Let $G$ be a group of order $24$ having no normal subgroups of order $3$. Show that $G$ has four subgroups of order $6$.

Question

Question

Let $G$ be a group of order $24$ having no normal subgroups of order $3$. Show that $G$ has four subgroups of order $6$.

Attempt

It is $24=3\cdot 2^3$ and $n_3\mid 2^3,\ n_3\equiv 1\pmod{3}\Rightarrow n_3=1,4$, $n_3\not=1$ from the hypothesis. Hence $n_3=4$. It is $n_2\mid 3$ and $n_2\equiv 1\pmod{2}\Rightarrow n_2=1,3$. If $n_2=1$ then there is a unique normal Sylow $2$-subgroup $P_2$ [and by taking the product with the $4$ Sylow $3$-subgroups we have the four subgroups of order $6$ needed.] {that's wrong}

Problem : If $n_2=3$ how could I proceed? EDIT: if $n_2=1 ?$

The problem does not require proving that G has exactly 4 subgroups of order 6. So when Sylow 2 is not normal we're guaranteed minimum 4 subgroups of order 6. — Lawrence Mano, Oct 01 '20 at 22:10
@Lawrence Mano I can't see why we have $4$ different subgroups of order $6$. Could help, please? — 1123581321, Oct 02 '20 at 04:39
The order of $P_2$ is 8, not 2. So the product of $P_2$ with any of the four Sylow 3-subgroups gives you all of $G$. — , Oct 02 '20 at 05:12
What people are saying is that your argument for the case $n_2=1$ is not correct. — Derek Holt, Oct 02 '20 at 07:36

score 3 · Accepted Answer · answered Oct 02 '20 at 07:55

We know that $n_3=4$. Consider the conjugation action of $G$ on the four Sylow 3-subgroups of $G$, and let $I$ be the image of this action. So $I$ is a subgroup of $S_4$.

Since the Sylow subgroups are all conjugate in $G$, the action is transitive. So $|I|$ is divisible by 4. Also, since no Sylow subgroup can normalize another, the action of a Sylow 3-subgroup $P$ is a fixed point (i.e. $P$ itself), together with a 3-cycle. So $|I|$ is divisible by 3.

So $I$ is a subgroup of $S_4$ of order divisible by 12, and it must be $A_4$ or $S_4$.

Case 1. $I=A_4$. So the kernel $K$ of the action has order 2. Since $A_4$ has no subgroups of order $6$, the subgroups of order 6 in $G$ must contain $K$, and so they are the inverse images of the four subgroups of order $3$ in $I$. So there are four such subgroups in total, which are cyclic. (This is the case $n_2=1$.)

Case 2. $I=S_4$, so $I \cong G$, and as HallaSurvivor pointed out, there are also exactly four subgroups of order 6 (isomorphic to $S_3$)in this case. (This is the case $n_2=3$.)

score 2 · Answer 2 · answered Oct 02 '20 at 05:02

It's a bit hacky, and relies on some knowledge I happen to have, but here is one solution:

It is known that if neither sylow subgroup is normal in a group of order $24$, then that group is $\mathfrak{S}_4$ (the symmetric group on $4$ letters). You can see a proof of this fact here.

Once you know this fact, then you can easily see that there are 4 copies of $\mathfrak{S}_3$ living in $\mathfrak{S}_4$ (fix one of the $4$ letters), which proves the claim.

Unfortunately, I don't know of an "elegant" way to know this fact about $\mathfrak{S}_4$. It's a theorem that I've seen in the past, and happened to remember. I'm leaving this as an answer, but I would also love to see a better motivated solution.

I hope this helps ^_^

Thanks for your time! In fact that's a very strong argument which solves immediately the problem but in fact I suppose there should be something simpler. I took the problem from http://math.uprrp.edu/asunacad/grad_exams/algebra/alg072.pdf which has relatively easy problems. — 1123581321, Oct 02 '20 at 05:33

Let $G$ be a group of order $24$ having no normal subgroups of order $3$. Show that $G$ has four subgroups of order $6$.

2 Answers2