(This question came up in a conversation with my professor last week.)
Let $\langle G,\cdot \rangle$ be a group. Let $x$ be an element of $G$.
Is there always an isomorphism $f : G \to G$ such that $f(x) = x^{-1}$ ?
What if $G$ is finite?
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Sign up to join this community(This question came up in a conversation with my professor last week.)
Let $\langle G,\cdot \rangle$ be a group. Let $x$ be an element of $G$.
Is there always an isomorphism $f : G \to G$ such that $f(x) = x^{-1}$ ?
What if $G$ is finite?
The Mathieu group $M_{11}$ does not have this property. A quote from Example 2.16 in this paper: "Hence there is no automorphism of $M_{11}$ that maps $x$ to $x^{−1}$."
Background how I found this quote as I am no group theorist: I used Google on "groups with no outer automorphism" which led me to this Wikipedia article, and from there I jumped to this other Wikipedia article. So I learned that $M_{11}$ has no outer automorphism. Then I used Google again on "elements conjugate to their inverse in the mathieu group" which led me to the above mentioned paper.
EDIT: Following Geoff Robinson's comment let me show that any element $x\in M_{11}$ of order 11 has this property, using only basic group theory and the above Wikipedia article. The article tells us that $M_{11}$ has 7920 elements of which 1440 have order 11. So $M_{11}$ has 1440/10=144 Sylow 11-subgroups, each cyclic of order 11. These subgroups are conjugates to each other by one of the Sylow theorems, so each of them has a normalizer subgroup of order 7920/144=55. In particular, if $x$ and $x^{-1}$ were conjugate to each other, then they were so by an element of odd order. This, however, is impossible as any element of odd order acts trivially on a 2-element set.
No, such an isomorphism does not always exist, and the smallest counterexample is $G=C_5\rtimes C_4$ with $C_4$ acting faithfully. It is not hard to see that the only automorphisms of $G$ are inner, and that they cannot map an element of order 4 to its inverse.
Here's a comment which might as well be written down. If $f$ is required to be an inner automorphism, then for $G$ finite this question can be understood using the character table of $G$:
$x$ is conjugate to its inverse if and only if $\chi(x)$ is real for all characters $\chi$.
Since $\chi(x^{-1}) = \overline{ \chi(x) }$, one direction is clear. In the other direction, if $\chi(x)$ is real then $\chi(x) = \chi(x^{-1})$ for all characters $\chi$, hence $c(x) = c(x^{-1})$ for all class functions $c$. One also has the following cute result: the number of conjugacy classes which are closed under inversion is equal to the number of irreducible characters all of whose values are real (equivalently, the number of self-dual irreps). Since there exist plenty of groups (even simple groups) whose character tables have complex entries, there are plenty of groups with elements not conjugate to their inverses.
This is one way to address the question for finite groups with no outer automorphisms.