# Does a Conjugacy Class always contain an element and its inverse?

The definition of a conjugate element

We say that $x$ is conjugate to $y$ in $G$ if $y = g^{-1}xg$ for some $g \in G$

Now for the group $G=Q_8$ , we have the group presentation $$Q_8 = \big<a,b: a^4 =1,b^2 = a^2, b^{-1}ab = a^{-1} \big>$$

Now the elements of $Q_8$ are $\{1,a,a^2,a^3,ab,a^2b,a^3b,b\}$ and after some calculation we would get $5$ different conjugacy classes, namely $a^G = \{a,a^3\}$ where $a^G$ denotes the conjugacy class of $a$ in $G = Q_8$,

also we have

$1^G = \{1 \}$, ${a^2}^G = \{ a^2 \}$, ${(a^2b)}^G = \{a^2b,b \}$ and ${(ab)}^G = \{ab,a^3b\}$

Of course , there is no surpise that for every element $x \in G$ we have $x \in x^G$ because $x = 1^{-1}x1$. However, we see that all the conjugacy classes for $Q_8$ contain the element and it's inverse. Like $a^{-1} = a^3$, ${(a^2)}^{-1} = a^2$, ${(a^2b)}^{-1} = b$ and so on.

My question is does this hold true for all groups ?

More formally , Is it true that for an element $x \in G$ then $x,x^{-1} \in x^G$ ?

• I once posed a question about the existence of some automorphsim (not necessarily inner) sending a given group element to its inverse; at least that excludes the Abelian counterexamples. The answer turns out to be still negative though (but with no really easy counterexamples). – Marc van Leeuwen Aug 11 '15 at 9:47

No, this does not hold: take any abelian group $G$, then $ab=ba$ for all $a,b\in G$, so $b^{-1}ab = a$ for all $a,b\in G$, so $a^G=\{a\}$ for all $a\in G$. So if $G$ contains an element of order different from $2$, it does not satisfy that $a,a^{-1}\in a^G$ for all $a$.
For a concrete example, take $G = \langle a: a^4=1\rangle$, then $a^{-1}=a^3\not\in a^G=\{a\}$.
No. Think of an abelian group. All the conjugacy classes contain a single element. Only elements of order ($1$ or) $2$ are their own inverse.
In a finite group $G$ of odd order $2n+1$, only the identity is conjugate to its inverse. For if $gxg^{-1} = x^{-1}$, then it is easy to see that $g^{2}xg^{-2} =x$, so that $g^{2}$ commutes with $x$. But $g = (g^{2})^{n+1}$ is a power of $g^{2}$, so that $g$ commutes with $x$. Hence $x = x^{-1}$, so that $x^{2} = 1_{G}$. But then $x = (x^{2})^{n+1} = 1_{G}$.
There is a connection to character theory of finite groups: a group is said to be ambivalent if every element is conjugate to its inverse. For a finite group, this is equivalent to every character of the group over complex numbers, being real-valued. It is easily seen that all symmetric groups $S_n$ are ambivalent. However, the alternating groups $A_n$ are ambivalent only for $n \in \{1,2,5,6,10,14\}$.