No simple group of order 1040

Question

Burnside (Proceedings of the London Mathematical Society, vol. 26; Collected Papers, vol. 1, p. 601) gave the following proof of the non-simplicity of groups od order 1040 :

"If simple, the group must have 26 sub-groups of order 5, each contained self-conjugately in a sub-group of order $2^{3}.5$. Such a sub-group necessarily contains an operation of order 10; and the corresponding operation of order 2, which is permuable with an operation of order 5, must, if expressed as an even substitution of 26 symbols, consist of 10 transpositions. It must therfore occur in 6 sub-groups of order $2^{ 3}.5$,"

So far, so good : we can find an element of order 2 commuting with an element of order 5 and such an element of order 2 normalizes exactly 6 subgroups of order 5. (I can give a detailed proof if anybody asks for it.) But Burnside continues :

"and be permutable with 6 sub-groups of order 5. But, since 6 is not a factor of the order of the group, this is impossible."

Well, the considered element of order 2 normalizes exactly 6 subgroups of oder 5, but Burnside seems to conclude from this that the considered element of order 2 centralizes exactly 6 subgroups of order 5. Why ?

So, my question is : could you explain the end of Burnside's proof or give another (correct) proof ? Thank you in advance.

Answer 1

Let $G$ be simple of order $1040$.

We have $|{\rm Syl}_{13}(G)|=40$, and the normalizer of a Sylow $13$-subgroup has order $26$. It must be dihedral by Burnside's Transfer Theorem*. Let $I_{13}$ be the conjugacy class in $G$ of involutions that normalize a $13$-subgroup. In the conjugation action on $ {\rm Syl}_{13}(G)$, elements $x \in I_{13}$ must be even permutations, and they fix at most one point in each of the three cycles of a $13$-element that they invert, so they fix four points altogether. Hence its centralizer in $|C_G(x)| \le 8$ and, in particular, $x$ is not central in a Sylow $2$-subgroup.

Now we consider the conjugation action of $G$ on the $26$ Sylow $5$-subgroups. Let $P \in {\rm Syl}_5(G)$ and $N = N_G(P)$, so $N=PS$ with $|S|= 8$. From Burnside's argument, there is an involution $s \in C_S(P)$, and $s$ fixes exactly $6$ points in the action. Since $C_S(P) \unlhd S$, we can choose $s \in Z(S)$, so $s \in Z(N)$. If $N < C_G(s)$ then $C_G(s)$ has more than one Sylow $5$-subgroup, and Burnside's argument gives a contradiction. So $C_G(s) = N$.

We can choose $ T \in {\rm Syl}_2(G)$ with $S \lhd T$, and we cannot have $s \in Z(T)$, so $\langle s \rangle$ is not normal in $T$.

Now we consider the possible isomorphism classes of $S$. We cannot have $S \cong C_8$, $Q_8$ or $D_8$, since they have a unique central subgroup of order $2$, which would be characteristic in $S$ and hence normal in $T$.

Suppose that $S \cong C_2 \times C_2 \times C_2$. It can be checked that any automorphism of $S$ of order $2$ must centralize a subgroup of $S$ of order $4$, so then we find that $|S \cap Z(T)| \ge 4$, and also, since ${\rm Aut}(P) \cong C_4$, we have $|C_S(P)| \ge 4$, and hence $|C_S(P) \cap Z(T)| \ge 2$, and we can choose $t \in Z(T)$, contradiction.

Finally we consider $S = C_4 \times C_2$. Then $s$ cannot be a square, or we would have $\langle s \rangle$ characteristic in $S$ again, so $S = \langle u \rangle \times \langle s \rangle$ with $|u|=4$, and $s$ is conjugate to $su^2$ in $T$.

The involutions in $I_{13}$ must be even permutations in their actions on ${\rm Syl}_5(G)$, so they act fixed-point-freely, and hence there exists $x \in I_{13} \cap S$. Since $x$ fixes at most one point in each of the two cycles of an element of order $13$ that it inverts, it fixes two points altogether, and so we cannot have $x=s$ or $x=su^2$ (which fix $6$ points). So $x = u^2$. But then $x\in Z(T)$, contrary to what we proved earlier: elements of $I_{13}$ are not in the centre of a Sylow $2$-subgroup.

• It would not be too hard to avoid using Burnside's Transfer Theorem.

Answer 2

Here's another proof using basic representation theory of the style that Burnside himself first introduced.

By Sylow III, $n_5 = 1,16,26$, so if $G$ is simple, then $n_5 = 16$ or $n_5 = 26$. We also have $n_{13} = 40$. Denote a $2$-Sylow and a $5$-Sylow and $13$-Sylow by $P_2$, $P_5$, and $P_{13}$ respectively.

If $n_5 = 16$, then $N_5 = N_G(P_5)$ has index $16$ and order $65$. But any group of order $65 = 5 \cdot 13$ is abelian, and so the order $13$ element is normal in this group, and hence $N_{13} = N_G(P_{13})$ has order dividing $5$. But $n_{13}=40$ so $N_{13}$ has order $26$, a contradiction.

We know that $N_{13}$ has order $26$. Since $P_{13}$ is not contained in $N_5$, any order $13$ element acts without fixed points on the $26$ $5$-Sylows. Thus the cycle decomposition of an element in $S_{26}$ corresponding to this action has cycle shape $13+13$. If $G$ contained an element of $26$ is would be conjugate to an element of $P_{13}$. But then (looking at its square) it would have to be a $26$-cycle. But the map from $G$ to $S_{26}$ lands inside $A_{26}$ since $G$ is simple, so it contains no such element. (This is the "not too hard" argument to avoid using Burnside's Transfer Theorem.) Similarly as above there is no element of order $65$, and so the only elements of order divisible by $13$ have order $13$. Each such element is conjugate to its inverse, and so there are six such conjugacy classes of elements each of order $1040/13 = 80$.

Let $V$ be the $26$-dimensional representation associated to the permutation representation on $5$-Sylows. This is just $\mathrm{Ind}^{G}_{N_5} \mathbf{Q}$. The corresponding character is defined over $\mathbf{Q}$. Any non-trivial element in $P_{13}$ has eigenvalues $1,\zeta,\ldots,\zeta^{12}$ each with multiplicity two where $\zeta^{13} = 1$. No other element has an eigenvalue in the field ${\mathbf{Q}}(\zeta)$ which is not $\pm 1$. Let us write

$$\chi_{V} = \sum a_W \chi_{W}$$

over the various irreducible representations $W$ of $G$. Since $\chi_V$ is defined over $\mathbf{Q}$, it follows that the RHS is also Galois invariant, and so $a_{W^{\sigma}} = a_{W}$ for any Galois automorphism $\sigma$. Since elements of order $13$ have exactly two eigenvalues $=1$ on $V$, and since the sum decomposition contains a single trivial representation ($\mathrm{Hom}_G({\mathbf{Q}},\mathrm{Ind}^{G}_{H}{\mathbf{Q}}) \simeq \mathrm{Hom}_{H}({\mathbf{Q}},{\mathbf{Q}}) \simeq \mathbf{Q}$), there will be a unique factor of dimension $> 1$ which also contains $1$ as an eigenvalue of an order $13$ element. It follows that this factor must also have traces over $\mathbf{Q}$. But the only sums of $p$th roots of unity which are rational are combinations of rational numbers and multiples of $\zeta+ \ldots + \zeta^{p-1}$. Thus this factor must have dimension exactly $13$.

This leaves a $12$-dimensional representation over $\mathbf{Q}$. The same argument shows this $12$-dimensional representation is irreducible over $\mathbf{Q}$. But it can't be absolutely irreducible because $12$ does not divide $1040$ and the dimension of any irreducible representation divides the order of the group (as proved by Burnside). It follows that this $12$-dimensional representation decomposes into a sum of Galois conjugate representations all of the same dimension $d$, where the constraints of dividing both $12$ and $1040$ imply that $d \in \{1,2,4\}$.

The case $d=1$ is ruled out by the fact $G$ is simple and hence perfect. If $d=2$, the trace of an order $13$ element will have to be defined over a subfield of $\mathbf{Q}(\zeta_{13})$ of degree at least $6$, and the trace of an order $5$ element will have to be defined over a subfield of $\mathbf{Q}(\zeta_5)$ of degree at least $2$ (in neither case could the eigenvalues both be one since then the representation would not be faithful contradicting the assumption that $G$ is simple). Hece any degree $d=2$ character is defined over a field of degree at least $12$, and so cannot live inside a $\mathbf{Q}$-representation of dimension less than $24$. So we arrive at a decomposition

$$\chi_V = \chi_{{\mathbf{Q}}} + \chi_{W_{13}} + \chi_U + \chi_{\sigma U} + \chi_{\sigma^2 U},$$

where $\mathrm{dim}(U)=4$ and $U$ is defined over the cubic subfield of $\mathbf{Q}(\zeta)$. Let $\chi$ be the character of $U$. Since, as proved above, the only elements of order divisible by $13$ are the elements of order $13$ themselves, It follows that the only values of $\chi_U$ which are not rational are on elements of order $13$. But we know that for some $\gamma$ of order $13$ we must have

$$\chi(\gamma^n) = \zeta^n + \zeta^{5n} +\zeta^{25n} + \zeta^{125n} = \zeta^n+ \zeta^{-n} + \zeta^{5n} + \zeta^{-5n},$$

since these are the only sums which lie in the cubic subfield of ${\mathbf{Q}}(\zeta)$. Recall that each element $\gamma \in P_{13}$ is conjugate to its inverse but no other order $13$ element. It follows that the conjugacy classes are given by $\gamma^n$ for $n=1,\ldots 6$, and they each have order $80 = 1040/13$. But now we find that, with $\sigma \zeta = \zeta^2$,

$$\begin{aligned} 1 - 0 = & \ \langle \chi_U,\chi_U \rangle - \langle \chi_U, \chi_{U^{\sigma}} \rangle \\ = & \ \frac{80}{1040} \sum_{n=1}^{6} ( \zeta^n+ \zeta^{-n} + \zeta^{5n} + \zeta^{-5n})^2 - ( \zeta^n+ \zeta^{-n} + \zeta^{5n} + \zeta^{-5n})( \zeta^{2n}+ \zeta^{-2n} + \zeta^{10n} + \zeta^{-10n}) \\ = & \ 2 \end{aligned}$$

This is a contradiction.

Answer 3

And yet another proof: Suppose $G$ is simple of order $1040=2^4\cdot 5\cdot 13$. By Frobenius $p$-normal complement theorem, there exists a $2$-subgroup $Q\le G$ such that $N_G(Q)/C_G(Q)\le\mathrm{Aut}(Q)$ is not a $2$-group. If $|Q|\le 8$, then $3$ and $7$ are the only odd prime divisors of $|\mathrm{Aut}(Q)|$. Hence, $Q$ is an elementary abelian Sylow $2$-subgroup and $|N_G(Q)|=2^4\cdot 5$. Since $13\not\equiv 1\pmod{8}$, there exists another Sylow $2$-subgroup $P$ such that $P\cap Q\ne 1$. Since $P$ is abelian, $N_G(P\cap Q)$ contains the Sylow subgroups $P$ and $Q$. Hence, $|N_G(P\cap Q)|=2^4\cdot 5$ ($2^4\cdot 13$ would be too big). Now a Sylow $5$-subgroup $R$ of $N_G(P\cap Q)$ must centralize $P\cap Q$. Therefore, $P\cap Q$ is contained in $N_G(R)$. Sylow's theorem applied to $N_G(P\cap Q)$ yields $N_G(P\cap Q)\subseteq N_G(R)$. But since $13\not\equiv 1\pmod{5}$, we must have $R\unlhd G$. Contradiction.