TITLE: Does a Conjugacy Class always contain an element and its inverse?
QUESTION [9 upvotes]: 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 $x,x^{-1} \in x^G$ ?

REPLY [2 votes]: 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 $n \in \{1,2,5,6,10,14\}$.