Well-definedness of induced representation

Question

I am sorry if this is too silly but I can't, for the life of me, understand how this works. This is the definition I am working with.

Definition. Let $G$ be an arbitrary group and $H$ be a subgroup with a representation $V$. Then the induced representation from $H$ to $G$ is the vector space $$Ind_H^G=\{f:G\rightarrow V∣f(gh^{-1})=hf(g) \;\;\forall g\in G, h\in H\}$$ with action defined by $$gf(x)=f(g^{-1}x).$$

Why should this be a well-defined action? Wouldn't this require $$f(abx)=(b^{-1}a^{1}f)(x)=f(bax)?$$

Edit: This was a stupid question, I was doing $$f(abx)=(ab)^{-1}f(x)=b^{-1}a^{-1}f(x)=b^{-1}f(ax)=f(bax)$$ and now I realize I was just applying it in the wrong order. Sorry for wasting your time.

Answer 1

Where did you get $f(abx)=(abf)(x)$ from? That's not how the action is defined. In fact $(abf)(x)=f((ab)^{-1}x)=f(b^{-1}a^{-1}x)$. And that's the reason why you have $g^{-1}$ inside: precisely for associativity to work.

In order to check correctness we need the following two things:

1. The operator is well defined, meaning it maps $Ind_H^G$ to itself, i.e. $gf\in Ind_H^G$ for $g\in G$ in $f\in Ind_H^G$. This holds, because for any $x\in G$ and $h\in H$ we have $$(gf)(xh^{-1})=f(g^{-1}xh^{-1})=hf(g^{-1}x)=h\big((gf)(x)\big)$$ Note the subtleties here: $gf$ is the action of $g$ on $f$ (the last one defined). Inside $f$ we have group mutliplication, while $hf(x)$ is actually the action of $h$ (as an element of $H$) on $f(x)$ (as an element of $V$). In particular $hf(x)$ is not the same as $(hf)(x)$.
2. The operator is a group action. It is trivial that for the neutral element $e\in G$ we have $ef=f$. While the associativity goes as follows: $$(abf)(x)=f\big((ab)^{-1}x\big)=f(b^{-1}a^{-1}x)=(bf)(a^{-1}x)=\big(a(bf)\big)(x)$$