Sylow theorems and groups of small order

In one sentence

Sylow's counting conditions usually pin down enough structure to classify a group from its order alone, and when they do not, counting elements of prime order finishes the job.

Why the exam asks it

Because "is every group of order $n$ abelian / cyclic / simple" is a question with a short mechanical answer, and because the two Sylow conditions have to be used together. Options routinely satisfy one and quietly fail the other.

The idea

With $|G| = p^k m$ and $p \nmid m$: Sylow $p$-subgroups have order $p^k$, they all exist, they are all conjugate, and

$$n_p \equiv 1 \pmod p, \qquad n_p \mid m .$$

$n_p = 1$ $\iff$ the Sylow $p$-subgroup is normal, since conjugacy of all of them means a solitary one is fixed.

The standard route to non-simplicity

1. Find a $p$ forcing $n_p = 1$. Often the congruence and the divisibility have only the solution $1$ in common. Then there is a proper non-trivial normal subgroup and $G$ is not simple.
2. If not, count elements. Distinct Sylow $p$-subgroups of prime order intersect trivially, so $n_p$ of them contribute $n_p(p-1)$ elements of order $p$. Add up over primes; if the total exceeds $|G|$, some $n_p$ must be $1$.
3. If that fails, act on the Sylow subgroups. $G$ acts by conjugation on the $n_p$ Sylow $p$-subgroups, giving $G \to S_{n_p}$. If $|G| \nmid n_p!$ the map has a non-trivial kernel, which is a normal subgroup.

Order 30. Suppose simple, so $n_3 > 1$ and $n_5 > 1$. Then $n_5 \mid 6$ with $n_5 \equiv 1 \pmod 5$ gives $n_5 = 6$, contributing $6 \times 4 = 24$ elements of order $5$. And $n_3 \mid 10$ with $n_3 \equiv 1 \pmod 3$ gives $n_3 = 10$, contributing $10\times2 = 20$ elements of order $3$. That is $44$ elements in a group of $30$. Contradiction.

Order $pq$, $p < q$ primes. $n_q \mid p$ and $n_q \equiv 1 \pmod q$; since $p < q$ the only option is $n_q = 1$, so the Sylow $q$-subgroup is normal and $G$ is a semidirect product.

• If $p \nmid q - 1$, then $n_p = 1$ too and $G \cong C_{pq}$ is cyclic.
• If $p \mid q-1$, there is exactly one non-abelian group as well.

So order $15$ ($3 \nmid 4$) is forced cyclic, and order $21$ ($3 \mid 6$) admits a non-abelian group.

No simple group of order less than $60$ except the primes. $A_5$ at $60$ is the first, which is why $60$ appears so often.

Groups of small order

Order $8$ has five groups: three abelian ($C_8$, $C_4\times C_2$, $C_2^3$) and two non-abelian ($D_4$, $Q_8$). Counting elements of order $2$ separates all five — $1, 3, 7, 5, 1$ respectively. $D_4$ and $Q_8$ have the same order, the same centre size, the same abelianisation, and differ exactly there: $Q_8$ has a unique element of order $2$, namely $-1$.

Useful general facts:

• Order $p$: cyclic. Order $p^2$: abelian, so $C_{p^2}$ or $C_p^2$.
• Order $2p$: $C_{2p}$ or $D_p$.
• A group whose every non-identity element has order $2$ is elementary abelian, hence $C_2^k$.
• Index of the smallest prime: a subgroup of index $p$, the smallest prime dividing $|G|$, is normal. This generalises the index-$2$ fact and closes several small-order arguments.

Where intuition breaks

"$n_p \equiv 1 \pmod p$, so $n_p = 1$." Both conditions are needed, and often several values survive. Order $12$ has $n_3 \in \{1,4\}$ and both occur.

"$|G| = pq$, so $G$ is cyclic." Only when $p \nmid q-1$. Order $21$ has a non-abelian group; order $15$ does not. Checking that divisibility is the question.

"A normal Sylow subgroup makes $G$ abelian." It makes $G$ a semidirect product. $D_4$ has a normal Sylow $2$-subgroup — itself — and is not abelian.

"Two groups with the same order, centre and abelianisation are isomorphic." $D_4$ and $Q_8$ agree on all three. Counting elements of each order is the invariant that separates them.

"$G$ has a subgroup of index $p$, so it is normal." Only when $p$ is the smallest prime dividing $|G|$. In $A_4$, index $3$ subgroups exist and $3$ is not smallest.

"Order $60$ means simple." $A_5$ is simple; $C_{60}$ is not. Order does not determine simplicity, and the correct statement is that $60$ is the smallest order admitting a non-abelian simple group.

"If $n_p > 1$ for every $p$, then $G$ is simple." It only means the Sylow subgroups are not normal. Normal subgroups can arise elsewhere — from the kernel of the action on Sylow subgroups, for instance.

The exam's angle

1. Factor $|G|$ and tabulate the possible $n_p$ using both conditions at once.
2. If any $n_p$ is forced to $1$, you are done — not simple, and you have a normal subgroup to build a semidirect product with.
3. Otherwise count elements of prime order. $n_p(p-1)$ each; exceeding $|G|$ is the contradiction.
4. Then try the action on Sylow subgroups and compare $|G|$ with $n_p!$.
5. For classification, count elements of each order. It is the cheapest complete invariant for small groups and it is what separates $D_4$ from $Q_8$.

The night before

• $n_p \equiv 1 \pmod p$ and $n_p \mid m$. $n_p = 1 \iff$ normal Sylow $p$-subgroup.
• Non-simplicity: force $n_p = 1$; else count $n_p(p-1)$ elements; else act on the Sylow subgroups and use $G \to S_{n_p}$.
• Order $30$: $n_5 = 6$ and $n_3 = 10$ need $24 + 20 = 44 > 30$ elements. Not simple.
• Order $pq$, $p<q$: $n_q = 1$ always. Cyclic $\iff p \nmid q-1$. Order $15$ cyclic, order $21$ not necessarily.
• Smallest non-abelian simple group is $A_5$, order $60$. Nothing simple below except primes.
• Order $8$: $C_8$, $C_4\times C_2$, $C_2^3$, $D_4$, $Q_8$. Elements of order $2$: $1,3,7,5,1$.
• A subgroup of index the smallest prime dividing $|G|$ is normal.