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.