3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $7$.

$\endgroup$
  • $\begingroup$ Thank you. However, I could not understand well, why we specifically considered a group of order $2.3.7=42$ .. Could you please explain? $\endgroup$ – MathMan Aug 26 '14 at 18:15
  • 1
    $\begingroup$ @VHP, I am making use of this proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory) $\endgroup$ – Andreas Caranti Aug 26 '14 at 18:20
  • $\begingroup$ 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 $7$? $\endgroup$ – MathMan Aug 26 '14 at 18:47
  • 1
    $\begingroup$ @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 $\lvert N/K \rvert \cdot \lvert K \rvert = 7 \cdot 4 = 28$. $\endgroup$ – Andreas Caranti Aug 26 '14 at 18:51
  • $\begingroup$ Thank you for the answer :-) $\endgroup$ – MathMan Aug 26 '14 at 19:00
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ 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) $\endgroup$ – Timbuc Aug 26 '14 at 17:53
  • 1
    $\begingroup$ Oh okay .. How shall we show that the group of order $28~~i.e.~~ HK$ is normal? $\endgroup$ – MathMan Aug 26 '14 at 17:57
  • 1
    $\begingroup$ For normality you may want to use Andreas Answer + the Correspondence Theorem for groups, @VHP $\endgroup$ – Timbuc Aug 26 '14 at 18:00
  • 1
    $\begingroup$ Thank you for the answer :-) $\endgroup$ – MathMan Aug 26 '14 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.