Let G be a group of order 80
a) Show that G has a normal Sylow 5-subgroup or a normal Sylow 2-subgroup b) Show that G is solvable
Approach: $80=5*2^4$
$Syl_5(G) \equiv 1mod \text{ } 5$ and $Syl_5(G) | 16$, so the only possibilities are 1 and 16
$Syl_2(G) \equiv 1 mod \text{ }2$ and $Syl_2(G)|5$, so the only possibilities are 1 and 5
How do we discard one of the possibilities to imply there exists a normal sylow p subgroup in G?
You only need to prove that the case when we have 5 Sylow-2 subgroups and 16 Sylow-5 subgroups is impossible, as otherwise we can use the fact that if a group has a single subgroup of a given order, it's a normal subgroup.
Now assume what I previously mentioned. Then two different Sylow-5 subgroups must have the identity element as the only shared element. Hence the 16 Sylow-5 subgroups give rise to at least $64$ elements of order $5$.
On the other hand take two Sylow 2-subgroups, $P_1$ and $P_2$. Obviously there is an element in $P_1$, not in $P_2$. Therefore there are 16 elements of order divisible by $2$ (The $15$ elemnts in $P_2$ and the one in $P_1$). Also we must have an element of order $1$, the identity. Therefore there are at least $64+16+1=81$ elements in a group of order $80$. A contradiction.