element algebraically distinguishable from its inverse - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:06:04Z http://mathoverflow.net/feeds/question/63633 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63633/element-algebraically-distinguishable-from-its-inverse element algebraically distinguishable from its inverse Ricky Demer 2011-05-01T19:20:03Z 2011-05-02T13:49:50Z <p>(This question came up in a conversation with my professor last week.)</p> <p>Let $\langle G,\cdot \rangle$ be a group. Let $x$ be an element of $G$. <br> Is there always an isomorphism $f : G \to G$ such that $f(x) = x^{-1}$ ? <br> What if $G$ is finite?</p> http://mathoverflow.net/questions/63633/element-algebraically-distinguishable-from-its-inverse/63642#63642 Answer by GH for element algebraically distinguishable from its inverse GH 2011-05-01T20:22:41Z 2011-05-02T13:49:50Z <p>The Mathieu group $M_{11}$ does not have this property. A quote from Example 2.16 in <a href="http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3895v1.pdf" rel="nofollow">this paper</a>: "Hence there is no automorphism of $M_{11}$ that maps $x$ to $x^{−1}$."</p> <p>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 <a href="http://en.wikipedia.org/wiki/Outer_automorphism_group" rel="nofollow">Wikipedia article</a>, and from there I jumped to this other <a href="http://en.wikipedia.org/wiki/Mathieu_group" rel="nofollow">Wikipedia article</a>. 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. </p> <p><strong>EDIT:</strong> 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 <a href="http://en.wikipedia.org/wiki/Mathieu_group" rel="nofollow">Wikipedia article</a>. 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.</p> http://mathoverflow.net/questions/63633/element-algebraically-distinguishable-from-its-inverse/63643#63643 Answer by Tim Dokchitser for element algebraically distinguishable from its inverse Tim Dokchitser 2011-05-01T20:25:22Z 2011-05-01T20:25:22Z <p>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.</p> http://mathoverflow.net/questions/63633/element-algebraically-distinguishable-from-its-inverse/63644#63644 Answer by Qiaochu Yuan for element algebraically distinguishable from its inverse Qiaochu Yuan 2011-05-01T20:26:21Z 2011-05-01T20:26:21Z <p>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 $G$:</p> <blockquote> <p>$x$ is conjugate to its inverse if and only if $\chi(x)$ is real for all characters $\chi$.</p> </blockquote> <p>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.</p> <p>This is one way to address the question for finite groups with no outer automorphisms. </p>