An Abstract Characterization of $S_5$ using involutions and their centralizers This is essentially an exercise from Jacobson's Basic Algebra I. (p.83, ex.10)

 I've managed to solve all the other part of the proof, except (vi) and (x). I've been thinking about this all day, but couldn't get an satisfactory answer.
In (vi), I don't see why there exists $H$ with $C(P)\subset H$ with 36 elements in it. What happens if $N(P)$ has 72 elements?
Finally, in (x), I don't see why the action of $G$ on the coset space is effective. For a given $g\neq 1$ does there exist some element $x$ of $G$ such that the coset $xN(V)$ is not fixed by $g$?. I don' understand why this is the case.
Some help would be nice. Thanks in advance!
 A: For item (vi):
Let $P=\{1,p,p^{-1}\}\subseteq C_1$. Because $C_1\cong Z_2\times S_3$, it is easy to see that $P$ is the unique subgroup of $C_1$ of order $3$, because there are only $3$ elements whose orders divide $3$, and that $C_1\nsubseteq C(P)$.
If $|N(P)|=72$, then $G=N(P)$, so for every $x\in G\setminus C(P)$ we have $xPx^{-1}=P$, but $xpx^{-1}\neq 1$ and $xpx^{-1}\neq p$, because $xpx^{-1}=p$ would imply $xp^{-1}x^{-1}=p^{-1}$, so $x\in C(P)$; therefore $xpx^{-1}=p^{-1}$. The converse is easy because $p^{-1}\neq p$. So $G\setminus C(P)$ is the set of $x\in G$ such that $xpx^{-1}=p^{-1}$. It is easy to see that $C(P)\neq G$, indeed, try to find an element $a\in C_1\setminus C(P)$.
Consider the function $F:G\rightarrow G$ such that $F(x)=axa^{-1}$. Then $F$ is a permutation of $G$, and $F[C(P)]=G\setminus C(P)$, so that $|C(P)|=\frac{72}{2}=36$.
For item (x):
Let $K$ the kernel of the action of $G$ on $G/H$. It easy to see that $|K|\mid|N(V)|=24$. We have some cases:

*

*If $|K|=24$, then $K=N(V)$, so $u_1\in K$ and $u_2\in K$, but $K$ is a normal subgroup, so $K$ has all conjugates of $u_1$ and $u_2$. There are $\frac{|G|}{|C(u_1)|}=\frac{120}{12}=10$ conjugates of $u_1$ and there are $\frac{|G|}{|C(u_2)|}=\frac{120}{8}=15$ conjugates of $u_2$. Moreover the conjugates of $u_1$ and the conjugates of $u_2$ form disjoint sets. Therefore $|K|\geq 10+15=25>24$, contradiction.


*If $2\mid|K|$, then $K$ has an involution $u$, so it has every conjugate of $u$. Because $u$ is conjugate to either $u_1$ or $u_2$, then either $u_1\in K$ or $u_2\in K$.
2,1) If $u_2\in K$, then because there are $15$ conjugates of $u_2$ we have $|K|\geq 15$, but $|K|\mid 24$, so $|K|=24$ and by case (1) we reach a contradiction.
2,2) If $u_1\in K$, then because of item (iii), $u_1u_2$ is conjugate to $u_1$, so $u_1u_2\in K$, so $u_2=u_1^{-1}(u_1u_2)\in K$, and by case (2,1) we reach a contradiction.


*If $|K|=3$, then $K$ is a normal Sylow $3$-subgroup of $G$, so because all Sylow $3$-subgroups are conjugate to each other (by Sylow's Theorems), then there is only one subgroup of $G$ order $3$. Looking at $C_1\cong Z_2\times S_3$, then there is a unique subgroup of $C_1$ of order $3$, so $K\subseteq C_1=C(u_1)$, so $u_1\in C(K)$, but by item (iii) the element $u_1u_2$ is conjugate to $u_1$, so there is $g\in G$ such that $u_1u_2=gu_1g^{-1}$, so $u_1u_2=gu_1g^{-1}\in gC(K)g^{-1}=C(gKg^{-1})=C(K)$, so $u_2\in C(K)$, but by (ii) we may assume that $u_2\in C_1$, but $u_2\neq u_1$, so looking again at the elements of $C_1\cong Z_2\times S_3$ then all involutions except for $u_1$ are outside $C(K)$, so $u_2\notin C(K)$, reaching a contradiction.

Therefore $|K|=1$, so the action is effective (faithful).
