I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,\Bbb C)$. In my course it is given for $(g,\phi)\in G\times F(X,\Bbb C) $ by:

$g*\phi(x\in X)\to \phi(g^{-1}x)$.

Is this a left group action?

I have: $h*(g*\phi(x))= h*\phi(g^{-1}x)= \phi(h^{-1}g^{-1}x) = \phi((gh)^{-1}x)= (gh)*\phi(x) \neq (hg)*\phi(x)$.

I must be missing something.


Note that by the definition $g*\phi$ is in $F(X,\mathbb C)$ defined by $(g*\phi)(x):= \phi(g^{-1}x)$, so the calculation is the following: $$\begin{align} (h*(g*\phi))(x)&= (g*\phi)(h^{-1}x)\\ &= \phi(g^{-1}h^{-1}x)\\ &= \phi((hg)^{-1}x)\\ &= (hg*\phi)(x). \end{align}$$ Thus $h*(g*\phi)= hg*\phi$.


In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*\phi)(x) := \phi(g^{-1}x)$, we have $g*\phi = x \to \phi(g^{-1}x)$, therefore

$$\begin{align} (h*(g*\phi))(x) &= (h*(t\to\phi(g^{-1}t)))(x)\\ &= (t\to\phi(g^{-1}t))(h^{-1}x)\\ &= \phi(g^{-1}(h^{-1}x))\\ &= (hg*\phi)(x). \end{align}$$


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