The second and third Sylow Theorems

We have seen that $p$-Sylow subgroup's exist. We now wish to show that any two $p$-Sylow subgroup's are conjugate. This is the content of the second Sylow Theorem.

Theorem 10.2.1 (Sylow II)   Let $G$ be a finite group and $p$ a prime such that $p\vert\, \vert G\vert$. Then all $p$-Sylow subgroup's of $G$ are conjugate. In other words, if $P_1$ and $P_2$ are any two $p$-Sylow subgroups of $G$ then there exists an $a \in G$ such that $P_1 = aP_2a^{-1}$.

Proof: Let $P_1$ and $P_2$ be two $p$-Sylow subgroup's of $G$, where $\vert P_1\vert = \vert P_2\vert = p^a$. We now decompose $G$ into double cosets with respect to $P_1$ and $P_2$ (see Section 8.2). Thus

$$G = P_1a_1P_2 \cup P_1a_2P_2\cup ... \cup P_1a_tP_2 \ \ \ \ {\rm (disjoint)}$$

and from equation (8.6)

$$\vert G\vert=\sum_{j=1}^t \frac{\vert P_1\vert\vert P_2\vert}{d_j},$$

where $d_j=\vert P_2\cap a_j^{-1}P_1a_j\vert$. Hence if $\vert G\vert = p^an'$, where $p\not\vert n''$, we have

$$p^a n'=\frac{p^ap^a}{d_1}+...+\frac{p^ap^a}{d_t},$$

or

$$n'=\frac{p^a}{d_1}+...+\frac{p^a}{d_t}.$$ (10.2)

Now $P_2\cap a_j^{-1}P_1a_j\leq P_2$, therefore $d_j\vert p^a$, so $d_j = p^b$, where $0 <b\leq a$. Thus each term on the right hand side of (10.2) is either $1$ or a power of $p$. Since $p\not\vert n'$, it follows that at least one term on the right hand side of (10.2) must equal $1$, say the $k^{th}$ term. This means $d_k=p^a$ so $\vert P_2\cap a_k^{-1}P_1a_k\vert=p^a$. Whence $P_2=P_2\cap a_k^{-1}P_1a_k\subset a_k^{-1}P_1a_k$. Since both $P_2$ and $a_k^{-1}P_1a_k$ have the same (finite) order and since one is contained in the other, they must be equal: $P_2=a_k^{-1}P_1a_k$. Hence the two $p$-Sylow subgroup's $P_1$ and $P_2$ are conjugate. $\Box$

We come now to the last of the three Sylow theorems. This one gives us information concerning the number of $p$-Sylow subgroup's. Let $n_p(G)$ designate the number of $p$-Sylow subgroup's of $G$

Theorem 10.2.2 (Sylow III)   Let $G$ be a finite group and $p$ a prime such that $p\vert\, \vert G\vert$. We have

$$n_p(G) \equiv 1 ({\rm mod}\, p),$$

i.e., $n_p(G)$ is of the form $1 + pv$ where $v \in\mathbb{Z}$. (We may write $n_p$ instead of $n_p(G)$ if it is clear which group $G$ we are working in.)

Proof: Let $P$ be a $p$-Sylow subgroup of $G$. Then by the second Sylow Theorem (Theorem 10.2.1) and by Theorem 6.1.1,

$$n_p(G)=[G:N_G(P)] =\frac{\vert G\vert}{n},$$

where $n = \vert N_G(P)\vert$. (From now on we drop the subscript $G$ and just write $N(P)$ and also just write $n_p$.) Now $P \subset N(P)$ and $P \lhd N(P)$; $\vert P\vert\,\vert n$, i.e., $p^a\vert n$, where $p^a =\vert P\vert$, so

$$n = p^an', \ \ \ \ \ {\rm where}\ gcd (n , p) = 1,$$

since $P$ is a $p$-Sylow subgroup of $G$ and $N(P) \leq G$. We now decompose $G$ into double cosets with respect to $P$ and $N(P)$. Thus

$$G = Pa_1N(P)\cup Pa_2N(P)\cup ... \cup Pa_tN(P),\ \ \ \ {\rm (disjoint)},$$

and using the numerical relation (8.6) yields

$$\vert G\vert=\frac{p^an}{d_1}+...+\frac{p^an}{d_t},$$ (10.3)

where $d_j=\vert N(P)\cap a_j^{-1} P a_j\vert$. Now the identity, $e$, of $G$ belongs to some double coset, and we may assume that, say $a_1 = e$. In this case, we have

$$Pa_1N(P) = PeN(P) = PN(P) = N(P),$$

hence, the first term on the right hand side of (10.3) becomes $\frac{p^an}{d_1}=n$. Now cancelling $n$ on both sides of (10.3) and recalling that $\vert G\vert = n[G : N(P)] = n\cdot n_p$, gives

$$n_p=1+\frac{p^a}{d_2}+...+\frac{p^a}{d_t}.$$ (10.4)

Next we observe that $N(P)\cap a_j^{-1} P a_j\subset a_j^{-1} P a_j$ and since $\vert a_j^{-1} P a_j\vert=p^a$, we must have $\frac{p^a}{d_j}=p^{b_j}$, where $0\leq b_j\leq a$ and $j = 2, ..., t$. If we can show that each such $b_j>0$, ( $2 \leq j \leq t$), then it will follow from (10.4) that $n_p$ is indeed, of the form $1 + pv$. Hence suppose on the contrary that for some $j$, say $j = s$ ( $2 \leq s \leq t$), that $p^a= ds$. But

$$N(P)\cap a_s^{-1} P a_s\subset a_s^{-1} P a_s$$

and

$$\vert N(P)\cap a_s^{-1} P a_s\vert=d_s=p^a,$$

so $N(P)\cap a_s^{-1} P a_s= a_s^{-1} P a_s$. But also $N(P)\cap a_s^{-1} P a_s\subset N(P)$. Thus both $P$ and $a_s^{-1} P a_s$ are $p$-Sylow subgroup's of $N(P)$. Hence by the second Sylow Theorem (Theorem 10.2.1, with $N(P)$ now playing the role of $G$ in that theorem), they must be conjugate in $N(P)$. But $P \lhd N(P)$, so we must have that $a_s^{-1} P a_s=P$. This means $a_s\in N(P)$, which implies that

$$Pa_sN(P) = PN(P) = N(P),$$

which contradicts the disjointness of the decomposition. Hence for $j = 2, ..., t$ every $b_j>0$ and this as already observed, completes the proof. $\Box$

The third Sylow Theorem tells us that $n_p(G)$, the number of $p$-Sylow subgroup's, is of the form $1 + pv$. However, we know as was used in the proof that $n_p(G) = [G : N_G(P)]$ from Theorem 6.1.1. Thus $n_p(G)\vert\, \vert G\vert$. This proves the following fact.

Corollary 10.2.3   Same hypothesis as in Theorem 10.2.2. Then $n_p(G)\vert\, \vert G\vert$.

As pointed out above this is really a corollary of the proof of the third Sylow Theorem. The two facts that $n_p(G) \equiv 1 ({\rm mod}\, p)$ and $n_p(G)\vert\, \vert G\vert$ are extremely useful. A few of their applications will be seen in the examples of the next section.

For the final theorems of this section, we turn our attention to prime power groups.

Theorem 10.2.4   Let $G$ be a group of order $p^n$. Then $G$ contains at least one normal subgroup of order $p^m$, for each $m$ such that $0 \leq m\leq n$.

Proof: The theorem is trivial for $n = 1$. We claim it is also true for $n =2$. Indeed, by Theorem 6.3.4, any group of order $p^2$ is abelian. This together with Theorem 10.1.2 establishes the claim.

We proceed now by induction on $n$. Thus we assume the theorem is true for all groups $G$ of order $p^k$ where $1 \leq k < n$, where $n > 2$. Let $G$ be a group of order $p^n$. Also let $N$ be a normal subgroup of order $p$. $N$ exists since $Z(G)$ is non-trivial (by Theorem 4.3.5) and is, of course, abelian. Thus again by Theorem 10.1.2, $Z(G)$ contains an element, say $z$, of order $p$. We can take $N = \langle z\rangle$ and so $N$ is a normal subgroup of $G$ of order $p$, since every subgroup of the center is normal in $G$ (WHY?). But then $G/N$ is of order $p^{n- 1}$, and therefore, contains (by the induction hypothesis) normal subgroups of orders $p^{m- 1}$, for $0\leq m-1 \leq n-1$. These groups are of the form $H/N$, where $H \lhd G$ contains $N$ (see the Corollary 8.3.3) and is of order $p^m$, $1 \leq m \leq n$, because $\vert H\vert=\vert N\vert[H:N] = \vert N\vert\cdot \vert N/H\vert$. $\Box$

We next introduce the concept of a $p$-group which generalizes the idea of groups of prime power order.

Definition 10.2.5   A $p$-group $G$ (where $p$ is any prime) is a group in which the order of every element is some power of $p$.

We observe that a $p$-group does not even have to be finite. But in the finite case, we have the following result.

Theorem 10.2.6   $G$ is a finite $p$-group if and only if $\vert G\vert = p^n$ for some $n \in \mathbb{N}$.

Proof: We leave the ``if'' part as an exercise, i.e., if $\vert G\vert = p^n$, then $G$ is a $p$-group. (See exercise 1 for this section.) Conversely, suppose that $G$ is a finite $p$-group. We would like to show that $\vert G\vert = p^n$. If there were a prime $q \not= p$ such that $q\vert\, \vert G\vert$, then by Cauchy's Theorem (Theorem 10.1.4) $G$ would contain at least one element of order $q$. This contradicts the fact that every element of $G$ has order a power of $p$, i.e., that $G$ is a $p$-group. Thus $\vert G\vert = p^n$. $\Box$