Question about Sylow's theorems and a particular group of order 60 I have a finite group $G$ with the following data:
Its order |$G$| is 60, it has exactly 6 Sylow-5-subgroups $P_i$$\ $ (i=1,...,6) and |$N_G(P_i)$|=10 $\forall$ i.
I have the following questions:

1) How can I deduce that any two of these 6 normalizers have to intersect in a subgroup  $K$ of order 2?
2) If the normalizers are cyclic, why is the group $K$ then normal in $G$?

Thanks for the help.
 A: 1) Consider an element of order $2$ $a \in N_{G}(P_{1})$, say. Let $K = \langle a \rangle$ act by conjugation on $P_{1}, \dots, P_{6}$. If $a$ does not fix a certain $P_{j}$, then it swaps it with another $P_{k}$. Since $a$ fixes $P_{1}$, and there is an even number of $P_{i}$, $a$ has to fix at least another $P_{i}$. Let us say that $a$ fixes both $P_{1}$ and $P_{2}$.
2) Let $b$ be an element of order $5$ in $N_{G}(P_{1})$. Then $H = \langle b \rangle$ permutes cyclically $P_{2}, \dots, P_{6}$, say $P_{2}^{b^{i}} = P_{i + 2}$, for $i = 0, \dots , 4$. So we have for $i \ge 0$
$$
P_{i+2}^{a} = P_{2}^{b^{i} a} =  P_{2}^{a b^{i}}=  P_{2}^{b^{i}} = P_{i+2}.
$$
So $a$ normalizes all $5$-Sylow subgroups, so
$$
a \in \bigcap_{i=1}^{6} N_{G}(P_{i}),
$$
where the RHS is a normal subgroup, as the $P_{i}$ form a conjugacy class of subgroups, and clearly then $K = \cap_{i=1}^{6} N_{G}(P_{i}) \triangleleft G$.
A: Hints: use the following
Theorem: An element $\,a\,$ of order a power of $\,p\,$ normalizes a Sylow $\,p$-subgroup $\,P\iff a\in P\,$
Hint-sketch of proof: show we can assume $\,ord_G(a)=p\,$ , and now justify the following steps:
$$a\notin P\implies aP\neq \bar1\in N_G(P)/P\implies p\,\mid\,[N_G(P):P] $$
and this is a contradiction (to what and why?)$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square\,$
Well, now it follows at once that
$$a\in\,N_G(P_i)\cap N_G(P_j)\implies a\in P_i\cap P_j\;\;\text{ or }\;\; ord_G(a)=2$$
