# 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?

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 $7$.

• Thank you. However, I could not understand well, why we specifically considered a group of order $2.3.7=42$ .. Could you please explain? – MathMan Aug 26 '14 at 18:15
• @VHP, I am making use of this proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory) – Andreas Caranti Aug 26 '14 at 18:20
• Ohkay .. So, I get that in the group $G/K$, there is a unique normal subgroup ( sylow $7$ subgroup ) of order $7$. Now, does this mean that in $G$ also, there exists a normal subgroup of order $7$? – MathMan Aug 26 '14 at 18:47
• @VHP, no, you obtain that in $G$ there is a normal subgroup of order $4 \cdot 7 = 28$, as you required. This is because every normal subgroup of $G/K$ is of the form $N/K$, where $N$ is a normal subgroup of $G$ containing $K$. Now you have found a normal subgroup $N/K$ of $H$ of order $7$, so by Legrange $N$ is a normal subgroup of $G$ of order $\lvert N/K \rvert \cdot \lvert K \rvert = 7 \cdot 4 = 28$. – Andreas Caranti Aug 26 '14 at 18:51
• Thank you for the answer :-) – MathMan Aug 26 '14 at 19:00

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.

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

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

$$|HK|=\frac{|H||K|}{|H\cap K|}$$

• No @VHP, you do not need that. You only need to show there's at least one such subgroup, which is given to you by Sylow theorems (or Cauchy's theorem) – Timbuc Aug 26 '14 at 17:53
• Oh okay .. How shall we show that the group of order $28~~i.e.~~ HK$ is normal? – MathMan Aug 26 '14 at 17:57
• For normality you may want to use Andreas Answer + the Correspondence Theorem for groups, @VHP – Timbuc Aug 26 '14 at 18:00
• Thank you for the answer :-) – MathMan Aug 26 '14 at 19:04