Show that if $G$ is a group of order $168$ that has a normal subgroup of order $4$ , then $G$ has a normal subgroup of order $28$ Show that if $G$ is a group of order $168$ that has a normal subgroup of order $4$ , then $G$ has a normal subgroup of order $28$.
Attempt: $|G|=168=2^3.3.7$
Then number of sylow $7$ subgroups in $G = n_7 = 1$ or $8$.
Given that $H$ is a normal subgroup of order $4$ in $G$. 

If we prove that $n_7$ cannot be $8$, then $n_7=1$ and as a result, the sylow $7$ subgroup $K$ is normal.Hence, $HK$ will also be a normal subgroup of $G$ and since, $H \bigcap K = \{e\} \implies |HK|=28$.

Now, suppose $n_7=8$ and hence, $K_1 \cdots K_8$ are the $8$ cyclic subgroups of order $7$.
Each $K_i$ has $\Phi(7)=6$ elements of order $7$ . Hence, total elements of orders $=7$ in the $K_i's$ are $6.8=48$
How do I move forward and bring a contradiction somewhere?
Thank you for your help. 
 A: If $K$ is a normal subgroup of $G$ of order $4$, you may well argue in $H = G/K$, and reduce to show, via the correspondence theorem, that in a group of order $2 \cdot 3 \cdot 7 = 42$ there must be a normal subgroup of order $7$.
To do this, just consider that the number of $7$-Sylow subgroups in $H$ must divide $42/7 = 6$, and be congruent to $1$ modulo $7$.
A: You can show it by using the quotient group as Andrea does, but you can also directly show that $n_7=1$.
A 7-group acts on the normal 4-group by conjugation. This action is an
automorphism. The automorphism group of 4-groups have order 2 or 6 and hence a
7-seven group must act trivially. This means that the 7-group commutes with the 4-group.
You could show the same by looking at the 28 group $P_4P_7$. By Sylow, there is 
one normal 7-group. The 4-group was assumed normal in G. Hence the 28-group is the direct product and the 4-group and 7-group commute.
The 4-group and $P_7$ are then both contained in $N_G(P_7)$ which must have index at most $\frac{168}{28}=6$. There can then be at most be 6 conjugates of the 7-group which rules out $n_7=8$.
Also note that the result of a normal 28-group implies one normal 7-group.
There is one 7-group in a 28-group by Sylow and it is hence characteristic.
A characteristic 7-group in a normal 28-group is normal in G.
A: If K  is a normal subgroup of G  of order 4 , you may well argue in $H=\frac{G}{K}$ , and reduce to show, via the correspondence theorem, that in a group of order $2\cdot 3\cdot 7=42$  there must be a normal subgroup of order 7 .
To do this, just consider that the number of 7 -Sylow subgroups in H  must divide $\frac{42}{7}=6$ , and be congruent to 1  modulo 7
A: Hints:
(a) If $\;H,K\;$ are subgroups of a group $\;G\;$ , then $\;HK:=\{hk\;:\; h\in H\,,\,k\in K\}\;$ is a subgroup of $\;G\;$ iff $\;HK=KH\;$
(b) If $\;H\;$ is a normal subgroup of $\;G\;$, then for any subgroup $\;K\;$ of $\;G\;$ we have that $\;HK=KH\;$
(c) In your question, you in fact only need to take any Sylow $\;7$-subgroup $\;K\;$ of $\;G\;$ to get what you want, since
$$|HK|=\frac{|H||K|}{|H\cap K|}$$
