba^-1. Then we can multiply on the LHS by a to get b = aba^-1. As shown above, b=ab^-1a^-1, so, aba^-1=ab^-1a^-1. By canceling the a's and a inverses we get b=b^-1, or b²=e.

Let me know your thoughts.
Thanks,
-Daon

 

[pka]

I hate these sorts of problems. There is always some obscure trick.

If \(\displaystyle a^{2k + 1} = e\) clearly k>1. As you found out: \(\displaystyle a^{2k} b = ba^{2k} .\)

This implies that
\(\displaystyle \begin{array}{rcl}
a^{2k + 1} ba & = & aba^{2k + 1} \\
\Rightarrow ba & = & ab \\
\Rightarrow b & = & aba^{ - 1} \\
\Rightarrow b & = & b^{ - 1} \\
\Rightarrow b^2 & = & e \\
\end{array}\)

 

[daon]

Ah. This has been by far the most difficult problem I've been assigned. I suppose I'd need to prove that lemma too.

Anyway, thanks for making it look so simple, haha.

 

[pka]

Well, you are the one who found the trick!
BTW: I found the text you are using among review copies I have received.
As I recall the review was not favorable.
I find it exceeding odd that the author puts this problem in the section on cyclic groups. As I also recall, this author does that in more than one case, sometimes even misplacing problems.