Let $f:G \rightarrow S_{G/H} $ the morphism induced by the action of $G$ by translation. Prove that $f(G)$ is not trivial.

Question

Question:

Let $G$ a finite group and $p$ the smallest prime number that divides $|G|$ . Let $H$ a sub group of indices $p$ of $G$. We write $f:G \rightarrow S_{G/H} $ the morphism induced by the action of $G$ by translation. Prove that $f(G)$ is not trivial.

My answer:

1-
By hypotesis we know that the index of $|G/H| = p >1 $ hence $\exists g \in G$ s.t. the equivalent class of $g$ written $[g]$ is different from the equivalent classe of $e_G$ written $[e_G]$.

2-
Now let define the following action translation of the group $G$ on $G$, $L_g(x)=[g] \circ [x]$ with the same $g$ as in "1-". Thus we see that this translation belongs to $f(G)$ by definition and that $L_g(e_G) =[g] \circ [e] = [g] $ that is not equal to the equivalent class of $[e_G]$ so $f(G)$ is not trivial.
Q.E.D.

I have a question concerning my part "2-" that depends directly to my understanding of the question. Indeed when in the question it is written:

We write $f:G \rightarrow S_{G/H} $ the morphism induced by the action of $G$ by translation.

It means the set of all possible action by translation of $G$ on $G$? Or does it mean something else?

Thank you for your help.

Answer 1

Let $G$ be a group, and let $H$ be any subgroup.

The set $X$ of (left) cosets of $H$ in $G$ is the set $$X = \{gH\mid g\in G\}.$$ It has $[G:H]$ elements.

The group $G$ has a natural action on $X$ by (left) translation: we define the function $\cdot\colon G\times X\to X$ by $$a\cdot (gH) = (ag)H.$$ This is well-defined: if $gH=g'H$, then $g^{-1}g'\in H$. Then $(ag)^{-1}(ag')=g^{-1}a^{-1}ag' = g^{-1}g'\in H$, so $(ag)H = (ag')H$.

This is an action: for all $gH\in X$ we have $e\cdot(gH) = (eg)H =gH$. And for all $a,b\in G$, $(ab)\cdot gH = (abg)H = (a(bg))H = a\cdot(bgH) = a\cdot(b\cdot(gH))$.

Like any action, the action of $G$ on $X$ by left translation induces a morphism $G\to S_X$, the group of permutations of $X$. This morphism is the morphism induced by the action of $G$ on $X$.

So that sentences refers to a specific, uniquely determined, action of $G$ on the set of left cosets of $H$. It does not mean "all possible actions" of anything, it does not mean any action of $G$ on itself (it is an action of $G$ on the cosets of $H$, not on $G$). So it definitely means "something else".