Show that every group of order $224 = 2^5*7$ has an element of order 14. I am reading a proof to the problem stated above and I have a few questions that I hope somebody helps me clarify. THe proof goes as follows:
Let $G$ be a subgroup of order 224 and $S$ be the set of $7$ Sylow subgroups of $G$. THe cardinality of $S$ divides 32 and is congruent to 1 mod 7, so has to be 1 or 8. Let $H$ be a 2-Sylow subgroup of $G$. It has 32 elements and acts on $S$ by conjugation.

This is my first question, how do we know that $H$ acts on $S$ by conjugation. My guess is that since $S$ is the set of $7$-sylow subgroups and the $7$ sylow subgroups are conjugate, then action by conjugation of an element $h \in H$, will just send one 7 Sylow subgroup to another one and so the conjugation of $S$ will just be $S$. Is this right or is there any other reason to have this action? If it is not right, how could we know when to use the action by conjugation?

Let $N\in S$ be a $7$-Sylow subgroup, and let $U$ be its stabilizaer subgroup in $H$. The $H$-orbit of $N \in S$ has at most 8 elements, so the stabilizer has $32/8=4$ elements.

Here I have another question, why does the stabilizer have just 4 elements? 

The group $U$ normalizes $N$. Let $\phi:U \rightarrow Aut(N)$ be the group homomorphism associated with the conjugation action of $U$ on $N$. Since $Aut(N)$ has $6$ elements and the number of elements of $U$ is divisible by $4$, the group homomorphism $\phi:U \rightarrow Aut(N)$ cannot be injective. 

It is still not clear for me why $\phi$ cannot be injective. Can someone explain the relationship between $Aut(N)$ having 6 elements, $U$ being divisible by $4$ and $\phi$ not being injective? 

We can choose a nontrivial element of order 2 in the kernel of $\phi$. We can also choose $g$ be a generator of $N$. Then $h$ and $g$ commute,$h$ has order 2 and $g$ has order 7. Then $hg$ has order 14.
 A: By Sylow's second theorem, the Sylow $p$-subgroups of $G$ are all
conjugate under $G$. Therefore they form one orbit under the action of
conjugation. Let $N$ be a given Sylow subgroup. Then, $NH=G$, and
$K$ fixes itself under conjugation, so the $H$-orbit of $N$ is the same
as the $G$-orbit of $N$ (in the conjugation action) so $H$ acts transitively
on the Sylow $7$-subgroups.
If there are $8$ Sylow subgroups, the stabiliser of $N$ in $H$ has order
$|H|/8=4$.
If there is an injective homomorphism $\phi:A\to B$ then $|A|\mid|B|$.
This is Lagrange's theorem, applied to $\phi(A)$.
ADDED IN EDIT
Here's another way of proving $G$ has an element of order $14$.
As in the question, we see $G$ has either $6$ or $48$ elements of order
$7$. The subgroup $H$ of order $32$ acts on these elements of order
$7$ by conjugation. It's impossible for all the orbits of this action
to have order $32$, so act least one has order a smaller power of $2$.
So there is $a\in G$ of order $7$ and a non-identity $b\in H$
with $bab^{-1}=2$. Some power $c$ of $b$ has order $2$. Then $cac^{-1}=a$.
Then $a$, $c$ commute and have orders $7$ and $2$ respectively.
Then $ac$ has order $14$.
A: 1) Any subset $\;X\;$ of $\;G\;$ acts on $\;S\;$ by conjugation since for any $\;x\in X,\,s\in S\,,\,\,x^{-1}sx\in S\;$ . This is true even without Sylow theorems, as conjugation (being an isomorphism) keeps order of subgroups.
2) We (must) know that for any action of a finite group of a finite set, $\;|\mathcal Orb(s)|=[G\,:\,Stab(s)]=\frac{|G|}{|Stab(s)|}\;$ . Since the left hand is either one or eight, the right hand is either one or eight, thus $\;|Stab(s)|\;$ is either $\;32\;$ or $\;4\;$.
3) You have a homomorphism $\;\phi: U\to\text{ Aut}\,N\;$ . If $\;|U|=32\;$ then $\;\phi\;$ clearly can't be injective, as the automorphism group of a group of order a prime is that prime minus one, thus $\;|\text{Aut}\,N|=7-1=6\;$ . Otherwise, $\;|U|=4\;$ , so it can't inject into a group of order 6 by Lagrange's theorem.
A: Question 1: Yes, that's exactly it. The entire group $G$ acts on $S$ by conjugation, so so does any subgroup of $G$.
Question 2: They do not say "just". What they say is "so the stabilizer has $32/8=4$ elements." There is no "exactly" in there. There is no "just". That means it is to be taken to mean "at least". They are saying that 4 distinct elements exist in that subgroup, but they make no implications about whether those are all the elements. They even say later "the number of elements in $U$ is divisible by $4$" not "is $4$".
Question 3: Lagrange's theorem. The image of $\phi$ is a subgroup of $Aut(N)$, and therefore its order must divide $6$. If $\phi$ were injective then the image of $\phi$ would be isomorphic to $U$ and therefore be divisible by $4$. No number divides $6$ and is divisible by $4$, so $\phi$ cannot be injective.
