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.
Best Answer
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 $7$.