Non simplicity of a group of order $p^{100}q$ given some conditions. Let $G$ be a group $p$, $q$ primes $p\neq q$, $|G|=p^{100}q$.
1.- Assume that $G$ has two different $p$-Sylow subgroups and intersection for each pair of $p$-Sylow subgroups is trivial. Show that there is a normal $p$-Sylow subgroup in $G$. (Probably this is a teacher mistake and ask for a $q$-Sylow subgroup)
2.- If that each $p$-Sylow subgroup of $G$ is commutative then $G$ is not simple.
Attempt: I tried 1.- and 2.- using Sylow theorem and counting (I think 1.- is a help for  2.-), I look cases when $p<q$, $p>q$ but I don't find a normal subgroup.
My attempt for 2.- is trying to prove for each pair of $p$-Sylow subgroup that the intersection is trivial and use 1.- to get no simplicity with a $q$-Sylow.
Edit: Im offering bounty for who proves the existence of a group of orden $p^{100}q$ where all pairs of $p$ Sylow subgroups intersects trivially, i.e. a non abelian group of order $p^{100}$ which is isomorphic to a semidirect product of one of his $p$ Sylow subgroups and the unique $q$ Sylow (normal) subgroup.
 A: Here are some examples. They only require matrix multiplication and the existence of finite fields to understand. Below them I also indicate why we need some sort of divisibility condition, and how to solve part 2.
Example: Let $p^a$ be a prime power dividing $q^b-1$, then there is a finite group $G$ of order $|G|=p^a q^b$ with $q^b$ abelian Sylow $p$-subgroups, each pair of which intersects trivially.
Proof: Let $F$ be a field with $q^b$ elements, and let $\alpha$ be an element of $F$ with multiplicative order $p^a$. Define $$G=\left\{ \begin{bmatrix} \alpha^i & \beta \\ 0 & 1 \end{bmatrix} : i \in \mathbb{Z}/p^a\mathbb{Z}, \beta \in F \right\} \leq \operatorname{GL}(2,F)$$ $$P = \left\{ \begin{bmatrix} \alpha^i & 0 \\ 0 & 1 \end{bmatrix} : i \in \mathbb{Z}/p^a\mathbb{Z} \right\} \qquad Q = \left\{ \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} : \beta \in F \right\}$$
Then $P$ is a Sylow $p$-subgroup, $Q$ is the unique Sylow $q$-subgroup. Let $x=\begin{bmatrix} \alpha^k & 0 \\ 0 & 1 \end{bmatrix} \in P \cap P^{g^{-1}}$ for $g=\begin{bmatrix} \alpha^j & \beta \\ 0 & 1 \end{bmatrix}$. Hence $xg =gy$ for some $y=\begin{bmatrix}\alpha^i & 0 \\ 0 & 1 \end{bmatrix} \in P^g \cap P$. However, direct calculation gives that $xg = \begin{bmatrix} \alpha ^{k+j} & \alpha^k \beta \\ 0 & 1 \end{bmatrix}$ while $gy=\begin{bmatrix} \alpha^{i+j} & \beta \\ 0 & 1 \end{bmatrix}$. Hence $\beta = 0$ so that $g \in P$ and $P^g=P$, or $\alpha^k=1$ so that $x=1$. Hence, if $P^g \neq P$, we get $x=1$ is the only element of $P^g \cap P = \{1\}$. More strongly we get that $P$ (and its conjugates) are the only $p$-local subgroups of $G$. Also $P$ is cyclic, and $Q$ is abelian. $\square$
However, the intersection is not forced to be trivial just by $P$ being abelian.
Example: Let $p^a$ be a prime power dividing $q^b-1$ and let $c$ be a non-negative integer. Then there is a finite group $H$ of order $|H|=p^{a+c} q^b$ with $q^b$ abelian Sylow $p$-subgroups, each pair of which intersects in a normal subgroup of order $p^c$.
Proof: Take $G$ form the previous example, and define $H=G \times C_{p^c}$. $\square$
If you don't have any kind of division condition, then there may not be any $G$ with more than one Sylow $p$-subgroup.
I assume familiarity with some early ideas of $p$-subgroups of a finite group, including Sylow's theorems, Burnside's basis theorem, Burnside's fusion theorem, and Frobenius's p-nilpotence criterion. Each of these is fairly directly related to normal $p$-subgroups of a finite group, and each has self-contained proofs that are only a total of a few pages. All could be found in Gorenstein's textbook for a course in Finite Groups.
Proposition: Suppose $G$ is a finite group of order $p^a q^b$ with $p,q$ distinct primes, $G$ has more than one Sylow $p$-subgroup, and every pair of distinct Sylow $p$-subgroups intersect trivially. Then $p$ divides $q^c-1$ for some $1 \leq c \leq b$.
Proof: Let $G$ be such a group. By Frobenius's normal $q$-complement theorem, if $N_G(Q)/C_G(Q)$ is a $q$-group for each $q$-subgroup $Q$, then $G$ has a normal subgroup $P$ of index $q^b$. However, $|G|=p^a q^b$, so $|P|=p^b$ is a Sylow $p$-subgroup. Since all Sylow $p$-subgroups are conjugate, and this one is normal, such a $G$ would have a unique Sylow $p$-subgroup, contrary to our hypothesis. Hence there is some $q$-subgroup $Q$ such that $p$ divides $[N_G(Q):C_G(Q)]$. Let $x \in N_G(Q)$ have order $p$
in the quotient $N_G(Q)/C_G(Q)$. Then $x$ normalizes $Q$ and its Frattini subgroup $\Phi(Q)$, so $x$ induces an automorphism of the elementary abelian $q$-subgroup $Q/\Phi(Q)$. By Burnside's basis theorem, $x$ continues to have order $p$ as an automorphism of $Q/\Phi(Q)$. Hence $p$ divides $|\operatorname{Aut}(Q)| = q^{c(c-1)/2}(q^c-1)(q^{c-1}-1)\cdots(q-1)$, or equivalently, $p$ divides $q^{c'} - 1$ for some $c'$ satisfying $1 \leq c' \leq c\leq b$. $\square$
Now we handle part (2).
Proposition: If $G$ has order $p^a q$ for $p,q$ distinct primes, and its Sylow $p$-subgroups are abelian, then either the Sylow $p$-subgroup is normal, or the Sylow $q$-subgroup is normal.
Proof: If the Sylow $p$-subgroup $P$ is not normal, then $N_G(P)=P$ since $P$ is a maximal subgroup. Since $P$ is abelian, $P \leq C_G(P) = N_G(P)$. Hence Burnside's normal $p$-complement theorem shows that $G$ has a normal subgroup $Q$ of index $p^a$, that is, of order $q$. Hence the Sylow $q$-subgroup of $G$, $Q$, is normal. $\square$
A: Just proffering an alternative solution to part 2 (not needing the normal $p$-complement theorem that Jack Schmidt described).
You know how to do it, if the Sylow $p$-subgroups intersect trivially. What if they don't? Let $x\neq1$ be an element in the intersection of at least two Sylow $p$-subgroups, say $P_1$ and $P_2$. Consider the centralizer of $x$
$$
C_G(x)=\{a\in G\mid ax=xa\}.
$$
It is known (or easy to see) that $C_G(x)\le G$. Because the abelian subgroups $P_1$ and $P_2$ are both contained in $C_G(x)$, we know that $C_G(x)$ has more than $p^{100}$ elements. Therefore Lagrange's theorem forces us to have $C_G(x)=G$. Therefore  $x$ is in the center of $G$, and thus the center forms a non-trivial normal subgroup of $G$.
A: Given 1, how many elements of order a power of $\;p\;$ we have? At least 
$$\left(p^{100}-1\right)q=p^{100}q-q$$
so the remaining elements must belong to the unique Sylow $\;q$-subgroup, which is then normal
