Applications

On the basis of the first Sylow Theorem, Theorem 10.1.3, and Theorem 10.2.4, we see that if $G$ is a finite group and if $p^k\vert\, \vert G\vert$, then $G$ must contain a subgroup of order $p^k$. One can actually show that, as in the case of Sylow $p$-groups, the number of such subgroups is of the form $1 + pt$, but we shall not prove this here.

We shall now consider a number of applications of the Sylow Theorems.

Example 10.3.1   There is no simple group of order $84$. Write $84 = 2^2 \cdot 3 \cdot 7$. If the $7$-Sylow subgroup is not normal, then it has $1 + 7v$ conjugates where $v \geq 1$ and $(1 + 7v)\vert 2^2 \cdot 3 \cdot 7$. Clearly $1 + 7v \not= 7n$, so the only possibilities are

$$1 + 7v = 2, 2^2, 3, 2 \cdot 3, 2^2 \cdot 3,$$

all of which are clearly impossible (none of these are $\equiv 1 ({\rm mod}\ 7))$. Hence the $7$-Sylow subgroup is normal and therefore, any group of order 84 is not simple.

Example 10.3.2   There is no simple group of order $12$. Write $12 = 2^2 3$. If the $3$-Sylow subgroup, which is a cyclic group of order $3$, $C_3$, is not normal, it has $1 + 3v$ conjugates where $v \geq 1$ and $(1 + 3v)\vert 12$. Clearly the only possibility is $v = 1$, in which case $C_3$ has $4$ conjugates. These groups have only the identity in common, and this accounts for $4\cdot (3-1) = 8$ nontrivial elements of $G$. This leaves then $4$ elements of $G$, which must constitute a $2$-Sylow subgroup of order $4$, which of course only has the identity in common with any $3$-Sylow subgroup. Thus this $2$-Sylow subgroup must be normal since it can have no distinct conjugate. If the reader goes back to the table for $A_4$ given in Section 6.1, it will be seen that exactly this situation prevails.

Example 10.3.3   There is only one group of order $15$ (up to isomorphism) the cyclic group. Let $\vert G\vert = 15$. It is clear that both the $5$-Sylow subgroup, $C_5$, and the $3$-Sylow subgroup, $C_3$, are normal (Why?). Since $C_5$ is cyclic of order $5$ and $C_3$ is cyclic of order $3$, $C_3 \cap C_5 = \{e\}$ and $G = C_3C_5$ since by Theorem 4.3.6, $\vert C_3C_5\vert = 15$. Hence $G = C_3 \times C_5= C_{15}$, a cyclic group of order $1$5 by Theorem 9.2.4.

Further applications along these lines are given in the exercises for this section. Deeper structural applications of the Sylow Theorems can be found in the more advanced literature on group theory (see for example [Sc]).

Theorem 10.3.4   Any finite abelian group G is a direct product of its Sylow subgroups.

Proof: Let $\vert G\vert=\prod_{i=1}^n p_i^{a_i}$, where the $p_i$ are distinct primes and $a_i\in \mathbb{N}$. Let $P_i$ be the $p_i$-Sylow subgroup. $P_i$ is unique by Theorem 10.2.1 (the second Sylow Theorem) and by the fact that $G$ is abelian. Of course each $P_i\lhd G$ and $\vert P_i\vert=p_i^{a_i}$. Since $P_i\subset P_1P_2...P_n$ ( $i = 1, 2, ..., n$), we have that $\vert P_1P_2...P_n\vert$ is divisible by $p_i^{a_i}$, and therefore $G = P_1...P_n$. Repeated application of Theorem 4.3.6 shows $\vert P_1P_2\vert=p_1^{a_1}p_2^{a_2}$, $\vert P_1P_2P_3\vert=p_1^{a_1}p_2^{a_2}p_3^{a_3}$, etc. from which it forms immediately that

$$P_i\cap P_1...P_{i-1}P_{i+1}...P_n = \{e\},$$

since the groups being intersected have coprime orders. $\Box$

Thus any finite abelian group is a direct product of its $p$-Sylow subgroup's. There are other finite groups, other than abelian groups, which are the direct products of their Sylow subgroups; such finite groups are called nilpotent. The notion of a nilpotent group can be extended to infinite groups by a consideration of various sequences of subgroups in such a way that for finite groups the notion reduces to the above characterization. However, we shall not go into these matters here.

Suppose finally that $G$ is a finite group and that $G$ is the direct product of its Sylow subgroups, say $\vert G\vert=\prod_{i=1}^n p_i^{a_i}$, and $G = P_1 \times P_2 \times ... \times P_n$, where $P_i$ is the $p_i$-Sylow subgroup of $G$. Let $d\vert\, \vert G\vert$, then $d= \prod_{i=1}^n p_i^{b_i}$, where $0 \leq b_i\leq a_i$ and $1 \leq i \leq n$ (WHY?). Since $P_i$ is a $p_i$-group of order $p_i^{a_i}$, it must contain a normal subgroup $N_i$ of order $p_i^{b_i}$ by Theorem 10.2.4 for every such $b_i$. Moreover, every such $N_i$ must actually be normal in $G$ by Theorem 9.2.3. Let $N = N_1N_2...N_n$. Then $N$ is a subgroup of $G$ since each $N_i \lhd G$ (by repeated application of Proposition 8.3.6). Also $N_i \cap (N_1...N_{i-1}N_{i+1}...N_n) = \{e\}$ since $P_i\cap (P_1...P_{i-1}P_{i+1}...P_n) = \{e\}$. (WHY is this true for the $P$'s?) and every $N_i \subset P_i$. Thus

$$N = N_1 \times N_2 \times ... \times N_n$$

and $\vert N\vert=\prod_{i=1}^n p_i^{b_i}=d$. Therefore we have proven that $G$ possesses a subgroup of order $d$ where $d$ was any positive divisor of $\vert G\vert$. As a matter of fact, $N$ is even a normal subgroup of $G$ (WHY?). Specializing to the case of an abelian group, which we know by Theorem 10.3.4 is a direct product of its Sylow subgroups, we obtain the converse of Lagrange's Theorem for such groups.

Theorem 10.3.5   If $G$ is a finite abelian group of order $n$, then for each $d\vert n$, $G$ has a subgroup of order $d$.