Group action on vector space of all functions G to $\mathbb{C}$ 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.
 A: The contragredient action is pretty tricky. This trickiness is actually seen all the way back in intermediate algebra, where the graph of $y=f(x+a)$ is the graph of $y=f(x)$ shifted $a$ to the left, and similarly the graph of $y=f(ax)$ ($a>1$) is the graph of $y=f(x)$ shrunk by a factor of $a$, even though $x\mapsto x+a$ shifts to the right and $x\mapsto ax$ stretches by a factor of $a>1$. This confusion has come up innumerable times in my experience as a T/A, and the fact that the contragredient action of permutations is indeed a right action similarly gave me serious pause at first.
Suppose we wish to compute where $\sigma_1(\sigma_2 f)$ sends $x$. Denote $g=\sigma_2 f$, so $g(x):=f(\sigma_2^{-1}x)$. Hence we arrive at $\sigma_1(\sigma_2 f)=\sigma_1 g:x\mapsto g(\sigma_1^{-1} x)=f(\sigma_2^{-1}(\sigma_1^{-1}x))=f(\sigma_1^{-1}\sigma_2^{-1}x)=f((\sigma_1\sigma_2)^{-1}x).$
My understanding of why this action works out, and how it can be motivated, also goes back to the basics with graphs. Suppose we have the graph of $y=f(x)$ on the real line, and want to shift it $1$ unit to the right. To shift the real line to the right, we would need to apply $\sigma:x\mapsto x+1$. However simply applying $\sigma$ after $f$ will shift the graph of $y=f(x)$ up to $y=f(x)+1$. In other words, the notion of left/right is very much attached to domain rather than codomain. Let $y=h(x)$ be the graph of $y=f(x)$ shifted to the right $1$ unit. Then where $f$ sends $x$, $h$ must send $x+1$: that is to say, $f(x)=h(x+1)$, and so $h(x)=f(\sigma^{-1}x)$.
The key to generalizing is this: if we think of the graph as the set $\{(x,y):y=f(x)\}$, then "shifting to the right" means shifting all of the individual points to the right, as in $\{(\sigma x,y):y=f(x)\}$, and this can be rewritten as $\{(x,y):y=f(\sigma^{-1}x)\}$. In more generality, we can thing of the "graph" of an arbitrary function $f:X\to Y$ as a subset of the points in $X\times Y$, and if $G$ acts on $X$ then it acts on the graph by "translating" in the same way: move each point $(x,y)$ to $(\sigma x,y)$.
Thus, $\sigma f$ sends $\sigma x$ to $f(x)$. Here is a visual way of appreciating this. Many tests and assignments include matching problems, where there are two columns each with a list of items and the student must match items from column A to those in column B. List off the elements of $X$ in column and and those of $Y$ in column B (might as well make everything finite for convenience sake). Think of a function $f:X\to Y$ as a set of arrows from points in the left column to those in the right - each item on the left only has one arrow connected, though items on the right can have any number, including, zero, connected. To apply $\sigma$, simply permute the base points of the arrows: for each $x\in X$, take the arrow connected to $x$, and reconnect it to $\sigma x\in X$ instead.
A: Substitute $g:=\sigma_2f$, $\tau':=\sigma_1^{-1}\tau$. By definition, then, we have $$(\sigma_1 g)(\tau)=g(\sigma_1^{-1}\tau)=g(\tau')=(\sigma_2 f)(\tau')=f(\sigma_2^{-1}\tau')=f(\sigma_2^{-1}\sigma_1^{-1}\tau).$$
A: Let's define an action of $G$ on $L(G)$ by
$$(\sigma\cdot f)(\tau)=f(T(\sigma) \tau)$$
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 $f_\sigma$
$$(\sigma\cdot f)(\tau)=f_\sigma(\tau)=f(T(\sigma) \tau)$$
on one hand
$$(\sigma_1\cdot (\sigma_2\cdot f))(\tau)=(\sigma_1\cdot f_{\sigma_2})(\tau)=f_{\sigma_2}(T(\sigma_1) \tau)=f(T(\sigma_2) (T(\sigma_1) \tau))=f((T(\sigma_2)T(\sigma_1)) \tau)$$
on another hand
$$(\sigma_1\cdot (\sigma_2\cdot f))(\tau)=((\sigma_1\sigma_2)\cdot f)(\tau)=f(T(\sigma_1\sigma_2) \tau)$$
so we should have
$$T(\sigma_1\sigma_2)=T(\sigma_2)T(\sigma_1)$$
That means transform
$T(\sigma)=\sigma^{-1}$ is ok.
