Groups of order 24.

Question

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?

Answer 1

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.

Answer 2

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.)

Answer 3

Offering a slightly different proof, similar to jimvb13's answer. All references are from Dummit & Foote, Abstract Algebra, 3e.

Suppose $n_2=3, n_3=4$, and fix any of Sylow-2 groups called $P_2$, and any of Sylow-3 groups $P_3$. Now, $|G|/|P_2| =3$, and the permutation of $P_2$ by left action gives a homomorphism $\varphi: G \mapsto S_3$. Denote $\tilde{V} :=\mathrm {ker}\; \varphi \lhd G$. Here, $\tilde{V} =4$ or $8$; indeed, for $\tilde{V} =12$ or $24$, the action would be not transitive. But 8 is not possible either, in which case, $\tilde{V}$ as a Sylow-2 would be normal. Thus $\tilde{V} =4$.

Now $\tilde{A} :=P_3 \tilde{V}_4$ is well defined (D&F, p.120, col.5), with $\tilde{A} =12$. Since $|G|/|\tilde{A}| =2$, the smallest prime factor of 24, we see $\tilde{A} \lhd G$ (D&F, p.120, col.5).

It is possible to find a order-2 element $a \in G -\tilde{A}$. In fact, even if all nonidentity elements in conjugates of $P_3$ are counted, we have only $3 \cdot 2 =6$ elements. The rest $2$ elements has order to be multiple of 2. So suppose $|\langle a \rangle| =2$, we see $G =\tilde{A} \rtimes \langle a \rangle$ (D&F, p.180, thm.12). For brevity, call it $G =\tilde{A} \rtimes Z_2$.

We know that there are two nonisomorphic order-12 groups, $A_4$ and $S_3 \times Z_2$ (D&F, p.183). Consider $(S_3 \times Z_2) \rtimes Z_2$, and I claim there is a contradiction. Actually, for $P_3'$ a Sylow-3 subgroup of $S_3 \times Z_2$, we have $P_3'\; \mathrm{char}\; (S_3 \times Z_2) \lhd G$ (D&F, p.144), thus (recognized as subgroup of $G$) $P_3 \lhd G$ (see D&F, p.143), contrary to assumption.

Thus $\tilde{A} =A_4$ the alternating group, now $G =A_4 \rtimes Z_2$. Notice the Klein group $V\; \mathrm{char}\; A_4 \lhd G$ and $V \lhd G$. And if the product is direct, the only possibility of $P_2$ is recognized to be $P_2 =V \times Z_2$, but that clearly $P_2 \lhd G$. Thus $G =A_4 \rtimes Z_2$ nontrivially, also we recognize $\tilde{V} =V$.

Consider $A_4 \leq S_4$, and pick any order-2 element in S_4 to form $A_4 \rtimes Z_2$. We know that a conjugation within $S_4$ is just renumbering, and such conjugation exhaust all automorphisms $\mathrm{Aut} (A_4)$. Considering the order, $A_4 \rtimes Z_2$ must be exactly $S_4$.

Answer 4

Edit 1: The verbose answer below has a gap/mistake.

We reason (correctly?) that if a group G of order 24 has both a normal subgroup of order 4, and also 4 subgroups of order 3, then G must be a semidirect product of Z_2 and (normal) A_4.

However, the muddled reasoning about A_4, trying to understand its involutions, may have tacitly assumed that all involutions of A_4 are inner.

The latter assumption is false, as shown (working now in S_4) by conjugation by a transposition.

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This answer is intended as elementary, as a useful supplement to the others, and also to more or less stand on its own.

In the bargain we will learn a bit about some other groups of order $24$.

The novelty, if there is one, is that we don't have to check that the kernel of the conjugacy map $G_{24}$-->$S_4$ is trivial, it suffices to show it has order 0 or 4.

We borrow some helpful ideas/observations/constuctions from the other answers, especially jimvb13 and Violapterin

For a natural number $k$ the notation $G_k$ means a group of order $k$.

We assume the Sylow theorems and the basics of semidirect products.

We will combine a few ingredients, more or less independent from each other, adequate to certify, up to isomorphism, the existence of at at most one group $G_{24}$ with four $G_3$ subgroups and three $G_8$ subgroups.

First some basics about an arbitrary group $G_{24}$ with 24 elements.

1. If there exists a $G_8$ normal in $G_{24}$ then by 2nd Sylow, $G_{24}$ has only one subgroup of order $8$.

2. If there exists a $G_4$ normal in $G_{24}$ then $G_{24}$ is a semi direct product of $G_{2}$ and a normal $G_{12}$ argued as follows.

The quotient $G_{24}$/$G_{4}$ has order $6$, and thus the quotient has a unique normal subgroup $G_{3}$ of order $3$. Thus, $G_{12}, $the pullback of $G_{3}$ in $G_{24}$, has order $12$, and is normal in $G_{24}$, since normality generally pulls back, or merely use the fact it has index $2$.

This shows $G_{24}$ is an internal semidirect product of subgroups $G_{2}$ and a normal $G_{12}$.

Thus, with the extra assumption that $G_{24}$ has four subgroups of order $3$ then they must also be subgroups of the normal $G_{12}$ in $G_{24}$, (since if $g$ has nontrivial coordinate in the $G_{2}$ semidirect product factor, then $g$ must have even order in $G_{24}$. There is likely an easier reason.)

This determines $G_{12}$ is isomorphic to the uniqe noncommutatove semidirect product of $G_{3}$ and a normal $G_{2}$ x $G_{2}$ subgroup of $G_{12}$. Thus $G_{12}$ is isomorphic to $A_{4}$

3. Summary so far beyond 1. If $G_{24}$ has a normal subgroup $G_{4}$ and also four subgroups isomorphic to $G_{3}$ then $G_{24}$ is a semidirect product of $G_{2}$ and $A_{4}$, the twelve orientation preserving isometries of a regular tetrahedron.

4. If the arbitrary group $G_{24}$ has a normal subgroup $G_{2}$ and also four subgroups isomorphic to $G_{3}$, then $G_{24}$ has a normal $G_{8}$ subgroup, argued as follows.

The quotient has order $12$ and must contain 4 subgroups isomorphic to $G_{3}$.

(By Sylow 3, the $G_{12}$ quotient has 1 or 4 subgroups isomorphic to $G_{3}$. The quotient map is 2-1 and thus cannot map a $G_3$ subgroup of $G_{24}$ to $0$ in $G_{12}$. The 4 $G_3$ subgroups of $G_{24}$ are all conjugate in $G_{24}$, and thus must map to conjugate $G_{3}$ subgroups of $G_{12}$. Thus there must be 4 such groups in $G_{12}$, since if there were only 1, the quotient map would be at least 4-1.)

Thus the quotient group has a normal subgroup of order 4, and whose pullback is normal subgroup of order 8 in $G_{24}$. (The quotient group is $A_{4}$ but this knowledge is not strictly necessary to establish 4.)

5. If the arbitrary group $G_{24}$ has 4 subgroups isomorphic to $G_{3}$, then the canonical conjugacy homomorphism, acting on the set of the 4 Sylow 3 subgroups, $G_{24}$-->$S_4$, has kernel of order 1,2,4, or 8, argued as follows.

(Edit: The key reason also follows from Sylow 2 and a standard proof of part of Sylow 3, If a Sylow-p subgroup P acts, via conjugation, on the set of all Sylow p subgroups, there is precisely one universal fixed point, P itself. Here's why: If Q is also a universal fixed point then P is in N(Q), and hence conjugate to Q in N(Q) and thus P=Q. Hence if g has order 3 in G_24, then g is NOT in the kernel.)

If $b$ in $G_{24}$ has order 1 or 3, then the corresponding conjugacy automorphism $f:G_{24}$--> $G_{24}$ has order 1 or 3, and permutes the 4 $G_{3}$ subgroups of $G_{24}$ of order 3.

This means either $f$ leaves invariant precisely one $G_{3}$ subgroup of $G_{24}$(and induces a 3 cycle on the other 3), or $f$ leaves invariant each $G_{3}$ subgroup of $G_{24}$.

In the former case $f$ is not in the kernel of our map.

In the 2nd case $f$ must fix each $G_{3}$ subgroup of $G_{24}$ pointwise, since $G_{3}$ has two automorphisms, and $f$ has odd order.

Thus, in the case at hand, $b$ commutes with each element of each of the four conjugate $G_{3}$ subgroups of $G_{24}$.

It follows that $b$ must be identity of $G_{24}$ since otherwise the union of the $G_{3}$ subgroups would form an (abelian) subgroup of order 9, in the group $G_{12}$, this is impossible since 9 does not divide 24.

6. If the arbitrary group $G_{24}$ has a subgroup isomorphic to $A_4$, then $G_{24}$ is either isomorphic to $S_4$, or to $G_2$ x $A_4$, the latter has a normal subgroup of order 8, argued as follows.

Since $A_4$ is a semidirect product of $G_3$ with a normal $Z_2$ x $Z_2$ subgroup, each element of order 2 in $A_4$ is an element of $Z_2$ x $Z_2$.

Any two elements of order 2 in $A_4$ are conjugate in $A_4$, thus, up to isomorphism there is a unique semidirect product determined by a nontrivial homomorphism $G_{2}$-->$A_4$.

The trivial homomorphism creates the product $Z_2$ x $A_4$, the subgroup $Z_2$ x ($Z_2$ x $Z_2$) is normal.

7. Putting it all together, if $G_{24}$ has 4 subgroups of order 3 and 3 subgroups of order 8, then $G_{24}$ is isomorphic to $S_{4}$

Suppose $G_{24}$ has 4 $G_{3}$ subgroups. The natural conjugacy homomorphism $G_{24}$--->$S_4$ has (normal) kernel of size 1,2,4, or 8.

If its has size eight or two then, by 1 and 4, $G_{24}$ has 1 subgroup of order 8.

If the kernel has size 4, then by 3, 5 and 6, then if $G_{24}$ is not isomorphic to ${S_4}$, then $G_{24}$ has a normal subgroup of order 8.

If the kernel has size 1, then the mentioned map is an isomorphism.