I am trying to prove the following as exercises.
Let the group $G$ act on the set $X$. We define the $kernel$ of this action as the normal subgroup $K = \{g \in G | \forall x \in X: g \cdot x = x\}$.
(a) Construct an action of $G/K$ on $X$ induced from the action of $G$.
(b) Show that the action $G/K$ on $X$ is faithful.
(c) Suppose $G$ acts non-trivially on a set of size $N$. Show that $G$ has a proper normal subgroup of index at most $n!$.
(d) Show that an infinite simple group has no proper subgroups of finite index.
My attempts for parts (a) and (b) are the following though I'm not confident about it. As for the rest, I am unsure how to begin.
For part (a), we show that there is an action $\cdot: G/K \times X \rightarrow X$. Well, we have $e_{G/K} = e_G$ and so $\forall x \in X (e_{G/K} \cdot x = x)$ and given $h \in G/K$, $h = gk$, $g \in G, k \in K$, and so $h \cdot x = (gk)\cdot x = g \cdot (k \cdot x) = g \cdot x$ so $gkx = gx$.
For part (b), suppose there exists $z \neq e_G \in K$, we have $\forall x (z \cdot x = x = e \cdot x)$. But this implies that $z \cdot x = e \cdot x$ and multiplying $x^{-1}$ on the right side of both equalities we have $z = e$ which is a contradiction so it must be the case that $z = e$.
In general, when we say that a group $G$ acts on a set $X$, we often times will give some sort of definition that looks like: $g.x = x'$. One way to think about this definition is to consider the group action as a mapping of the form $G \times X \rightarrow X$. However, we can also think of the the group action as a homomorphism that takes $G$ to $S_X$.
Let's call this homomorphism $\rho: G \rightarrow S_X$. Then, for any given $g \in G$, $\rho (g)$ is a bijection to and from our set $X$. Then consequently, the we can express the notation $g.x$ as the following: $$g.x = \rho(g) (x)$$
This is a very useful way to think about group actions and you should take some time to convince yourself that both definitions are equivalent.
Problem (a)
When we say that $G$ acts on the set $X$, we are claiming that there exists a homomorphism $\rho: G \rightarrow S_X$. Notice that for any point $g \in K$, it holds that $\rho$ maps $g$ to the identity map in $S_X$. Hence, $K$ is the kernel of the group homomorphism $\rho$.
Hint: Use the first isomorphism theorem.