(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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$\begingroup$ I wonder if (in a non abelian maybe finite group) $x^{-1}$ is conjugate to $x$ ... $\endgroup$– Olivier BégassatCommented May 1, 2011 at 19:33
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$\begingroup$ You might try thinking of G as a permutation group (via Cayley's Theorem). $\endgroup$– Dan RamrasCommented May 1, 2011 at 19:45
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1$\begingroup$ @Olivier: No, in the alternating group $Alt(4)$, the 3-transposition $(1,2,3)$ is not conjugate to its inverse. (But it is in $Sym(4)$...) $\endgroup$– Alain ValetteCommented May 1, 2011 at 19:48
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2$\begingroup$ Olivier, see crazyproject.wordpress.com/2010/05/14/… $\endgroup$– Andreas ThomCommented May 1, 2011 at 19:51
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8$\begingroup$ Incidentally, this is important and irritating if you were say, creating a library of character tables (ctbllib), since an element and its inverse are (a) basically indistinguishable to the average eye or from the ATLAS, but (b) are actually distinguishable by nit-picking details of things you are in fact storing about groups. This complicates the consistency checking of tables, especially modular tables where (indeed in one of the Mathieu groups) an element and its inverse may actually be distinguished by the modular table but not the ordinary, or maybe by the homology but not the character. $\endgroup$– Jack SchmidtCommented May 2, 2011 at 0:51
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3 Answers
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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.
answered May 1, 2011 at 20:22
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31$\begingroup$ I like that this answer gives an explanation of how you can learn new math by clever searching. It's an important skill and one that people don't talk about so much. $\endgroup$ Commented May 1, 2011 at 22:58
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6$\begingroup$ It would have been nice to say in the answer itself which element of $M_{11}$ the element $x$ is. I think one choice is an element of $x$ order $11$, since a Sylow $11$-normalizer in $M_{11}$ has order $55$, so no element of order $11$ is conjugate to its inverse within $M_{11}$, which gives what you want since $M_{11}$ has no outer automorphisms. $\endgroup$ Commented May 2, 2011 at 7:46
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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.
answered May 1, 2011 at 20:25
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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.
answered May 1, 2011 at 20:26