# Groups of order 24.

I supposed $$n_3=4$$ and $$n_2=3$$, and then I made $$G$$ act by conjugation on $$\text{Syl}_3 (G)$$.

I want to show that $$G\cong S_4$$ (looking at all order 24 groups here I saw the only one that has $$n_3=|\text{Syl}_3(G)|=4$$ and $$n_2=|\text{Syl}_2(G)|=3$$ is $$S_4$$) so I want to show that the kernel of this conjugation action is trivial. So this action defines a representation $$\varphi:G\rightarrow S_4$$. I was thinking of using the order of $$G/\ker(\varphi)$$ but this didn't solve anything. So I thought about this kernel as the set that normalizes all three Sylow $$3-$$subgroups, the intersection of the normalizers.

If we say $$\text{Syl}_3 (G)=\left\{P_1,P_2,P_3\right\}$$,then I had, by the Orbit-stabilizer theorem, that the order of the stabilizers is $$6$$. I saw that in $$S_4$$ any two of these normalizers intersect in two elements, but when I intersect all of them the intersection is trivial. I want to prove that. So I was doing the following.

By Lagrange, the intersection of all, has order $$1, 2, 3$$ or $$6$$. It can't have order $$3$$, it'd be a normal Sylow $$3-$$subgroup and that's impossible. So it has order $$1,2$$ or $$6$$, and I want to discard $$2,6$$. How could I show that the normalizers are different (This would discard $$6$$)? And how could I show that the intersection of all of them has more than two elements?

I was thinking about saying that, if the intersection had order $$2$$, then there would be a normal subgroup of order $$2$$, and then a subgroup of order $$6$$ isomorphic to $$\mathbb{Z}_2\rtimes_\phi \mathbb{Z}_3\cong \mathbb{Z}_6$$. Could I get somewhere this way?

And how could I do the $$2$$ and $$6$$ parts?

• Notice that by saying $n_3=4$ you already said that the group of order $3$ is not normal. – mesel Nov 18 '14 at 23:45
• I think you should write $n_2=3$ not $n_4=3$ as $4$ is not a prime. – mesel Nov 18 '14 at 23:46
• Yes, my mistake. – David Molano Nov 18 '14 at 23:50
• And I know the group of order 3 isn't normal (The paragraph about the semidirect product had a mistake). – David Molano Nov 18 '14 at 23:51
• It is very hard to find what your question is here. Please clarify. – Pedro Tamaroff Nov 19 '14 at 0:22

It's a bit harder to rule out $|K|=2$ (where $K$ is the kernel). In that case $G/K$ would have order $12$ and would also have $4$ Sylow $3$-subgroups, and so $G/K \cong A_4$. But $A_4$ has a normal Sylow $2$-subgroup, and hence so does $G$, so $n_2=1$ in this case. (Note that the group ${\rm SL}(2,3)$ has $n_3=4$, $n_2=1$. It is often hard to rule out cases for which there is a group with similar parameters.)

• Or would it be in this particular case, because we're supposing kernel has size 2? – David Molano Nov 19 '14 at 1:39
• Thanks, that's it. I already understood why. – David Molano Nov 19 '14 at 3:39

if stabilizer are same then you have an normal subgroup $N$ of order $6$. Let $H$ be subgroup of $N$ of order $3$. Notice that $H$ is uniqe in $N$ so $H$ must be normal in $G$ contradiction.

For other case, you have an normal subgroup of order $2$. Notice that any normal subgroup of order $2$ must be contained $Z(G)$ as $Z(G)$ is trivial contradiction.

• Thanks for the part of order 6 (I already solved it in a similar way). – David Molano Nov 19 '14 at 0:42
• For the part of order 2, why would it be contained the center? And why would the center be trivial? How can I say that the center is trivial in some, or all non-abelian groups of order 24? I know it is in $S_4$, but I don't know anything else in these kinds of groups. – David Molano Nov 19 '14 at 0:45

This is how I see it. $$G$$ has 3 2-Sylow subgroups, and 4 3-Sylow subgroups. So $$G$$ acts on the 3-Sylow subgroups by conjugation, resulting in a map $$\phi:G → S_4$$. Let $$K$$ be the kernel of this map. Let $$P$$ be a 3-Sylow subgroup, and restrict $$\phi$$ to P; call the map $$\phi_P$$. Then the Sylow Permutation Theorem* states that the only fixed point of $$\phi_P$$ on the 4 conjugates is $$P$$ itself. This means that $$\phi_P$$ acts nontrivially on the other three 3-Sylow subgroups, implying that $$K$$ cannot contain an element of order 3, so that the order of $$K$$ is not divisible by 3, so it has to be 1, 2, 4, or 8.

Let $$\psi: G → S_3$$ be the action of $$G$$ on the 2-Sylow subgroups. Then $$G$$, order 24, acts on $$S_3$$ of order 6, implying a normal subgroup $$V$$ in $$G$$ of order 4. This subgroup then combines with a 3-Sylow subgroup, by semi-direct product, to form a subgroup $$A$$ of $$G$$ of order 12. There can be only one such subgroup of order 12, as there is no room for any more. So $$A$$ is of order 12 without normal 3-Sylow subgroups, so it is isomorphic to $$A_4$$ and $$V$$ is isomorphic to the Klein four-group. The 2-Sylow subgroups are of order 8, and each one of them must contain $$V$$ as a subgroup. Hence we have 1 identity element, 3 elements in $$V$$, 4 x 3 or 12 elements in the 2-Sylows outside of $$V$$, and 8 elements of order 3, accounting for all 24 elements of G. There is no more room for any more.

$$K$$ cannot be of order 8, because the subgroups of order 8 are not normal. $$K$$ cannot be of order 2, since in that case $$K$$ would be normal in $$G$$, so it combines with $$P$$ to form a subgroup of order 6 with subgroups of order 2 which are normal, hence is $$Z_6$$, containing elements of order 6, for which there is no room. Finally, $$K$$ can't be of order 4, since the normalizer of $$P$$ and its conjugates is of order 6, so the intersection of all these normalizers has to have order 1, 2, 3, or 6 and be contained in $$K$$, so it can't have order 4.

Therefore, $$K$$ is the trivial group, so that $$\phi$$ is an isomorphism between $$G$$ and $$S_4$$.

*The Sylow Permutation Theorem says that if a group $$G$$ has $$n$$ $$p$$-Sylow subgroups, then $$G$$ acts on $$Syl_p(G)$$ by conjugation, and the only fixed point of this action restricted to $$P$$ is $$P$$ itself. This occurs throughout the literature, mainly in proving that there are no simple groups of some specified order, but not as a theorem by itself.