AAR Some structure and classification theorems

Section 2.4 Some structure and classification theorems

Here we state with few proofs some structure theorems which advance the goal of classifying finite groups. We also include a few examples. We begin with a major result whose proof relies on group actions.

We have defined a \(p\)-group as any group whose order is a power of a prime \(p.\) Suppose that \(G\) is a finite group of order \(n\) and \(p\) is prime dividing \(n.\) Write \(n = p^k n_0\) where \(p \nmid n_0\text{,}\) so \(k\) is the largest exponent so that \(p^k \mid n.\) A \(p\)-subgroup \(H\) of \(G\) is simply any subgroup which is a \(p\)-group. A Sylow \(p\)-subgroup of \(G\) is a subgroup whose order is the largest power of \(p\) dividing the order of the group, in this case \(p^k\text{.}\) We are about to state a result showing that such subgroups always exist.

Theorem 2.4.1. Sylow theorems.

Let \(G\) be a group of order \(n = p^k n_0\) with \(p\) a prime and \(p \nmid n_0\text{.}\) Then

There exist subgroups of \(G\) of all orders \(p^\ell\) with \(1 \le \ell\le k\text{.}\) In particular, Sylow \(p\)-subgroups exist for all primes \(p\) dividing \(|G|\text{.}\) Every \(p\)-subgroup of \(G\) is contained in a Sylow \(p\)-subgroup of \(G\text{.}\)
For a fixed prime \(p\text{,}\) if \(P\) and \(Q\) are two Sylow \(p\)-subgroups of \(G\text{,}\) they are conjugate, i.e., there exists \(g \in G\) with \(P = gQg^{-1}\text{.}\)
Let \(n_p\) equal the number of Sylow \(p\)-subgroups of \(G\text{.}\) Then \(n_p \equiv 1 \pmod p\) and \(n_p \mid n_0 = |G|/p^k\text{.}\)}

Underlying concepts.

Let \(p\) be a prime dividing \(|G|\text{,}\) and let \(X\) be the set of all Sylow \(p\)-subgroups in \(G.\) The first Sylow theorem says that \(X\) is non-empty. Note that \(G\) acts on \(X\) by conjugation: if \(P\in X\text{,}\) then since \(x\mapsto gxg^{-1}\) is an (inner) automorphism, \(gPg^{-1}\) is a subgroup of \(G\) having the same order as \(P\text{,}\) so is again an element of \(X.\)

Now that we have a group action, we can talk about orbits and stabilizers. So let \(P \in X\) be a Sylow \(p\)-subgroup. The second Sylow theorem says that the orbit \(G\cdot P = X\text{;}\) \(G\) is said to act transitively on \(X\text{.}\)

But now \(n_p\) is the number of Sylow \(p\)-subgroups in \(G\text{,}\) but that is simply the size of \(X.\) Thus

\begin{equation*} n_p = |X| = |G\cdot P| = [G:N_G(P)]\text{,} \end{equation*}

where the last equality comes from the orbit-stabilizer theorem. Now we always have the inclusions

\begin{equation*} P \le N_G(P) \le G \end{equation*}

and

\begin{equation*} n_0 = \frac{|G|}{|P|} = [G:P]= [G:N_G(P)][N_G(P):P]= n_p [N_G(P):P]\text{.} \end{equation*}

In particular \(n_p \mid n_0\text{,}\) part of the third Sylow theorem.

The normal subgroups of a group provide insight into its structure. A group \(G\) whose only normal subgroups are \(G\) and \(\{e\}\) is called a simple group and are fundamental to the so-called Hölder program. For groups which are not simple, their normal subgroups often lead to their characterization as a product of smaller groups, which we investigate shortly. A corollary of the Sylow theorems provides a simple way to determine if a Sylow \(p\)-subgroup is normal.

Corollary 2.4.2.

In the notation of the Sylow theorem, \(n_p=1\) if and only if the Sylow \(p\)-subgroup is normal.

Proof.

Suppose that \(n_p=1\) and \(P\) is given Sylow \(p\)-subgroup. For any \(g\in G,\) \(gPg^{-1}\) is also a Sylow \(p\)-subgroup, and since there is only one, \(P = gPg^{-1}\) for any \(g\in G,\) so \(P\normal G.\)

Conversely, suppose that \(P\) is a normal Sylow \(p\)-subgroup, and let \(Q\) be any Sylow \(p\)-subgroup. By the second Sylow theorem, \(Q = gPg^{-1}\) for some \(g\in G.\) But since \(P\) is normal, \(Q=P\text{,}\) hence \(n_p=1.\)

Recall that given two groups \(G_1\) and \(G_2\text{,}\) we can make their Cartesian product, \(G_1\times G_2\text{,}\) of ordered pairs into a group under component-wise operations. What we would like is to characterize when a given group is isomorphic to a direct product of groups.

Proposition 2.4.3.

Let \(G\) be a group and \(H\text{,}\)\(K\) subgroups. Suppose that

\(H\) and \(K\) are both normal subgroups.
\(\displaystyle H\cap K = \{e\}\)
\(\displaystyle G = HK (= KH)\)

Then the map \(H\times K \to HK=G\) given by \((h,k) \mapsto hk\) is an isomorphism, and \(G\) is called the (internal) direct product of the subgroups \(H\) and \(K\text{.}\)

Proof.

Let \(\varphi: H\times K \to G\) be defined by \(\varphi((h,k)) = hk.\) The map \(\varphi\) is surjective by the third assumption, and it is one-to-one by the second assumption:

\begin{equation*} hk = h'k' \iff (h')^{-1}h = k'k^{-1}, \end{equation*}

but since \(H\cap K = \{e\}\text{,}\) \((h')^{-1}h = k'k^{-1} =e\text{,}\) thus \((h,k) = (h', k').\)

Showing that \(\varphi\) is a homomorphism is we need to exercise a bit of care.

\begin{align*} \varphi( (h_1, k_1)(h_2,k_2))\amp = \varphi((h_1h_2, k_1k_2)) = h_1h_2k_1k_2\\ \varphi((h_1, k_1))\varphi((h_2, k_2)) \amp = h_1k_1h_2k_2\text{.} \end{align*}

So we need to show that

\begin{equation*} h_1h_2k_1k_2 = h_1k_1h_2k_2 \iff h_2k_1 = k_1h_2 \end{equation*}

for any \(h_i \in H\) and \(k_i \in K.\) While in general \(HK=KH\) being a subgroup of \(G\) does not imply that the elements commute, but when both subgroups are normal (and have trivial intersection) we gain some added power:

\begin{equation*} hk = kh \iff h^{-1}k^{-1}hk = e, \end{equation*}

but using the normality of each subgroup, we see

\begin{align*} h^{-1}k^{-1}hk \amp = (h^{-1}k^{-1}h) k \in K \\ h^{-1}k^{-1}hk \amp = h^{-1}(k^{-1}h k) \in H \end{align*}

and since

\begin{equation*} H\cap K = \{e\}\text{,} \end{equation*}

the elements commute, and our map \(\varphi\) is a homomorphism.

Before giving an example, we state an important, but simple result which can be viewed as one version of the Chinese Remainder Theorem, though we give a direct proof.

Proposition 2.4.4.

Let \(Z_n\) denote a cyclic group of order \(n,\) so \(Z_n \cong \Z/n\Z\text{.}\)

\begin{equation*} Z_m \times Z_n \cong Z_{mn} \text{ iff } \gcd(m,n) =1. \end{equation*}

Proof.

Let \(d=\gcd(m,n)\text{,}\) and note that \(\ds \frac{mn}{d} = m \frac{n}{d} = n\frac{m}{d}\) is a product of integers. It follows that every element of \(Z_m\times Z_n\) has exponent \(mn/d\text{.}\) So if \(d > 1,\) there is no element of order \(mn\) in \(Z_m\times Z_n\text{,}\) so the group is not cyclic.

Conversely, suppose that \(d=1.\) Let \(x\in Z_m\) have order \(m\) and \(y\in Z_n\) have order \(n,\) and put \(z=(x,y) \in Z_m\times Z_n.\) Since

\begin{equation*} z = (x,y)= (x,e)(e,y) = (e,y)(x,e) \end{equation*}

it is easy to see that \(z\) has exponent \(mn.\) We want to show that the order of \(z\) is \(mn.\)

So suppose that \(\ell\) is any exponent for \(z.\) So

\begin{equation*} z^\ell =(x^\ell, y^\ell) = (e,e). \end{equation*}

Since \(m\) is the order of \(x,\) we know that \(m\mid \ell,\) and since \(n\) is the order of \(y,\) we know that \(n \mid \ell.\) But \(d=\gcd(m,n) = 1 \) which implies that \(mn \mid \ell.\) Thus \(\ell = mn\) is the smallest exponent, hence the order.

Thus \(Z_m\times Z_n\) is cyclic of order \(mn\text{,}\) so

\begin{equation*} Z_m\times Z_n \cong Z_{mn} \end{equation*}

as there is a unique cyclic group of any given order (up to isomorphism).

Example 2.4.5.

Let \(p \lt q\) be primes with \(p \nmid (q-1)\text{.}\) Then every group of order \(pq\) is cyclic.

Solution.

We first apply the Sylow theorems to \(G.\) Let \(H_p\) and \(H_q\) be (respectively) Sylow \(p\) and \(q\)-subgroups of \(G\text{.}\) Because they have prime order, we know that \(H_p \cong Z_p\) and \(H_q\cong Z_q.\) We want to know that \(G\) is the direct product of \(H_p\) and \(H_q.\) That they have trivial intersection is immediate from Lagrange since they have relatively prime orders.

Given their trivial intersection, Proposition 2.1.3 tells us that \(|H_pH_q| = pq\text{,}\) so necessarily \(G=H_pH_q.\) All the remains is for us to show that each of the Sylow subgroups is normal.

Proposition 2.3.14 tells us that since \([G:H_q]=p\) is the smallest prime dividing the order of \(G\text{,}\) that it must be normal, though we give an independent proof using the Sylow theorems.

In the notation of the Sylow theorems, the subgroups will both be normal iff \(n_p = n_q = 1.\) By the Sylow theorems, we know that

\begin{equation*} n_p \equiv 1\pmod p \text{ and } n_p \mid q,\text{ while } n_q\equiv 1 \pmod q \text{ and } n_q \mid p. \end{equation*}

Since \(n_q\mid p,\) \(n_q = 1\) or \(p.\) But if \(n_q = p,\) then \(n_q = p \equiv 1 \pmod q\) which says that \(q \mid (p-1).\) But \(p \lt q\) by assumption, so that is impossible. Thus \(n_q=1\) implying \(H_q\) is a normal subgroup.

Similarly, \(n_p = 1\) or \(q.\) If \(n_p = q,\) then \(n_p = q \equiv 1\pmod p,\) implying that \(p\mid (q-1),\) contrary to assumption.

Thus both subgroups are normal, have trivial intersection, and their product is \(G\text{,}\) so by Proposition 2.4.3,

\begin{equation*} G\cong H_p\times H_q \cong Z_p \times Z_q \cong Z_{pq} \end{equation*}

the last isomorphism by Proposition 2.4.4.

Remark 2.4.6.

The condition that \(p \nmid (q-1)\) was absolutely critical in the previous example. Consider groups of order \(6 = 2\cdot 3\text{.}\) Since \(2\mid (3-1)\) the argument we gave does not show the Sylow 2-subgroup is normal. Indeed it need not be.

If \(G=S_3\text{,}\) the symmetric group, we know that the Sylow group \(H_3\) is normal (generated by either 3-cycle), however there are 3 Sylow 2-subgroups, each generated by a different transposition, so \(n_2=3\text{,}\) and Sylow 2-subgroups are not normal.

Moreover, it is clear that \(S_3 \not\cong H_2\times H_3\) since the later is abelian, while \(S_3\) is not. Of course if the group had been \(G=\Z/6\Z\text{,}\) both Sylow subgroups would be normal, and \(G\cong H_2\times H_3\text{.}\)

Finally, this is not an isolated situation. After all for any odd prime \(q\text{,}\) \(2\mid (q-1)\text{,}\) so any group of order \(2q\) can have this issue. Indeed, we know the problem will arise, since there are dihedral groups (non-abelian of order \(2n\)) for every \(n \ge 3.\)

Remark 2.4.7.

It is also worth noting that when a group \(G\) has subgroups \(H,K\) with \(G=HK\text{,}\) \(H\cap K = \{e\}\) and only one of \(H\) or \(K\) is normal, there is still something that can be said, namely that \(G\) is a semi-direct product of \(H\) and \(K\text{.}\) The structure is more complicated since that map \(\varphi: H\times K \to G\) given by \((h,k)\mapsto hk\) is not a homomorphism.

One of the remarkably pretty results from group theory is the classification of finite abelian groups. While relatively easy to state, the proof is rather long. You will find direct proofs in textbooks focused on just on groups, or more general proofs which apply to finitely generated modules over PIDs of which finite abelian groups are a special case.

We begin with an intermediate result which we shall use to give the full result.

Theorem 2.4.8.

Let \(G\) be a finite abelian group whose order \(n\) has prime factorization \(n = p_1^{e_1} \cdots p_r^{e_r}.\) Let \(H_{i}\) be a Sylow \(p_i\)-subgroup of \(G\text{.}\) Then

\begin{equation*} G \cong H_{1}\times \cdots \times H_{r}\text{.} \end{equation*}

Proof.

The proof is by induction on \(r.\) First note that all subgroups of \(G\) are normal since \(G\) is abelian. If \(r=1\) there is nothing to prove, and the case of \(r=2\) is a direct application of Proposition 2.1.3 and Proposition 2.4.3.

Now consider \(r=3\text{.}\) By Proposition 2.4.3, \(H:= H_{1}H_{2} \cong H_{1}\times H_{2}.\) Moreover \(H\) is normal since \(G\) is abelian, and \(H\cap H_{3} = \{e\}\) by Lagrange. Again by Proposition 2.1.3 and Proposition 2.4.3, we have

\begin{equation*} G=HH_{3} \cong H\times H_3\cong H_1\times H_2\times H_3. \end{equation*}

Now assume that \(r\ge 4\text{,}\) and as above we have constructed

\begin{equation*} H:= H_1\cdots H_{r-1} \cong H_1\times \cdots \times H_{r-1}\text{.} \end{equation*}

The same arguments now show that \(G = HH_r \cong H\times H_r\) which finishes the proof.

Having reduced the structure of a finite abelian group to a direct product of its Sylow \(p\)-subgroups, we now characterize all the isomorphism types of an abelian \(p\)-group.

Theorem 2.4.9.

Let \(G\) be a finite abelian group of order \(p^n\) for some prime \(p\) and \(n \ge 1\text{.}\) Then \(G \cong Z_{p^{a_1}}\times Z_{p^{a_2}} \times \cdots \times Z_{p^{a_r}}\) with \(a_1\ge a_2\ge \cdots \ge a_r\ge 1\) and \(\sum_{i=1}^r a_i = n\text{.}\)

Moreover, if \(H \cong Z_{p^{b_1}}\times Z_{p^{b_2}} \times \cdots \times Z_{p^{b_s}}\) with \(b_1\ge b_2\ge \cdots \ge b_s\ge 1\) and \(\sum_{i=1}^s b_i = n\text{,}\) then \(G \cong H\) iff \(r=s\) and \(a_i = b_i\) for all \(1\le i\le r\text{.}\) The powers, \(p^{a_i}\text{,}\) are called the elementary divisors of \(G.\)

Remark 2.4.10.

The integers \(a_1\ge a_2\ge \cdots \ge a_r\ge 1\) with \(\sum_{i=1}^r a_i = n\) is said to form a partition of \(n\text{.}\)

One defines the partition function, \(p(n)\text{,}\) which counts the number of partitions of the positive integer \(n\text{.}\) The first few values are easy to compute:

\begin{equation*} p(0) := 1; \quad p(1) = 1; \quad p(2) = 2;\quad p(5)=7; \quad p(10) = 42,\dots \text{.} \end{equation*}

On the other hand, larger values can be more challenging:

\begin{align*} p(20)\amp = 627\\ p(40)\amp =37338\\ p(100)\amp = 190,569,292\\ p(200)\amp = 3,972,999,029,388 \end{align*}

There are many theorems about the partition function including asymptotics:

\begin{equation*} p(n) \sim \frac{e^{\kappa \sqrt n}}{4n\sqrt 3} \end{equation*}

as \(n\to \infty\) where \(\kappa = \pi\sqrt{2/3}\text{,}\) or generating functions:

\begin{equation*} \prod_{m=1}^\infty \left(\frac{1}{1-x^m}\right) = \sum_{n=0}^\infty p(n) x^n\text{.} \end{equation*}

There are also recurrence formulas which produce the exact numbers listed above.

Example 2.4.11.

Up to isomorphism find all abelian groups of order \(p^5\text{,}\) that is find a set of representatives of all the isomorphism classes of abelian groups of order \(p^5\text{.}\)

Solution.

We begin by listing the partitions of 5:

\begin{equation*} 5 = 4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1 \end{equation*}

so \(p(5)=7\text{,}\) so there will be 7 isomorphism classes. The integers in the partition correspond to the elementary divisors \(p^a\) in the decomposition.

Thus \(G\) is isomorphic to precisely one of the following abelian groups:

\begin{align*} \amp Z_{p^5}\\ \amp Z_{p^4}\times Z_p \\ \amp Z_{p^3}\times Z_{p^2}\\ \amp Z_{p^3}\times Z_p\times Z_p\\ \amp Z_{p^2}\times Z_{p^2}\times Z_p \\ \amp Z_{p^2}\times Z_p\times Z_p\times Z_p\\ \amp Z_p\times Z_p\times Z_p\times Z_p\times Z_p\text{.} \end{align*}

Now we would like to combine Theorem 2.4.8 and Theorem 2.4.9 into one theorem, which characterizes finite abelian groups by their invariant factors.

Theorem 2.4.12.

Every finite abelian group is isomorphic to exactly one group of the form \(Z_{n_1}\times Z_{n_2}\times \cdots \times Z_{n_r}\) where and \(n_1\mid n_2 \mid \cdots \mid n_r\ge 2\text{.}\) It follows that \(|G| = n_1n_2\cdots n_r.\) The \(n_i\) are called the invariant factors of \(G.\)

(Main Idea).

We know that every finite abelian group is a direct product of its Sylow \(p\)-subgroups, each one of which has a decomposition in terms of elementary divisors. To combine these products we use the Chinese Remainder theorem for groups joining the largest powers of elementary divisors for each of the primes into the first invariant factor, then the second largest into the second invariant factors, and so on.

An example translating between the two types should make things clear.

Example 2.4.13.

To go from the invariant factor decomposition to the product of Sylow subgroups:

\begin{align*} Z_{20}\times Z_{300} \amp \cong (Z_4\times Z_5)\times (Z_4\times Z_3\times Z_{25})\\ \amp \cong (Z_4\times Z_4)\times Z_3\times (Z_5\times Z_{25})\\ \amp \cong H_2\times H_3\times H_5 \text{ (Sylow subgroups) } \end{align*}

Now we go from Sylow/elementary divisor to invariant factor decomposition collecting the largest powers of elementary divisors, then second largest and so on.

\begin{align*} (Z_p\times Z_{p^2}\times Z_{p^3}) \times (Z_q\times Z_{q^5}) \amp \cong (Z_p)\times (Z_{p^2}\times Z_q)\times (Z_{p^3}\times Z_{q^5})\\ \amp \cong Z_p\times Z_{p^2q}\times Z_{p^3q^5} = Z_{n_1}\times Z_{n_2}\times Z_{n_3}. \end{align*}

Remark 2.4.14.

While the Sylow/elementary divisor method may seem more natural, and is certainly the easier way to list all the isomorphism classes of an abelian groups of a given order, the invariant factor decomposition reveals deeper structure of the group. For example, while we know that if

\begin{equation*} G \cong Z_{n_1}\times Z_{n_2}\times \cdots \times Z_{n_r} \end{equation*}

with \(n_1\mid n_2 \mid \cdots \mid n_r\ge 2\text{,}\) then the group has order \(n = n_1\cdots n_r\) and that every element in the group has order dividing \(n.\) The invariant factor decomposition tells us more, namely that the largest order of an element in the group is \(n_r.\)

We do one last example listing the isomorphism classes in both elementary divisor and invariant factor forms

Example 2.4.15.

Up to isomorphism, classify all abelian groups of order \(6125 = 5^3\cdot 7^2\text{.}\)

Solution.

First we decompose \(G\) into a product of its Sylow subgroups: \(G \cong H_5\times H_7\) with \(|H_5| = 5^3\) and \(|H_7| = 7^2\text{.}\) Next for each Sylow subgroup we need to compute its possible elementary divisors, and to do so, we need to compute the partitions of 2 and 3:

\begin{align*} \text{Partitions of 3: }\amp 3=2+1=1+1+1\\ \text{Partitions of 2: }\amp 2 = 1+1 \end{align*}

This translates to the following possibilities for the elementary divisor decomposition of each Sylow subgroup:

\begin{align*} H_5 \cong Z_{5^3} \amp \text{ or } Z_{5^2}\times Z_5, \text{ or }Z_5\times Z_5\times Z_5\\ H_7 \cong Z_{7^2} \amp \text{ or } Z_7\times Z_7\text{.} \end{align*}

Finally we assemble all the data we have computed to arrive at all the possible isomorphism classes both in terms of elementary divisors and invariant factors.

\begin{align*} \amp \Z_{5^3}\times \Z_{7^2}\amp \amp \Z_{6125}\\ \amp \Z_{5^3}\times \Z_7\times \Z_7 \amp \amp \Z_7 \times \Z_{875}\\ \amp \Z_5\times \Z_{5^2}\times \Z_{7^2} \amp \amp \Z_5 \times \Z_{1225}\\ \amp \Z_5\times \Z_{5^2}\times\Z_7\times \Z_7\amp \amp \Z_{35}\times \Z_{175}\\ \amp \Z_5\times \Z_5\times \Z_5\times \Z_{7^2}\amp \amp \Z_5\times \Z_5 \times \Z_{245}\\ \amp \Z_5\times \Z_5\times \Z_5\times \Z_7\times \Z_7\amp \amp \Z_5 \times \Z_{35}\times\Z_{35}\\ \amp \mbox{elementary divisors}\amp \amp \mbox{invariant factors} \end{align*}

Checkpoint 2.4.16.

The group \(G=Z_{25}\times Z_{245}\) is also an abelian group of order 6125, but does not appear in the lists given in the example above. To which groups in the above list is it isomorphic?

Hint.

Decompose the given direct product into elementary divisors

Answer.

By Proposition 2.4.4 \(Z_{245} \cong Z_{5}\times Z_{49}\text{,}\) so that

\begin{equation*} G=Z_{25}\times Z_{245} \cong (Z_5\times Z_{25})\times Z_{49} \cong Z_{5}\times Z_{1225}. \end{equation*}