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Turning the Sylow conditions into proofs: forcing a count to be one, counting the elements that the alternatives would need, and ruling out simple groups of orders such as $15$, $30$ and $56$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to list the possible Sylow counts for a given order, force a count to be one where the conditions allow it, carry out an element-counting argument using the fact that distinct subgroups of prime order meet trivially, use the action on cosets when element counting is unavailable, and prove that no group of order $30$ is simple.
The last lesson stated the Sylow theorems and proved that a unique Sylow subgroup is normal. That corollary is the whole engine of this lesson: everything here is an attempt to force some $n_p$ to equal $1$, either directly from the two conditions or by showing that the alternatives would need more elements than the group has.
A group is simple when its only normal subgroups are the trivial one and itself. The normaliser $N_G(P) = \{g : gPg^{-1} = P\}$ is the stabiliser of $P$ under conjugation, so $n_p = [G : N_G(P)]$. Element counting is the technique of adding up the elements of each order that a set of Sylow counts forces to exist, and comparing the total with $|G|$.
Every argument in this lesson has the same two moves.
Move one: the conditions alone. $n_p \equiv 1 \pmod p$ and $n_p \mid m$. List the divisors of $m$, keep the ones congruent to $1$, and hope only $1$ survives. It often does, and then that Sylow subgroup is normal and the group is not simple.
For $|G| = 20 = 2^{2} \cdot 5$: $n_5 \mid 4$ and $n_5 \equiv 1 \pmod 5$. The divisors of $4$ are $1, 2, 4$, and none but $1$ is $1$ modulo $5$. Done in one line.
Move two: count the elements. When several values survive, suppose the group is simple — so every count takes a value above $1$ — and count what that forces.
The key fact is that two distinct subgroups of prime order meet only at the identity: their intersection is a subgroup of each, so its order divides $p$ and is not $p$. Therefore $n_p$ subgroups of order $p$ hold $n_p(p-1)$ different elements of order $p$, and elements of different orders are never confused. Adding over the primes gives a lower bound on $|G|$, and when it exceeds $|G|$ the assumption of simplicity fails.
$|G| = 30$: simplicity would force $n_3 = 10$ and $n_5 = 6$, needing $20 + 24 = 44$ elements in a group of $30$.
A third move, when neither works. Sometimes the largest counts just fit, and then what fails is the room left over. For $|G| = 56 = 2^{3} \cdot 7$: $n_7 \in \{1, 8\}$, and $n_7 = 8$ uses $48$ elements of order $7$, leaving exactly $8$ — which must be the single Sylow $2$-subgroup, so that one is normal instead. Either branch ends with a normal subgroup.
And a fourth: the index action. If $G$ is simple with a subgroup $H$ of index $k > 1$, then acting on the cosets gives an injective map $G \to S_k$, so $|G|$ divides $k!$. That rules out, for instance, a simple group of order $80$ with a Sylow subgroup of index $5$, since $80 \nmid 120$ would have to be checked — and more usefully it bounds how small an index a simple group can have.
A second course in algebra counts before it constructs. An orbit size divides the order of the group, a Sylow count is pinned between a divisibility and a congruence, and a degree multiplies up a tower — and each of those numbers rules out structures nobody has to go looking for.
Another way: picture
Think of the group as a room with $|G|$ seats. Each Sylow $p$-subgroup of prime order claims $p - 1$ seats that nobody else may use, and the identity has its own. Assuming simplicity forces each family to be as large as it can be, and the families then demand more seats than the room has. The contradiction is literally that the people do not fit.
Another way: steps
To show no simple group has order $n$: 1. Factor $n$ and list the candidate values of $n_p$ for each prime. 2. If some prime forces $n_p = 1$, stop — there is a normal subgroup. 3. Otherwise assume simplicity, so every count is its largest candidate. 4. Count $n_p(p-1)$ elements for each prime whose Sylow order is prime. 5. Compare the total with $n$; if it is too big, or leaves too little room for a Sylow subgroup that must still fit, the group is not simple. 6. If everything still fits, try the action on the cosets of a subgroup of small index.
The method does not always work, and where it stops is informative.
$|G| = 60$. Sylow gives $n_5 \in \{1, 6\}$, $n_3 \in \{1, 4, 10\}$, $n_2 \in \{1, 3, 5, 15\}$. Assuming simplicity, $n_5 = 6$ costs $24$ elements and $n_3 = 10$ costs $20$, which is $44$ of the $60$ — tight, but not yet impossible. Sharper arguments are needed, and the reason they fail is that $A_5$ exists: order $60$ is the smallest order of a non-abelian simple group.
$|G| = p^{a}$. The class equation already showed the centre is non-trivial, so a $p$-group is never simple unless it has prime order.
$|G| = pq$ with $p < q$ prime. Then $n_q \mid p$ and $n_q \equiv 1 \pmod q$; since $p < q$, the only possibility is $n_q = 1$. Every group of order $pq$ has a normal subgroup, and if additionally $q \not\equiv 1 \pmod p$ it is cyclic.
What the exercise is really doing. Each such result removes one order from the list of possible sizes of a finite simple group. The classification of finite simple groups is the completion of that programme, and it runs to thousands of pages — but the first few hundred orders are settled by exactly the counting in this lesson.
Counting $n_p \cdot p$ instead of $n_p(p-1)$. The identity is shared by every subgroup and must be counted once, not $n_p$ times.
Using the clean count when the Sylow order is not prime. Two subgroups of order $p^{2}$ can meet in a subgroup of order $p$, so they do not contribute $n_p(p^{2}-1)$ new elements. The argument has to be made by hand there.
Forgetting that only one count needs to be $1$. The conclusion is not simple, so a single normal Sylow subgroup finishes it. There is no need to force all of them.
*Concluding therefore abelian or therefore cyclic.* Non-simplicity is much weaker. $S_4$ is not simple and is nothing like abelian.
Assuming the method must succeed. Simple groups of order $60$, $168$ and $360$ exist, and no counting argument can rule them out — because they are there.
$56 = 2^{3} \cdot 7$, so $n_7 \mid 8$ and $n_7 \equiv 1 \pmod 7$: the candidates are $1$ and $8$.
Two possibilities.
If $n_7 = 8$, the eight subgroups of order $7$ contribute $8 \times 6 = 48$ elements of order $7$.
Element counting.
That leaves $56 - 48 = 8$ elements, and a Sylow $2$-subgroup needs all eight — so there is exactly one of those, and it is normal. If instead $n_7 = 1$, that subgroup is normal. Either way, not simple.
Both branches end the same way.
$255 = 3 \cdot 5 \cdot 17$. For $p = 17$: $n_{17} \mid 15$ and $n_{17} \equiv 1 \pmod{17}$, and none of $1, 3, 5, 15$ but $1$ qualifies.
A normal subgroup of order $17$.
The quotient has order $15$, and there is only one group of order $15$, the cyclic one. So $G$ has a normal subgroup of order $17$ with cyclic quotient.
Pass to the quotient.
A short further argument — the automorphism group of $\mathbb{Z}_{17}$ has order $16$, coprime to $15$ — makes the extension trivial, so $G \cong \mathbb{Z}_{255}$.
One group of that order.
$36 = 2^{2} \cdot 3^{2}$, so $n_3 \mid 4$ and $n_3 \equiv 1 \pmod 3$: the candidates are $1$ and $4$.
Two possibilities.
If $n_3 = 4$, let $G$ act by conjugation on those four Sylow $3$-subgroups. That is a homomorphism $G \to S_4$.
An action on a small set.
A simple group has trivial kernel, so $36$ would have to divide $|S_4| = 24$, which it does not. So $n_3 = 1$ and that subgroup is normal. Element counting was not available here, because the Sylow order $9$ is not prime — the action on the cosets is what replaces it.
$A_5$ has $60 = 2^{2} \times 3 \times 5$ elements. The order of a Sylow subgroup for each prime is given. Fill in how many there are and how many elements of order exactly that prime they contribute between them, then the total with the identity.
| Order of a Sylow subgroup | How many there are | Elements of that order they hold | |
|---|---|---|---|
| The prime $5$ | 5 | ||
| The prime $3$ | 3 | ||
| The prime $2$ | 4 | ||
| All of them, plus the identity | — | — |
How many Sylow $3$-subgroups does $S_4$ have?
Answer:
Is there a simple group of order $168$ that is not of prime order?
Select every statement that may be used in a Sylow element-counting argument.
This task has no paper form; do it on a device.
Put in order the steps of showing that no simple group has a given order.
Number the steps in order (write the number in the box):
Build the proof that no group of order $30$ is simple.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Is there a simple group of order $168$ that is not of prime order?
You can decide whether the Sylow conditions rule out a simple group of a given order, and carry out the element count when they do not settle it alone. Say in your own words why two different subgroups of prime order share only the identity. Next: chains of normal subgroups, and what the pieces of a group are.
9. Your turn: no simple group of order $36$, step 3