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Prove that if $G$ is a finite group of odd order, then no $x\in$$G$ , other than $x=1$, is conjugate to its inverse.

This question is from Advanced Modern Algebra (exer 2.79) by Joseph J. Rotman.

The hint states that if $x$ and $x^{-1}$ are conjugate, how many elements are in $x^{G}$?

What I know so far:

  1. $\left\lvert x^{G}\right\rvert$ is odd (greater than 1, otherwise it's in the center) and is a divisor of |$G$|
  2. |Z($G$)| has a common factor (other than 1) with size of the orbit $\left\lvert x^{G}\right\rvert$ so that the center is not just the identity. This is from the class equation.
  3. The centralizer of $x$ has odd size and $\lvert C_{G}(x) \rvert \cdot \lvert x^{G}\rvert=\lvert G\rvert$

I don't see the implication of $x$ and its inverse being conjugate has other than their orbits having the same size and laying in the same conjugacy class.

Is the info I have useful? Any help would be appreciated.

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Your second observation is false. Groups of odd order can have trivial center. Even if they have non-trivial center, the size of a class is not related to the size of the center: the non-abelian group of order 105 with non-trivial center has classes of sizes 1, 3, 7, but a center of size 5. – Jack Schmidt Sep 17 '12 at 2:06
Wouldn't a conjugacy class of size one just mean that element is in the center? Thanks for catching my false observation. – user41442 Sep 17 '12 at 2:31
up vote 4 down vote accepted

Suppose $x \neq 1$ is conjugate to $x^{-1}$. Let $C(x)$ be the conjugate class containing $x$. If $x = x^{-1}$, then $x^2 = 1$. Hence $|G|$ is divisible by $2$. This is a contradiction. Hence $x \neq x^{-1}$. Since $|C(x)|$ is odd, $C(x)$ contains $y$ which is neither $x$ nor $x^{-1}$. Since $x$ is conjugate to $y$, $x^{-1}$ is conjugate to $y^{-1}$. Hence $y^{-1} \in C(x)$. Since $y \neq y^{-1}$, we must conclude that $|C(x)|$ is even. This is a contradiction.

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Thank you for the elegant response. – user41442 Sep 17 '12 at 2:19

Your point 1. is very useful. I don't think I need the other points; here's how I would proceed:

Try listing the elements of $x^G$. Well, we know $x$ is in there, and $x^{-1}$. If these two are distinct, then we know they're not the only two, because you told us that $|x^G|$ was odd. So some $y$ is in there too. But now what about $y^{-1}$?

Can you come up with a good reason why $x$ and $x^{-1}$ are distinct?

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x and its inverse are distinct otherwise (as mentioned in another suggestion) there would be an element of even order, but if that was true then the order of that element should divide the order of G. That can't happen as G has odd size. Thank you. – user41442 Sep 17 '12 at 2:11
Right, the answer Makoto Kato gave was pretty much what I was hinting at. – Ben Millwood Sep 17 '12 at 17:17

$|G|$ is odd implies that no element (other that 1) has even order. Suppose $x$ conjugate to its inverse, then $x=g(gx)^{-1}$ implies $gx=g^2(gx)^{-1}$, so we obtain that $g^2=1$, a contradiction is found.

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This is not correct. $g^2$ centralizes $x$, but it need not be the identity. – Jack Schmidt Sep 17 '12 at 2:02
Yeah, this argument confuses $gx$ on the LHS with $(gx)^{-1}$ on the right. – Ben Millwood Sep 17 '12 at 17:16

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