Using sylow's theorem to show a group is non-simple problem
Prove any group G, such that |G|=48 is not simple.
My solution:
$|G|=48=2^4*3=3*16$ so we can choose the prime number to be 2 or 3.
$n_2 | 3$ and $n_2 = 1 mod(2)$ ==> $n_2=1$ or $n_2=3$
$n_3 | 16$ and $n_3 = 1 mod(3)$ ==> $n_3=1$ or $n_3=4$  or $n_3=16$
I want to show that $n_2=1$ or $n_3=1$
so suppose that $n_2=3$ and $n_3=4$
elements in $syl_2(G)=3*15+1=46$
elements in $syl_3(G)=4*2+1=9$
but $46+9-1>48$  this is contradiction, so either $n_2=1$ or $n_3=1$
can anybody explain that if my solution is right or wrong? the part I am not sure about is
elements in $syl_2(G)=3*15+1=46$,
elements in $syl_3(G)=4*2+1=9$
 A: I might be wrong, but it doesn't seem that a pure Sylow counting argument will crack this one, because we don't know how the $2$-sylow subgroups intersect. However, there is a technique which lets us exploit that intersection, as follows.
Let $H$ be one of the $2$-sylow subgroups. Then $G$ acts on $G/H$, the set of left cosets of $H$, by left multiplication. The kernel of this action consists of the elements $g \in G$ which act as the identity, meaning that $g(aH) = aH$ for every $a \in G$, or equivalently, $g \in aHa^{-1}$ for every $a$, or equivalently, $g \in \bigcap_{a \in G}aHa^{-1}$, which is called core of $H$. Let's call this core $K$ for brevity.
Since $K$ is the kernel of an action, it is a normal subgroup of $G$. If $G$ is simple, then $K$ must be either $1$ or $G$. It can't be $1$, because then the group action induces an injection from $G$ to $S_3$ (since $|G/H| = 3$). This is impossible because $|G| = 48$ whereas $|S_3| = 6$. So $K$ must be all of $G$. But this is also impossible since $K \leq H$.
We conclude that $G$ cannot be simple.

We can in fact say a bit more. Since $G/K$ must be isomorphic to a subgroup of $S_3$, which has order $6$, this means that $|G/K| = |G|/|K| \leq 6$ and so $|K| \geq |G|/6 = 8$. Since $|K|$ also divides $|H| = 16$, this means that $|K|$ is either $8$ or $16$. In the latter case, all of the conjugates of $H$ equal $K$, so $H$ is the unique (hence normal) $2$-sylow subgroup. So, if $n_3 > 1$, we conclude that $G$ has a normal subgroup $(K)$ of order $8$.

A further note about the core $K = \bigcap_{a \in G}aHa^{-1}$: recall that all of the Sylow subgroups of a given order are conjugate, so in fact $K$ is the intersection of all of the $2$-sylow subgroups. Since this is canonically defined, $K$ is in fact a characteristic subgroup of $G$, not merely normal. (We didn't need that fact here, which is why it is a side note.)
