What is a group action, and how can we apply it to Sylow theory 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 $(g_1g_2)x=g_1(g_2x)$
 A: Note that another way to understand group actions is by "currying" the action into a function $G\to(S\to S)$; in this language a group action is just a homomorphism from $G$ to the permutation group on $S$. Thus in many ways the theory of group actions does not add anything more than you will get from understanding permutation groups and homomorphisms. Nonetheless, it can sometimes be useful to view group actions the "usual" way as a function $G\times S\to S$.
One great way to motivate the usage of group actions is to read the proof of Sylow's theorems. The "fundamental theorem" of group actions is the orbit-stabilizer theorem, and it is this that leads to most of the divisibility constraints in Sylow's theorems.
A: A group action on a finite set of k elements is nothing but a homomorphism from the group to $S_{k}$. Now, since all the 7 sylow subgroups are conjugate, then G acts by conjugation on them. This gives rise to a homomorphism from $G$ to $S_{8}$. Let K be the kernel, then by the first isomorphism theorem $[G:K]$ divides 8!. Since 49 does not divide 8!, then 49 does not divides [G:K].
For the other part, you might want to use the transitivity of the action and the orbit stabilizer theorem to show that $[G:N]$ is divisible by 8.
A: A group $G$ acting on a set $X$ means each element of $G$ leads to a permutation of the set. The effect of a permutation corresponding to $g\in G$ on an element $x\in X$ is written as $g.x$ or simply as $gx$. We can compose permutations and also apply group law on two elements of $G$. The permutations associated to these elements have to be such that an 'associative property' holds: that is what your last sentence means. and id element of the group should correspond to identity permutation.  Another way of saying the same this that we have a group homomorphism from $G$ to the group of permutations of the set $X$. Note that two different elements of $G$ may lead to the same permutation.
A: 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 $G$.
