proving Caley's Theorm using the conjugacy operation instead of left multiplication

Question

I need to prove Caley's theorem that each group G is isomorphic to a sub group of S(n). Wherever I check it is proven using the operation of multiplying from left side. that means $f_g(x) = g*x$

I proved it using the conjugacy operation -

$f_g(x) = g*x*g^{-1}$

I have been told it is wrong and can't figure out why..

any help will be aprriciated

Answer 1

The induced (by the conjugation action) morphism is one-to-one if and only if the center of the group is trivial.

Answer 2

The reason the left multiplication action works is because for fixed $g\in G$, the map $\psi_{g}: G\to G$ given by $x\mapsto gx$ is bijective (only as a set theoretic map!). Hence, the correspondence $G\to S_{|G|}$ given by $g\mapsto \psi_{g}$ is a group homomorphism (which then you easily check has a trivial kernel, so that completes Cayley's theorem).

Now, what goes wrong with using conjugation? Let's try the same thing: for each fixed $g\in G$, the map $\phi_{g}: G\to G$ is given by $x\mapsto g x g^{-1}$. Once again $\phi_{g}$ is bijective as a set map (in fact, it is a group homomorphism in this case). So then consider $G\to S_{|G|}$ given by $g\mapsto \phi_{g}$. This is no longer injective in general. Think about what happens when $g\in Z(G)$.