Is it possible to show that there is no natural group structure on conjugacy classes in a group?

Alternatively, for a path connected space $ X $, the set $ [S^1,X] $ of free (unpointed) homotopy classes of loops in $ X $ is in bijection with the set of conjugacy classes of $ \pi_1(X) $. Is it possible to show that this has no natural group structure?


I interpret your question to mean the following: there is a functor, which I'll write $G \mapsto LG$, sending a group $G$ to its set of conjugacy classes. Is it possible to lift this functor to a group-valued functor?

The answer is no. To prove this, it suffices to find an inclusion $H \to G$ of groups inducing an inclusion $LH \to LG$ of conjugacy classes, but such that the order of $LH$ doesn't divide the order of $LG$ (so it can't be a group homomorphism with respect to any group structure on $LH$ or $LG$). A very small example suffices: the inclusion $C_2 \to S_3$ induces an inclusion $LC_2 \to LS_3$, but the former has order $2$ and the latter has order $3$.

(However, $LG$ is naturally a groupoid-valued functor; namely, it should really be thought of as returning the action groupoid $G/G$ of $G$ acting on itself by conjugation. The geometric realization of this thing is, among other things, the free loop space of $BG$.)

  • 1
    $\begingroup$ I love it when you start telling us what we should be doing ;-) $\endgroup$ – Mariano Suárez-Álvarez Feb 17 '15 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.