Groups of order $2520$

Question

Suppose that $G$ is a group of order $2520 = 2^3 \cdot 3^2 \cdot 5 \cdot 7$. The property that I want to check is this:

Must $G$ contain an abelian subgroup of order at least $12$?

If $G$ is soluble then, yes. A Hall $\{5,7\}$-subgroup of $G$ (which exists by Hall's theorem) is necessarily abelian. On the other hand, if $G$ is simple, then $G \cong A_7$, since $A_7$ is the only simple group of that order (in fact, it is the only perfect group of that order). Also, $A_7$ has an abelian subgroup of order $12$.

Some further observations:

• If $G$ is a candidate counterexample to the (implicit) assertion, then for $p \in \{3,7\}$ a Sylow $p$-subgroup of $G$ must be self-centralising. Using the $N/C$ theorem and the standard $n_p \equiv 1\,(\operatorname{mod} p)$, we arrive at the only possibilities $n_7 = 2^3 \cdot 3 \cdot 5$ and $n_3 = 2 \cdot 5 \cdot 7$.

• For $p=5$ we cannot argue that a Sylow $5$-subgroup of $G$ must be self-centralising, because the possibility that its centraliser has order $2 \cdot 5$ cannot be excluded. At least not immediately. In any case though, $|C_G(P):P| \in \{1,2\}$ and $n_5 = 2 \cdot 3^2 \cdot 7$. Here $P$ is a Sylow $5$-subgroup of $G$.

• Since $G$ cannot be soluble, it must have a composition factor isomorphic to one of $\{A_5, \operatorname{PSL}_3(2), \operatorname{PSL}_2(8), A_6\}$. (We have already argued the case $G \cong A_7$.) That composition factor, however, cannot be direct.

Thoughts?

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MatheinBoulomenos notices that I had missed one possibility for a non-abelian composition factor of $G$, namely $\operatorname{PSL}_2(8)$. I have now included this in the list.

Answer 1

Here is a simpler approach. In all cases, we have a unique nonabelian composition factor $T$. Let $R$ be the soluble radical of $G$. Then $G/R$ has trivial soluble radical, and a unique nonabelian composition factor $T$, so it must be almost simple with socle $T$.

If $T=PSL(2,7)$, then $G/R=PSL(2,7)$. (It cannot be $PGL(2,7)$ by order considerations.) That means that $|R|=15$ and so $R$ is cyclic.

If $T=A_5$, then $R$ has order $21$ or $42$ (depending whether $G/R$ is $A_5$ or $S_5$). In any case, $R$ has a normal Sylow $7$-subgroup $P$. $P$ is characteristic in $R$, so normal in $G$. Now, a Sylow $5$-subgroup of $G$ together with $P$ generate a (cyclic) group of order $35$.

If $T=A_6$, then $R=C_7$ and, as in the last case, we get a (cyclic) group of order $35$.

Finally, if $T=PSL(2,8)$, then $R=C_5$ and again we get a (cyclic) group of order $35$.

(Note: In the above, I'm assuming knowledge of the size of outer automorphism groups of the relevant simple groups, but I don't need to know about Schur multipliers, or Schur-Zassenhaus.)

Answer 2

I will make a case distinction on different composition series.

If $G$ has $A_6$ as a composition factor, then the sequence $1 \to C_7 \to G \to A_6\to 1$ or $1 \to A_6 \to G \to C_7 \to 1$ splits by Schur-Zassenhaus. As $A_6$ is simple, there is no non-trivial homomorphism $A_6 \to \operatorname{Aut}(C_7)$, also as $\operatorname{Aut}(A_6)$ has order $1440$, there is no-nontrivial homomorphism $C_7 \to \operatorname{Aut}(A_6)$ so we get $G \cong C_7 \times A_6$, this obviously has an abelian subgroup of order $42$

If $G$ has $\mathbf{PSL}_3(2)$ as a composition factor, then one case is that we have a surjective homomorphism $G \to \mathbf{PSL}_3(2)$, then the kernel has order $15$, and hence is cyclic.

One other case is that we have an exact sequence $1 \to H \to G \to C_3 \to 1$, where $H$ is of order $840$ and has $C_5$ and $\mathbf{PSL}_3(2)$ as composition factors. Schur Zassenhaus gives us that $H$ is a semidirect product of $C_5$ and $\mathbf{PSL}_3(2)=\mathbf{PSL}_2(7)$. $\operatorname{Aut}(\mathbf{PSL}_3(2))=\operatorname{Aut}(\mathbf{PSL}_2(7))=\mathbf{PGL}_2(7)$ has order $336$ and thus no element of order $5$, so we have no non-trivial homomorphism $C_5 \to \operatorname{Aut}(\mathbf{PSL}_2(7))$. As $\mathbf{PSL}_2(7)$ is simple, there is also no nontrivial homomorphism $\mathbf{PSL}_2(7) \to \operatorname{Aut}(C_5)$. So we get that this semidirect product is split in either case, so $H \cong C_5 \times \mathbf{PSL}_2(7)$ and from there obviosuly $H$ and thus $G$ have an abelian subgroup of order $35$.

The remaining case is that we have an exact sequence $1 \to H \to G \to C_5 \to 1$, where $H$ has composition factors $\mathbf{PSL}_2(7)$ and $C_3$.
One case is that we have an exact sequence $1 \to C_3 \to H \to \mathbf{PSL}_2(7) \to 1$. Because $\mathbf{PSL}_2(7)$ is simple, the associated homomorphism $\mathbf{PSL}_2(7) \to \operatorname{Aut}(C_3)$ is trivial, thus $C_3 \subset H$ is central, so we can take an element $x$ of order $7$ in $H$ and consider the subgroup generated by that element and the central $C_3$ to get an abelian subgroup of order $35$ in $H$ and hence in $G$.
The other case is that we have an exact sequence $1 \to \mathbf{PSL}_2(7) \to H \to C_3 \to 1$. In this case, as $\mathbf{PSL}_2(7)$ is normal in $H$, we get $H \leq N_G(\mathbf{PSL}_2(7)$. If we consider the conjugation homomorphism $H \to \operatorname{Aut}(\mathbf{PSL}_2(7))$, the kernel must have an element of order $3$, because $9 \mid |H|$, but $9 \not \mid |\operatorname{Aut}(\mathbf{PSL}_2(7))|=|\mathbf{PGL}_2(7)|=336$, thus there is an element of order $3$ which centralizes the copy of $\mathbf{PSL}_2(7)$. Taking the subgroup generated by this element and an element of $\mathbf{PSL}_2(7)$ gives an abelian subgroup of order $21$.

There is also $\mathbf{PSL}_2(8)$ of order $504$. We can use Schur-Zassenhaus again to get a semidirect product of $\mathbf{PSL}_2(8)$ and $C_5$. Since $\operatorname{Aut}(\mathbf{PSL}_2(8))=\mathbf{P\Gamma L}_2(8)$ has order $1512$, we get a direct product, which finishes this case.

It remains to treat the case that $G$ has $A_5$ as a composition factor:

One possibility is that we have a surjective homomorphism $p:G \to A_5$. Consider a copy of $A_4$ inside of $A_5$, then the preimage $H := p^{-1}(A_4) \leq G$ is a subgroup of order $2520/12=210=2 \cdot 3 \cdot 5 \cdot 7$, so as $H$ has squarefree order, it is solvable, thus by Hall's theorem, there is a Hall $\{5,7\}$-subgroup of $H$ which must be abelian.

Another possibility is that we have an exact sequence $1 \to H \to G \to C_7 \to 1$, where $H$ has order $360$. By Schur-Zassenhaus, this sequence splits, so $G$ is isomorphic to a semidirect product $H \rtimes C_7$. As $360 \equiv 3 \pmod{7}$, any action of $C_7$ on a set of $360$ elements must have at least $3$ fix points. Choose $x \in H$, not the identity element that is fixed by the action of $H$, then by the construction of the semidirect product, elements in $C_7$ commute with $H$, so the subgroup in $H \rtimes C_7$ generated by $x$ and $C_7$ is abelian (and it has at least order $14$.)

We also have the possibility that we have an exact sequence $1 \to H \to G \to C_3$, where $H$ has order $840$ and composition factors $A_5$, $C_7$ and $C_2$.
One subpossibility is that we have a surjective homomorphism $p:H \to A_5$, then we can take (again) the preimage $p^{-1}(A_4)$ which will be a subgroup of order $840/12=70$. Every group of order $70$ has a subgroup of order $35$, which must be abelian.
Another subpossibility is that we have an exact sequence $1 \to K \to H \to C_7$, where $K$ has order $120$ and has composite factors $A_5$ and $C_2$.
This sequence splits by Schur-Zassenhaus, so $H \cong K \rtimes C_7$.
It's possible to show that the only groups with these composition factors are $A_5 \times C_2$ and $S_5$, since this argument is already long enough, I'll skip that part. (See for example here.) If $K= A_5 \times C_2$, then we can take a cyclic subgroup of $A_5$ that generates together with $C_2$ and abelian subgroup of order $12$. If $K=S_5$, then $\operatorname{Aut}(K)=S_5$, which doesn't have elements of order $7$, so the homomorphism $C_7 \to \operatorname{Aut}(K)$ is trivial and we get $H \cong S_5 \times C_7$, which obviously has an abelian subgroup of order at least $12$.
The last subpossibility on this level of case distinctions is that we have an exact sequence $1 \to K \to H \to C_2 \to 1$, such that $K$ has order $420$ and has composition factors $A_5$ and $C_7$.
There are no non-trivial homomorphism $A_5 \to \operatorname{Aut}(C_7)$ or $C_7 \to \operatorname{Aut}(A_5)=S_5$, because $A_5$ is simple and $S_5$ doesn't have an element of order $7$, so yet another Schur-Zassenhaus argument shows that $K \cong A_5 \times C_7$, which has an abelian subgroup of order at least $12$.

There's the case left where we have an exact sequence $1 \to H \to G \to C_2 \to 1$, where $H$ has order $1260$ and has composition factors $A_5$, $C_3$ and $C_7$.

In the first subcase, we have an exact sequence $1 \to K \to H \to C_7 \to 1$, where $K$ has order $180$. Schur-Zassenhaus gives us that $H \cong K \rtimes C_7$. Because $180 \equiv 5 \pmod{7}$, every action of $C_7$ on a set of $180$ elements has at least $5$ fix points, so we may choose a non-identity element in $H$ which is fixed by the action of $C_7$, this element together with $C_7$ generates an abelian subgroup of order at least $14$ in the semidirect product.

Another subcase is that we have a surjective homomorphism $p:H \to A_5$, then taking the preimage $p^{-1}(A_4)$ gives a subgroup of $H$ of order $1260/12=105=3\cdot5\cdot 7$. As $105$ is squarefree, groups of order $105$ are solvable, so they contain a $\{5,7\}$-Hall subgroup by Hall's theorem which must be abelian and of order $35$.

The last subcase is that we have an exact sequence $1 \to K \to H \to C_3 \to 1$ where $K$ has order $420$ and composite factors $A_5$ and $C_7$. We have already shown in another part of the argument that this implies that $K \cong A_5 \times C_7$ which has a large enough abelian subgroup.

Done!
I hope I made no errors in this long argument, but it's only a handful tricks applied multiple times.

Answer 3

If you take a closer look at the $3$-subgroups, you do not have to rely on the classification of finite simple groups of order dividing $2520$:

Let's assume that $G$ is a group of order $2520$ without any abelian subgroups of order $\ge 12$.

We'll frequently use the fact that

(*) elements of order $5$ or $7$ cannot normalize any non-trivial $3$-subgroup,

as they would centralize it yielding an abelian subgroup of order $\ge 12$.

In particular, $G$ has at least two different Sylow $3$-subgroups.

Claim: The intersection $S\cap T$ of any two (different) Sylow $3$-subgroups $S$ and $T$ of $G$ is trivial.

Otherwise $S\cap T$ has order $3$ and its centralizer $C = C_G(S\cap T)$ has at least two different Sylow $3$-subgroups ($S$ and $T$, as they are abelian). As $C$ cannot have elements of order $5$ or $7$, its order $|C|$ divides $2^3\cdot 3^2$ and $C$ has four Sylow $3$-subgroups. So it has an abelian subgroup of order $4$ centralizing $S\cap T$ leading to a contradiction.

Claim: The number $n_3$ of Sylow $3$-subgroups of $G$ is $280$.

By (*) the number $n_3$ is a multiple of $35$, so it is either $70$ or $280$ by Sylow.

In case $n_3 = 70$ take a look at the action by conjugation of a fixed Sylow $3$-subgroup $S$ on the set of all Sylow $3$-subgroups. If an element $1\ne s\in S$ fixes a Sylow $3$-subgroup $T$, i.e., $T^s = T$ then it is contained in $T$, $s\in T$, as otherwise $s$ and $T$ generate a $3$ subgroup $\langle s, T\rangle$ properly containing the Sylow $3$-subgroup, which is absurd. As therefore every non-trivial element of $S$ fixes only $S$, all other orbits have length $9$ contradicting $63 \ne 1 \bmod 9$.

Final contradiction: $n_3$ cannot be $280$ either.

As you already noted, one can obtain a contradiction using the $N/C$-theorem.

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An alternative argument (with three different endings) for the final contradiction can be given using Frobenius groups:

As $n_3 = 280$ implies that the Sylow $3$-subgroups are self-normalizing (i.e., equal to their normalizers) and intersect trivially by the first claim, $G$ is a Frobenius group with Frobenius complement any Sylow $3$-subgroup. By Frobenius $G$ has a normal subgroup $K$ (the Frobenius kernel) of order $2^3\cdot 5\cdot 7$.

Ending 1: By John Thompson's famous thesis $K$ is nilpotent, which implies that $K$ has an abelian subgroup of order $140$.

Ending 2: One can repeat the argument just given by looking at the Sylow $5$-subgroups of $K$ to show that $K$ is a Frobenius group with Frobenius complement any Sylow $5$-subgroup and Frobenius kernel $N$ of order $2^3\cdot 7$. $N$ is normal in $G$ and has either a normal Sylow $7$-subgroup or is a Frobenius group with Frobenius complement any Sylow $7$-subgroup and Frobenius kernel the normal Sylow $2$-subgroup. Both cases easily lead to abelian subgroups of order $\ge 12$.

Ending 3: An even easier way to finish the proof is to look at the action by conjugation of a Sylow $3$-subgroup $S$ of $G$ on the Sylow $5$-subgroups of $K$ ($G$'s Frobenius complement). As the number of Sylow $5$-subgroups of $K$ is not a multiple of $3$, there is at least one fixed point $U$, which is normalized and hence centralized by $S$ giving an abelian subgroup $US$ of order $45$.