From Dummit & Foote, Abstract Algebra, $\S6.2$, Exercise 17(a).

Prove that there is no simple group of order 420.

Suppose not; label such group $G$. the number of Sylow 7-subgroups of $G$ is 15. Let $G$ act on the set of Sylow 7-subgroups (denoted hereon by "letters") by conjugation. There is only one orbit of size 15 (by Sylow 2nd), thus each stabilizer on one letter has size 420/15 = 28.

The action induces an injective homomorphism of $G$ into $A_{15}$. Each stabilizer on a single letter should have an element of order 7 (by Cauchy), permuting the remaining 14 letters. Naturally, this element then generates the unique Sylow 7 within the stabilizer.

I now assume that the element of order 7 is a product of 2 disjoint 7-cycles. Is this valid, and why? In particular, am I able to eliminate the possibility of the element being a single 7-cycle?

If the above assumption is valid, then I am now able to eliminate the possibility of order 14 and 28 elements, since order 14 implies single 14-cycle (odd) or product of single 7-cycle and some 2-cycles (2nd power is single 7-cycle), likewise for order 28.

Now use the fact that the Sylow 7 is normal within the stabilizer: the permutations that sends a 7-cycle to its power by conjugation is either a product of 3 2-cycles (sends to inverse), 2 3-cycles (sends to 2nd/4th power), or a 6-cycle (sends to 3rd/5th power). By similar calculations, permutations that switch between the two 7-cycles are either 7 2-cycles, a 2-cycle and 3 4-cycles, a 2-cycle and 2 6-cycles, a 2-cycle and a 12-cycle, or a 14-cycle. Since the remaining elements are either order 2 and 4, our choices are either 6 2-cycles, 7 2-cycles (odd), or a 2-cycle and 3 4-cycles (third power breaks the pairing of the two 7-cycles).

Are my calculations and reasoning correct here?

The pair of fixed letters in the 2 7-cycles then determines the whole of the permutation, but noting that the two different fixed letter 2-cycles (per 1 7-cycle) result in a product of a 7-cycle, we conclude that the pairings of fixed letters must be disjoint for two different elements. Then, the number of possible remaining elements is 7 out of a required 21; contradiction.

In general, is there a cleaner way to go about this exercise, or proving nonexistence of groups of some highly composite order? I only know the method of embedding the group in an alternating group and trying to derive a contradiction from there (outside the usual repertoire).

  • 1
    $\begingroup$ I've updated my answer to include an answer to the main question itself which doesn't need the permutation calculations so much, but it is much longer (at least when covered in detail) and pretty much required classifying all order 28 groups, so I'm not sure if it's better or worse, but it's the way I would go about it personally. $\endgroup$ Feb 8 '20 at 17:09
  • $\begingroup$ You can do this slightly differently for a quick solution. First, an element of order $7$ in $A_15$ must be a product of two 7-cycles: otherwise it would normalize another Sylow 7-subgroup (each fixed point of the permutation is a normalized Sylow subgroup) and that would give a subgroup of order $49$. Now the centralizer of the product of two 7-cycles in $A_15$ has order$49$. So there's no element of order $2$ centralizing. However, in your group the normalizer of a Sylow 7-subgroup has order $28$, and the normalizer/centralizer theorem says there has to be an order 2 centralizing element. $\endgroup$
    – Steve D
    Nov 3 '20 at 3:25

As an answer for you first questions: let me call the homomorphism $\phi:G\rightarrow A_{15}$, and the order 7 element $g$, generating the Sylow 7-subgroup $P$, which has normaliser (the stabilisers of your action are called normalisers) $N_G(P)$ with order 28. First, you state that $P$ is the unique order 7 subgroup of $N_G(P)$, which is true, because an order 28 group has only 1 Sylow 7-subrgroup. Note that $\phi(g)^7=\phi(g^7)=\phi(1)=\text{id}$, so $\phi(g)$ (which permutes 14 letters as you say) has order dividing 7, i.e. either 1 or 7. Therefore it is either the identity permutation, a single 7-cycle, or 2 disjoint 7-cycles. But in any case apart from the last case, $\phi(g)$ fixes at least 7 other letters, in other words it normalises some Sylow 7-subgroups that are not $P$; call one of these $Q$. But now $N(Q)$ contains 2 distinct order 7 subgroups, namely $P$ and $Q$, which contradicts your point that an order 28 group can have only 1 order 7 subgroup. Therefore we get the result that you want; $\phi(g)$ is 2 disjoint 7-cycles.

EDIT: As an answer to the main problem: I don't think I would go down your route directly (mainly because I don't like composing permutations, which I find very fiddly), so here is a way to continue with minimal fiddly permutation computations.

Let's look at the structure of $N_G(P)$; we know it has at least one order 4 (Sylow 2) subgroup, so call this $H$. now we have $P\unlhd N_G(P)$ and $H\le N_G(P)$, so $PH\le N_G(P)$. But $P,H\le PH$ so the size of $PH$ is divisible by 4 and 7, so it has size (at least) 28, and hence $PH=N_G(P)$; we can represent all elements of $N_G(P)$ as the product of an element of $P$ with one of $H$, and this representation is unique because there are only $7\cdot4=28$ possible representations. $P\cong C_7$, and either $H\cong C_4$ or $H\cong V_4$. Then we have (by $P$ normal in its normaliser) that $N_G(P)$ is isomorphic to a semidirect product, $C_7\rtimes C_4$ or $C_7\rtimes V_4$, determined by a homomorphism $\psi:H\rightarrow\text{Aut}(P)\cong\text{Aut}(C_7)\cong C_6$. $\ker(\psi)$ has size 1, 2 or 4. But ${H\over\ker(\psi)}\cong\text{im}(\psi)\le C_6$, so it cannot have size 1. If it has size 4, then the homomorphism is trivial, the semidirect product is direct, and $N_G(P)$ is abelian, and hence either $C_{14}\times C_2$ or $C_{28}$. But as you say, it cannot have an order 14 or order 28 element, contradiction. So $\vert\ker(\psi)\vert=2$, and so the image is the order 2 subgroup of $\text{Aut}(P)$, namely the trivial automorphism and the inversion automorphism (call this automorphism $\varphi$).

First, suppose $H\cong V_4$, and call the 3 non-identity elements $x$, $y$ and $xy$. If $\psi(x)=\psi(y)=\varphi$, then $\psi(xy)=\varphi^2=\text{id}$, so in any case we have non-identity element $h\in H$ such that $\psi(h)=\text{id}$, i.e. $h$ commutes with $g$. But then (easily verified) $gh$ has order 14, contradiction again.

So we must have $H\cong C_4$, so there exists $h\in H$ with order 4. If $\psi(h)=\text{id}$, then $\varphi$ is not in the image of $\psi$ ($h$ generates $H$), contradiction. So $\psi(h)=\varphi$.

EDIT 2: as pointed out in the comment, at this point I was already done, because $h^2$ now commutes with $g$, so as before we have an order 14 element $gh$ which is a contradiction. If you want to read a far more convoluted solution, here it is:

By semidirect product, $hgh^{-1}=g^{-1}$. Then finally we do have to look at permutations again; let $\phi(h)=\sigma$, which should be an order 4 permutation by the injection $\phi$. We have $\sigma\phi(g)\sigma^{-1}=\phi(g)^{-1}$, where w.l.o.g. $\phi(g)$, two disjoint 7-cycles, sends $i$ to $i+1$ (mod 7) and $i'$ to $i'+1'$ (mod 7') as a permutation of the letters 0 through 6 and 0' through 6'. Working modulo 7 and 7':

If $\sigma(i)=j$ then $\sigma(i+1)=[\sigma\phi(g)\sigma^{-1}](j)=[\phi(g)^{-1}](j)=j-1$, and by induction $\sigma(i+k)=j-k$ for all $k$ modulo 7. Let $k=j-i$, and we get $\sigma(j)=i$. Similarly, if $\sigma(i')=j'$, then $\sigma(j')=i'$.

By the same inductive step we get that if $\sigma(i)=j'$ then $\sigma(i+k)=j'+k'$ for each $k$. But now $\sigma(j')=i+n$ for some particular $n$, which by the same induction again gives $\sigma(j'+k')=i+n+k$ for each $k$. Now $\sigma(i+n)=j'+n'$, and then by having order 4, $i=\sigma^{4}(i)=\sigma^{3}(j')=\sigma^{2}(i+n)=\sigma(j'+n')=i+n+n=i+2n$ so in fact we must have $n=0$, by inverse of 2 existing modulo 7. Hence $\sigma(i)=j'$ gives $\sigma(j')=i$. Again similarly, we also get that if $\sigma(i')=j$ then $\sigma(j')=i$.

This covers all cases for where $\sigma$ sends the letters 0 through 6 and 0' through 6', so $\sigma$ is composed only of disjoint transpositions, and hence has order 2, the final contradiction we needed, therefore $G$ is not simple.

  • 1
    $\begingroup$ In the last case ($H \cong C_4$), can't you just say that $\psi(h^2) = \text{id}$, so $g$ commutes with $h^2$ and $gh^2$ has order $14$? This would save doing any "fiddly permutation computations". $\endgroup$
    – math54321
    Feb 8 '20 at 21:23
  • $\begingroup$ Yep... now I feel pretty stupid going through all off that and missing that obvious fact. I'll make an edit. I always do this; I work on a problem until I find any solution, and then just stick with it, without thinking about how I could simplify it. Thanks $\endgroup$ Feb 8 '20 at 22:05

A15 does have subgroups of order 14 or 28. Actually an odd cycle is an even permutation, so 7-2-2 and 7-cycle are in A15. Without futher deduction that Sylow 2 group is isomophic to V4, the group of 28 could be <(1,2,3,4,5,6,7),(8,9,10,11)(12,13,14,15)>, i.e. A group generated by a 7-cycle and a disjoint 4-cycles. By contradiction assume that G is simple, |G|=4 x 3 x 5 x 7, so no subgroup of index less than 7, making n5=21, n7=15. Denote N7 be the normalizer of a sylow-7 group, so |N7|=28 and |N5|=20.

  • We first prove that n3=70 and n2=35
  • show that N2=A4 (N2 isomorphic to A4) and P2=V4 (Klein 4 group)
  • There is an element of order 14 in N7 and an element of order 10 in N5.
  • G acts transitively on all subgroups of order 2 by conjugation.
  • There is a subgroup U of order 2, s.t. <P2, P5, P7> <= NG(U), whose index is less than 7.

By sylow's theorem, n3=10, 28 or 70, n2=7, 15, 21, 35 or 105. If n3=7 (or |N3|=60), since N3 is not simple group of order 60, n5=1. That means a P5 normalizes a P3, contradicting to |N5|=20. By the same method, we rule out |N2|=60.
If n3=10 (or |N3|=42), then there is a P7 commute with P3 in N3, contradict to |N7|=28.
Leaving only n3=70.
Next we rule out n2=15 (or N2=28). If it were, we first prove that N7=N2=C28, i.e. N7 and N2 are in the same conjugation class, then contradict by counting elements. If N7 is not isomorphic to C28, we can conjugate any N2 so that it contains the same P7 subgroup. The the unqiue sylow-2 subgroup Q2 of N2 normalizes P7 but distincts from P2 in N7, since N7 is not isomorphic to C28. So P7 is normal in <P7, P2, Q2>, contradicts to |N7|=28. So N7=N2=C28. The total elements of order 3 and 5 is 2x70+4x21=224, and the total unique elements of all conjugations of N2 (or N7) is greater than 196, overflowing |G|.
Now we rule out n2=21 (or N2=20). By the same way, we argue that N2 and N5 are in the same conjugation class and isomorphic to C20. We first find out U of order 2 commute with a P7, then find a P5 and P2 commute with U, so U=<P7, P5, P2>, whose order is greater than 140.
Next we rule out n2=105, leaving only n2=35 (or N2=12).
Now we show that N2=A4. If not, then it is isomorphic to an abelian group, contradicts to N3=6.
Since A4 act transitively on subgroups of order 2 in V4 (the order 4 subgroup in A4), and G act transitively on all P2, by conjugation, completing the proof that G acts transitively on all order 2 subgroups of G.
Since P2=V4, a homomorphism from V4 to Aut(P7) cannot be injective, so there is an element of order 2 compute with element of order 7 in N7. The same reason, There is an element of order 2 compute with element of order 5 in N5.
Now that G acts transitively on any subgroup U of order 2, we can choose a P5, P7, P2 normalizing U, making order of NG(U) bigger or equal to 70. QED.


Firstly, we have the following factorization $420=2^$. I want to work on Sylow $5-$subgroups. I will use the $P_q$ for Sylow $q-$subgroups of G, and $n_q$ for the number of Sylow $q-$subgroups.

From the Sylow Theorems we have $n_q \equiv 1\pmod q $ and $n_q ||G| $. Thus, we have following possibilities for $n_5$:

$$n_5 = \begin{cases} 6 \\[2ex] 21 \end{cases}$$

In each case we have $3|n_5$, and this means that $N_{G}(P_5)$ is not multiple by $3$. Thus, we cannot have an element of order $15$, as if we have such an element then we should have an element of order $3$ in the $N_{G}(P_5)$. This consequence can be improved.

Lemma 1: Simple group $G$ of order 420 does not have any subgroup of order $15$.
This is because every group of order $15$ is cyclic as it is with the form of $pq$ and $q $ is not $ 1\pmod p $.

Now we want to use Sylow $3-$subgroups.
$$n_3 = \begin{cases} 4 \\[2ex] 7 \\[2ex] 28 \end{cases}$$

We know that $n_3$ could not be $4$, since we don't have any subgroups with index less than $7$. In both other cases we will have a subgroup of order $15$. We know that $P_3$ is a normal subgroup of $N_G(P_3)$, and in both cases $5|N_G(P_3)$. Therefore, we have a subgroup of order $5$ in $N_G(P_3)$, and actually its is the Sylow $5-$subgroup of $G$. But, $P_3$ is a normal subgroup in $N_G(P_3)$; as a result, $P_3P_5$ is a subgroup of $N_G(P_3)$. This means $G$ has a subgroup of order $15$. Contradiction

Edited Part: I missed $n_3=10$. In such condition, $P_7$ is a normal subgroup itself. As if $n_3=10$ this means that $|N_G(P_3)|=2.3.7$. In this group $P_7$ is normal in $N_G(P_3)$ and $P_3P_7$ is a cyclic subgroup. Thus, $3.7| |N_G(P_3)|$. Moreover, $n_7|2^2.5$ and $n_7$ is $1 \pmod 7$. Therefore, $n_7 = 1$.

  • $\begingroup$ you missed the case that $n_3=10$ $\endgroup$ Oct 3 '21 at 5:11
  • $\begingroup$ @Minglingmaster Thanks. I added that part. $\endgroup$
    – Janbazif
    Oct 3 '21 at 14:12

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