TITLE: What is a group action, and how can we apply it to Sylow theory
QUESTION [5 upvotes]: I am studying Sylow theorems at the moment, more specifically trying to solve the following problem that I recently posted:
Let G be a finite group which has exactly eight Sylow 7 subgroups. Show that there exits a normal subgroup N of G such that the index [G:N] is divisible by 56 but not by 49
The link is here: If we have exactly 1 eight Sylow 7 subgroups, Show that there exits a normal subgroup $N$ of $G$ s.t. the index $[G:N]$ is divisible by 56 but not 49.
My initial response was to use Sylow's theorems to understand the order of the group. Though I am still very new at this. I was then told to look at this problem through group actions, which I know considerably less in. After taking a look at group actions again, I know just a few more things. Considering that I am just beginning at understanding this theory, I was wondering if someone can explain to me in laymen terms, the application of group actions and how to use them to solve problems. Of course, I know that I could just look up book definition, but I would like a little more insight than that. 
How would you explain group actions to someone with just calculus level knowledge maybe? What is the meaning of group actions, using language that is easy for a beginner to understand?
The definition of group action that I have found comes from Hungerford's Algebra. An action of a group $G$ on a set $S$ is a function $GxS \to S$ such that for all $x \in S$ and $g_1, g_2 \in G$: $ex=x$ and $(g_1g_2)x=g_1(g_2x)$

REPLY [2 votes]: In your case, $G$ acts transitively by conjugation on the set $\operatorname{Syl}_7(G)$, where $|\operatorname{Syl}_7(G)|=8$. Therefore, there's a homomorphism $\varphi\colon G \rightarrow \operatorname{Sym}(\operatorname{Syl}_7(G))$, whence (First Homomorphism Theorem):
$$G/\operatorname{ker}\varphi \cong \operatorname{im}\varphi \le \operatorname{Sym}(\operatorname{Syl}_7(G)) \tag 1$$
and thence (Lagrange):
$$[G:\operatorname{ker}\varphi]=|G/\operatorname{ker}\varphi| \mid 8! \tag 2$$
Now, by contrapositive, let's suppose $49 \mid [G:\operatorname{ker}\varphi]$; thence, by $(2)$, $49\mid8!$: contradiction. Therefore, $49 \nmid [G:\operatorname{ker}\varphi]$. So, if we prove that $56 \mid [G:\operatorname{ker}\varphi]$, then take $N=\operatorname{ker}\varphi$, and we are done. To this aim, let's remind that:
$$\ker\varphi=\bigcap_{P\in {\rm{Syl}_7(G)}}{\rm{Stab}}(P) \tag 3$$
Therefore:
\begin{alignat}{1}
56 \mid [G:\operatorname{ker}\varphi] &\iff 56 \mid \frac{|G|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}{\rm{Stab}}(P)|} \\
\tag 4
\end{alignat}
Now, by the Orbit-Stabilizer Theorem:
$$8\cdot|{\rm{Stab}}(Q)|=|G|, \forall Q \in \operatorname{Syl}_7(G) \tag 5$$
and thence:
\begin{alignat}{1}
56 \mid \frac{|G|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}{\rm{Stab}}(P)|} &\iff \\
56 \mid \frac{|G|}{|{\rm{Stab}}(Q)|}\cdot\frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}{\rm{Stab}}(P)|} &\iff \\
56 \mid 8\cdot\frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}{\rm{Stab}}(P)|} &\iff \\
7 \mid \frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}{\rm{Stab}}(P)|} \\
\tag 6
\end{alignat}
But this latter holds, as a corollary of this general claim (take $p=7$). So $\operatorname{ker}\varphi$ is the sought normal subgroup, $N$, of $G$.