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 […]

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 […]

Group Generated by Commutators of Two Normal Subgroups is a Normal Subgroup
Let $G$ be a group and $H$ and $K$ be subgroups of $G$.
For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$.
Let $[H,K]$ be a subgroup of $G$ generated by all such commutators.
Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup […]