Sylow $7$-subgroup of a group of order $4\cdot3\cdot5\cdot7$ is normal How to show that the sylow $7$-subgroup of a group of order $420$ is normal. I Know that it is true using GAP. But how to show it without using GAP. I don't know how to start this. Thanks for any help.
 A: So we need to eliminate the possibility of $15$ Sylow $7$-subgroups in the group $G$ with $|G|=420$. Assume that happens. Then, for $P \in {\rm Syl}_7(G)$,we have $|N_G(P)|=28$, and so there exists $t \in N_G(P)$ of order $2$ centralizing $P$.
Now consider the action of $G$ by conjugation on the set of $15$ Sylow $7$-subgroups. Then the point stabilizer $N_G(P)$ has an element of order $7$ which must act as two $7$-cycles.
Now there are two possibilities for the action of $t$. Firstly, it could fix all $15$ Sylow $7$-subgroups. (Thanks for comment from @jpvee for pointing that out.) In that case it is in the kernel of the action. Since the image of the action in $S_{15}$ is transitive and has stabilizer of order divisible by $7$, its order is $105$ or $210$. We shall see below that groups of these orders have normal (and hence unique) Sylow $7$-subgroups, which is a contradiction, because we are assuming that there are $15$.
The second possibility is that $t$  interchange the two $7$-cycles, so $t$ acts as an odd permutation. Now by intersecting with the alternating group, we find that $G$ has a normal subgroup $H$ of index $2$, so $|H|=210$, and it is sufficient to prove that $H$ has a normal Sylow $7$-subgroup.
Since $H$ has twice odd order, by considering its regular permutation representation and intersecting with the alternating group, we find that $H$ also has a normal subgroup $K$ of index $2$, and it is sufficient to solve the same problem for $K$, which has order $105$.
Now if $K$ has $15$ Sylow $7$-subgroups, then we have $90$ elements of order $7$, so there are only $15$ more elements in total. Hence $K$ must have a unique and therefore normal Sylow $5$-subgroup, and since this has no automorphism of order $7$, it is centralized by a Sylow $7$-subgroup $P$, so $|N_K(P)| \ge 35$, contradiction.
A: Here's another way of ruling out the case that $G$ contains $15$ Sylow $7$-subgroups.
Assume the contrary, and let $S$ be one of those $15$ Sylow $7$-subgroups. Then $N:=N_G(S)$ has order$28$, in particular, no element of order 5 normalizes (much less centralizes) $S$. This implies that no Sylow $5$-subgroup of $G$ is normal in $G$, so by Sylow's theorem, the number of Sylow $5$-subgroups must be $21$. Let $F$ by a Sylow $5$-subgroup of $G$. Then similarly, any element of order $3$ centralizes neither $S$ nor $F$, so again by Sylow, $35$ must divide the number of Sylow $3$-subgroups which therefore has to be $70$.
In particular, $G$ contains $15*6=90$ elements of order $7$, $21*4=84$ elements of order $5$ and $70*2=140$ elements of order $3$.
Next we look at the structure of $N$. Since $|Aut(S)|=6$, $C_N(S)$ contains (at least) one element $t$ of order $2$ and therefore $N$ contains at least $6$ elements of order $14$. If one such element $x$ were contained in a conjugate $N^g$ other than $N$, then $S=<x^2>\leq N^g=N_G(S^g)$ and therefore $S=S^g$ contradicting $N^g\neq N$. Therefore $G$ contains at least $15*6=90$ elements of order $14$.
Likewise, by looking at the structure of the normalizer of $F$, we see that $G$ contains at least $21*4=84$ elements of order $6$.
In total, the number of elements of $G$ of orders $1$, $7$, $5$, $3$, $14$ or $6$ is $1+90+84+140+90+84=489>420=|G|$, a clear contradiction.
