Back to the on-screen lesson ·
The converse to Lagrange that does hold: for the full power of each prime a subgroup exists, all such subgroups are conjugate, and their number is squeezed between a divisibility and a congruence.
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 split a group order as a prime power times a coprime part, state the three Sylow theorems precisely, give the order of a Sylow subgroup, list the values the count may take from the divisibility and congruence conditions, say what the theorems do not claim, and prove that a Sylow subgroup is normal exactly when it is the only one.
Lagrange's theorem says the order of a subgroup divides the order of the group. Its converse is false: $A_4$ has twelve elements and no subgroup of six. So which divisors are achieved was left open by the first course, and the Sylow theorems close the most important part of that question — the prime power divisors, where the answer is always yes.
A $p$-group is a group whose order is a power of the prime $p$. Writing $|G| = p^{a}m$ with $p \nmid m$, a Sylow $p$-subgroup is a subgroup of order exactly $p^{a}$ — the largest possible $p$-group inside $G$. The number of them is written $n_p$, and $\operatorname{Syl}_p(G)$ is the set of them.
Let $G$ be finite with $|G| = p^{a}m$, where $p$ is prime and $p \nmid m$.
First theorem (existence). $G$ has at least one subgroup of order $p^{a}$. More is true — there is a subgroup of order $p^{k}$ for every $k \le a$, and each sits normally inside one of order $p^{k+1}$ — but the top one is what the name is reserved for.
Second theorem (conjugacy). Any two Sylow $p$-subgroups $P$ and $Q$ satisfy $Q = gPg^{-1}$ for some $g \in G$. In particular they are isomorphic, and every $p$-subgroup of $G$ lies inside some Sylow $p$-subgroup.
Third theorem (counting).
$$n_p \equiv 1 \pmod p \qquad \text{and} \qquad n_p \mid m.$$
The second condition holds because $G$ acts transitively on $\operatorname{Syl}_p(G)$ by conjugation, so $n_p$ is an index; the first comes from letting one Sylow subgroup $P$ act on the set, where $P$ itself is the only fixed point and every other orbit has size a positive power of $p$.
The corollary that matters. $P$ is normal in $G$ exactly when $n_p = 1$. Conjugation carries a Sylow subgroup to a Sylow subgroup, so if there is only one it is sent to itself; and conversely a normal one has no other conjugates, which by the second theorem means no other Sylow subgroups at all.
Why this is a strong tool. Two arithmetic conditions on one integer, both read off the order of the group, frequently leave a single possibility — and that possibility is a structural statement. A great many classification results about groups of small order are one line of divisor arithmetic followed by this corollary.
What Sylow does not say. Nothing about non-prime-power divisors: the converse of Lagrange stays false. Nothing about uniqueness in general: $S_3$ has three Sylow $2$-subgroups. And nothing that distinguishes the isomorphism types of the Sylow subgroups themselves — order $p^{2}$ leaves two possibilities, and the theorems are silent about which occurs.
Another way: picture
Picture the Sylow $p$-subgroups as a small set of objects the group permutes by conjugation. The second theorem says the group can carry any one of them to any other, so the set is a single orbit; the third theorem is then orbit-stabiliser, which makes $n_p$ an index. And the congruence comes from letting just one of them do the pushing: it cannot move itself, and everything else it moves travels in orbits whose sizes are powers of $p$.
Another way: steps
To apply the theorems to a group of known order: 1. Factor $|G| = p^{a}m$ for the prime of interest. 2. A Sylow $p$-subgroup has $p^{a}$ elements, and exists. 3. List the divisors of $m$. 4. Keep those congruent to $1$ modulo $p$: those are the possible $n_p$. 5. If only $1$ survives, that subgroup is normal and $G$ is not simple. 6. If more survive, count elements: each Sylow subgroup of prime order contributes $p - 1$ elements of order $p$ that no other one holds.
Both halves of the third theorem are actions, which is why this unit put actions first.
Divisibility. Let $G$ act on $\operatorname{Syl}_p(G)$ by conjugation. The second theorem says this action is transitive, so the whole set is one orbit and $n_p = [G : N_G(P)]$, where $N_G(P)$ is the stabiliser — the normaliser of $P$. Since $P \le N_G(P)$, that index divides $|G| / p^{a} = m$.
The congruence. Now restrict the action to $P$ itself. A fixed point is a Sylow subgroup $Q$ with $xQx^{-1} = Q$ for all $x \in P$, and a short argument shows the only such $Q$ is $P$. Every other orbit has size dividing $|P| = p^{a}$ and greater than $1$, so is a positive power of $p$. Adding: $n_p = 1 + (\text{a sum of positive powers of } p)$, which is $1$ modulo $p$.
That is the class equation's argument again, on a different set. It is worth noticing how little machinery all of this needs: an action, orbit-stabiliser, and the observation that orbit sizes divide the order.
A worked count. $|G| = 12 = 2^{2} \cdot 3$. For $p = 3$: $n_3 \mid 4$ and $n_3 \equiv 1 \pmod 3$, so $n_3 \in \{1, 4\}$. Both occur — $\mathbb{Z}_{12}$ has one, $A_4$ has four. For $p = 2$: $n_2 \mid 3$ and $n_2$ odd, so $n_2 \in \{1, 3\}$; both occur too. So order $12$ is not settled by Sylow alone, and indeed there are five groups of order $12$. The theorems narrow; they do not always decide.
Taking a Sylow $p$-subgroup to have order $p$. It has order the full power $p^{a}$. In a group of order $24$ the Sylow $2$-subgroup has eight elements, not two.
Assuming a Sylow subgroup is normal. Only when it is unique. Three of the six elements of $S_3$ generate three different Sylow $2$-subgroups.
*Reading the second theorem as there is only one.* Conjugate is not equal. What it gives is that they are all alike, which is weaker and still very useful.
Inverting Lagrange. Sylow gives subgroups of prime power order only. A group of order $12$ need have no subgroup of order $6$.
Forgetting that $n_p$ must satisfy both conditions. A divisor of $m$ that is not $1$ modulo $p$ is not a candidate, and the congruence is usually the one that does the work.
Expecting the count to be determined. Often two values survive, and then more than one group of that order exists — which is information, not a failure.
$15 = 3 \cdot 5$. For $p = 5$: $n_5 \mid 3$ and $n_5 \equiv 1 \pmod 5$. The divisors are $1$ and $3$, and $3$ is $3$ modulo $5$, so $n_5 = 1$.
The congruence kills the other divisor.
For $p = 3$: $n_3 \mid 5$ and $n_3 \equiv 1 \pmod 3$. The divisors are $1$ and $5$, and $5$ is $2$ modulo $3$, so $n_3 = 1$.
Both counts forced.
Both Sylow subgroups are normal, they intersect trivially by Lagrange, and their product has $15$ elements — so $G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}$. There is exactly one group of order $15$.
A classification from two divisor lists.
$|S_4| = 24 = 2^{3} \cdot 3$. A Sylow $2$-subgroup has $8$ elements and a Sylow $3$-subgroup has $3$.
Split the order.
$n_3 \mid 8$ and $n_3 \equiv 1 \pmod 3$, so $n_3 \in \{1, 4\}$; the eight three-cycles fill four subgroups of order $3$, so $n_3 = 4$.
Counted directly.
$n_2 \mid 3$ and $n_2$ is odd, so $n_2 \in \{1, 3\}$; there are three copies of the square's symmetry group inside $S_4$, so $n_2 = 3$. Neither is normal, and indeed the normal subgroups of $S_4$ are $1$, $V_4$, $A_4$ and $S_4$.
Both counts above one.
$56 = 2^{3} \cdot 7$, so a Sylow $7$-subgroup has $7$ elements and $n_7$ divides $8$.
Split the order.
The divisors of $8$ are $1, 2, 4, 8$, and of those only $1$ and $8$ are congruent to $1$ modulo $7$.
Two candidates.
So $n_7$ is $1$ or $8$. If it is $1$ the subgroup is normal; if it is $8$ then the eight subgroups meet only at the identity and contribute $8 \times 6 = 48$ elements of order $7$, leaving exactly $8$ elements for a single Sylow $2$-subgroup, which is then normal. Either way the group is not simple.
A group has $360 = 2^{3} \times 3^{2} \times 5$ elements. For each prime, fill in the order of a Sylow subgroup for that prime and the part of $360$ left when that power is removed.
| Order of a Sylow subgroup | The part left over | |
|---|---|---|
| The prime $2$ | ||
| The prime $3$ | ||
| The prime $5$ |
What is the order of a Sylow $3$-subgroup of $A_5$?
Answer:
For a finite group $G$ with $|G| = p^{a}m$ and $p$ not dividing $m$, select every statement the Sylow theorems make.
This task has no paper form; do it on a device.
Match each question about Sylow subgroups to the statement that answers it.
| Always — that is the first theorem | They are conjugate to one another | A number dividing $m$ and congruent to $1$ modulo $p$ | Exactly when there is only one of them | Only when $G$ is abelian | |
|---|---|---|---|---|---|
| Does a subgroup of order $p^{a}$ exist? | |||||
| How are two Sylow $p$-subgroups related? | |||||
| How many Sylow $p$-subgroups are there? | |||||
| When is a Sylow $p$-subgroup normal? |
Put in order the steps of working out what the Sylow theorems say about a group whose order is given.
Number the steps in order (write the number in the box):
Build the proof that a Sylow $p$-subgroup is normal exactly when it is the only one.
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.
A group has $360 = 2^{3} \times 3^{2} \times 5$ elements. For each prime, fill in the order of a Sylow subgroup for that prime and the part of $360$ left when that power is removed.
| Order of a Sylow subgroup | The part left over | |
|---|---|---|
| The prime $2$ | ||
| The prime $3$ | ||
| The prime $5$ |
You can state the three Sylow theorems and list the possible values of a Sylow count from the order of a group. Say in your own words why a unique Sylow subgroup has to be normal. Next: turning those counts into proofs that certain orders admit no simple group.
9. Your turn: the Sylow $7$-subgroups of a group of order $56$, step 3