TITLE: Group action on vector space of all functions G to $\mathbb{C}$
QUESTION [6 upvotes]: I have a simple question about this following action:
Let $L(G)$ be the vector space of all functions from $G$ to $\mathbb{C}$. Define an action of $G$ on $L(G)$ by 
$$(\sigma f)(\tau) = f(\sigma ^{-1}\tau)$$
for all $\tau \in G$.
To show its a group action we need to show $ef=f$ and $(\sigma _1 \sigma _2)f=\sigma _1 (\sigma _2 f)$, but
$((\sigma_1\sigma _2) f)(\tau) = f(\sigma _2 ^{-1} \sigma _1 ^{ -1} \tau)$
$(\sigma _1 (\sigma _2 f))(\tau) = f(\sigma _1 ^{-1} \sigma _2 ^{-1} \tau)$
In general, these won't be the same. What am I misunderstanding? I'm fairly certain this is really a group action, as my professor stated it.

REPLY [2 votes]: Let's define an action of $G$ on $L(G)$ by
$$(\sigma\cdot f)(\tau)=f(T(\sigma)\cdot \tau)$$
where T - some transform and $\cdot$ used to denote action. The key observation is: action $\sigma$ sends function $f$ to some another function which we denote $f_\sigma$
$$(\sigma\cdot f)(\tau)=f_\sigma(\tau)=f(T(\sigma)\cdot \tau)$$
on one hand
$$(\sigma_1\cdot (\sigma_2\cdot f))(\tau)=(\sigma_1\cdot f_{\sigma_2})(\tau)=f_{\sigma_2}(T(\sigma_1)\cdot \tau)=f(T(\sigma_2)\cdot (T(\sigma_1)\cdot \tau))=f((T(\sigma_2)T(\sigma_1))\cdot \tau)$$
on another hand
$$(\sigma_1\cdot (\sigma_2\cdot f))(\tau)=((\sigma_1\sigma_2)\cdot f)(\tau)=f(T(\sigma_1\sigma_2)\cdot \tau)$$
so we should have
$$T(\sigma_1\sigma_2)=T(\sigma_2)T(\sigma_1)$$
That means transform
$T(\sigma)=\sigma^{-1}$ is ok.