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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?

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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 $BG$.)

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

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