algebra

Applications of Sylow Theorems

Manjil Saikia

Jan 10, 2012 • 2 min read

Sylow Theorem’s is Group Theory are an important mathematical tool. Below we give a simple application of the theorems.

Theorem: If $H$ and $K$ are subgroups of $G$ and $H l e q N_{G} (K)$ , then $H K$ is a subgroup of $G$ . In particular, if $K$ is normal is $G$ then $H K l e q G$ for any $H l e q G$ .

Proof: We prove $H K = K H$ . We let $h i n H, k i n K$ .
By assumption, $h k h^{- 1} i n K$ , hence $h k = (h k h^{- 1}) h i n K H$ .
This proves that $H K l e q K H$ .
Similarly, $k h = h (h^{- 1} k h) i n H K$ , proving the reverse.
We know that if $H$ and $K$ are subgroups of a group then $H K$ is a subgroup if and if only $H K = K H$ .
This proves the result.

Theorem: Let $G$ be a group of order $p q$ , where $p$ and $q$ are primes such that $p < q$ .

If $p m i d (q - 1)$ , there exists a non-abelian group of order $p q$ .
Any two non-abelian groups of order $p q$ are isomorphic.

Proof: We let $Q$ be a Sylow $q$ -subgroup of the symmetric group of degree $q$ , $S_{q}$ . We know that if $p$ is a prime and $P$ is a subgroup of $S_{p}$ of order $p$ , then $m i d N_{S_{p}} (P) m i d = p (p - 1)$ .
We know that every conjugate of $P$ contains exactly $p - 1$ $p$ -cycles and computing the index of $N_{S_{p}} (P)$ in $S_{p}$ we can prove the above result.
So, $m i d N_{S_{q}} (Q) m i d = q (q - 1)$ .
Since $p m i d q - 1$ and by Cauchy’s theorem $N_{S_{q}} (Q)$ has a subgroup $P$ of order $p$ .
Using the previous theorem we can see that $P Q$ is a group of order $p q$ .
Since $C_{S_{q}} (Q) = Q$ so, $P Q$ is non-abelian.
This proves the first result.

Let $G$ be any group of order $p q$ , let $P i n S y l_{p} (G)$ and let $Q i n S y l_{q} (G)$ .
We have $p m i d q - 1$ and let $p =< y >$ .
Since $A u t (Q)$ is cyclic, it contains a unique subgroup of order $p$ , say $< g a m m a >$ , and any homomorphism $p h i : P r i g h t a r r o w A u t (Q)$ must map $y$ to a power of $g a m m a$ .
There are therefore $p$ homomorphisms $p h i_{i} : P r i g h t a r r o w A u t (Q)$ given by $p h i_{i} (y) = g a m m a^{i}, 0 l e q i l e q p - 1$ .
Now each $p h i_{i}$ for $i$ not equal to $0$ give rise to a non-abelian group $G_{i}$ , of order $p q$ .
It is straightforward to check that these groups are all isomorphic because for each $p h i_{i}, i > 0$ , there is some generator $y_{i}$ of $P$ such that $p h i_{i} (y_{i}) = g a m m a$ .
Thus up to a choice for the arbitary generator of $P$ , these semi-direct products are all the same.

Sign up for more like this.