# Permutation isomorphism issue with inverse

It is known that any group is isomorphic to a group of permutations. But if $\phi: G\to T$ is an isomorphism defined as $\phi_g(x) =gx$ for elements $g, x \in G$, then how can it be that $T_{g^{-1}}=[T(g)]^{-1}$? Because $T_{g^{-1}}(x)=g^{-1}x$, but $[T(g)]^{-1}=(gx)^{-1}=x^{-1}g^{-1}\neq g^{-1}x$.

What am I missing?

• My bad. Let me edit the post. Commented Nov 9, 2015 at 2:19

Claim: If $\phi: G \to T$ is an isomorphism, then $\phi(1_G) = 1_T$.

Proof: For any $g \in G$, $\;\phi(g) \phi(1_G) = \phi(g\, 1_G) = \phi(g) \in T$; therefore, $\phi(1_G) = 1_T$.

For this reason, left translation (or left multiplication) does not give an isomorphism except when $g = 1$, in which case $\phi_1 = \operatorname{id}_G$, the identity map on $G$.

What you are missing is the meaning of $[T(g)]^{-1}$. Indeed in what you wrote you say that :

$$[T(g)]^{-1}\text{ is the function from }G\text{ to } G \text{ sending } x\in G \text{ to } [T(g)(x)]^{-1}=(gx)^{-1}=x^{-1}g^{-1}$$

But $[T(g)]^{-1}$ does not mean this, it is the inverse mapping of $T(g)$. By definition (if it exists) it sends $T(g)(x)$ on $x$ for any $x\in G$.

Now $T(g)^{-1}(T(g)(y))=y$ so that $T(g)^{-1}(g.y)=y$ this is true for any $y\in G$, in particular if $y=g^{-1}.x$ :

$$g^{-1}.x=T(g)^{-1}(g.(g^{-1}.x))=T(g)^{-1}((gg^{-1}).x)=T(g)^{-1}(x)$$

If it exists then $T(g)^{-1}(x)=g^{-1}.x$. There is no choice. One can see that this is indeed the inverse function for $T(g)$.