Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. \tag{*}\]
Then we compute
\begin{align*}
(xg)^2&=(gx^{-1})(xg)=g^2\\
(xg)^3&=(xg)(xg)^2=(xg)(g^2)=xg^3\\
(xg)^4&=(xg)^2(xg)^2=g^2g^2=g^4.
\end{align*}

In general, we have by induction
\begin{align*}
(xg)^k=
\begin{cases}
xg^k & \text{ if } k \text{ is odd}\\
g^k & \text{ if } k \text{ is even}. \tag{**}
\end{cases}
\end{align*}

Let $m$ be the order of $g$. Since $G$ is a finite group of odd order, $m$ is odd.
We have by (**)
\[(xg)^{m-1}g=g^{m-1}g=g^m=1.\]
Thus we have $g^{-1}=(xg)^{m-1}$. Multiplying by $xg$ from the right, we have
\begin{align*}
g^{-1}xg &=(xg)^m\\
&=xg^m \text{ by (**)}\\
&=x.
\end{align*}

Thus we have $gx=xg$, and combining with (*) we obtain $gx=gx^{-1}$.
This implies that $x^2=1$, and since the order of $G$ is odd, this implies $x=1$.

This contradicts our choice of $x$.
Therefore $x$ cannot be conjugate to $x^{-1}$.

Finite Group and a Unique Solution of an Equation
Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]
We give two proofs.
Proof 1.
Since $m$ and $n$ are relatively prime […]

Are Groups of Order 100, 200 Simple?
Determine whether a group $G$ of the following order is simple or not.
(a) $|G|=100$.
(b) $|G|=200$.
Hint.
Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow's […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

A Group with a Prime Power Order Elements Has Order a Power of the Prime.
Let $p$ be a prime number. Suppose that the order of each element of a finite group $G$ is a power of $p$. Then prove that $G$ is a $p$-group. Namely, the order of $G$ is a power of $p$.
Hint.
You may use Sylow's theorem.
For a review of Sylow's theorem, please check out […]

If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group
Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.
Proof.
Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]
Multiplying the equality by $yx$ from the right, we […]

If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself
Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.
Then show that $N_G(H)=H$.
Hint.
Use the conjugate part of the Sylow theorem.
See the second statement of the […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set
Let $X$ be a subset of a group $G$. Let $C_G(X)$ be the centralizer subgroup of $X$ in $G$.
For any $g \in G$, show that $gC_G(X)g^{-1}=C_G(gXg^{-1})$.
Proof.
$(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$.
Take any $h\in C_G(X)$. Then for […]