I am trying to prove Cayley's theorem in a way which avoids any particular maps from $G$ to a group of permutations on any particular set (contrary to the standard way of proving it, which involves associating every element $a$ with a permutation $ax$ as $x$ spans $G$). Here's an outline for Abelian groups of non-prime order:
- If groups $X$ and $Y$ are each isomorphic to a group of permutations, $X \times Y$ is also isomorphic to a group of permutations. (this is generally true for non-Abelian groups as well)
- We use strong induction on the order of $G$, assuming that all groups of order $(<k)$ are isomorphic to groups of permutations: the case for $1$ is trivially true.
- By the basis theorem for finite Abelian groups, every finite Abelian group $G$ is isomorphic to the product of cyclic subgroups (of $G$). Thus, by the inductive hypothesis, and $(1)$, we reach the desired conclusion. (I guess this fails when $G$ is of prime order)
However, the same approach fails when I try to apply it to non-Abelian groups. I have tried using the same techniques used in the proof of the basis theorem for finite Abelian groups. That entailed constructing a group $\langle x_1\rangle\times\langle x_2\rangle\times...\times\langle x_n\rangle$ of all or some elements of $G$ (I have also briefly tried selecting elements of prime power order). My approach was to prove the theorem from the fact that if a group $S$ is isomorphic to a group of permutations, then any subgroup of $S$ is also. So if we manage to find a group bigger than $G$ that satisfies the statement, we are done. But the one thing where this approach seems to fail is in proving that $\langle x_1\rangle\times\langle x_2\rangle\times...\times\langle x_n\rangle$ actually contains an isomoporphic copy of $G$. If you try to simply associate each product $(x_1, ..., x_n)$ with he result of multiplying all the elements in $G$, that does not generally produce a homomorphism precisely because of the non-Abelian nature of $G$:
if $f$ is the function that associates every element of $\langle x_1\rangle\times\langle x_2\rangle\times...\times\langle x_n\rangle$ with the result of multiplying all the individual elements of $(x_1, ..., x_n)$ in $G$,
$$f((x_1, ..., x_n)\times (y_1, ..., y_n)) = x_1y_1...x_ny_n$$ which is not generally equal to $$f(x_1, ..., x_n)\cdot f(y_1, ..., y_n) = x_1...x_ny_1...y_n$$
So my question is, is there a way of extending this construction-less approach to proving Cayley's theorem to all groups or is there no way of doing so? I think it might just be possible to do so because the basis theorem used in the Abelian approach has stronger implications than are needed for proving just this theorem. If it were possible to prove that $G$ is just equal to some product of proper subgroups, we would be mostly done. Only the prime-ordered cases would remain.