## 10 Jan Applications of Sylow Theorems

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 \$\$Hleq N_G(K)\$\$, then \$\$HK\$\$ is a subgroup of \$\$G\$\$. In particular, if \$\$K\$\$ is normal is \$\$G\$\$ then \$\$HKleq G\$\$ for any \$\$Hleq G\$\$.

Proof: We prove \$\$HK=KH\$\$. We let \$\$hin H, k in K\$\$.
By assumption, \$\$hkh^{-1}in K\$\$, hence \$\$hk=(hkh^{-1})h in KH\$\$.
This proves that \$\$HKleq KH\$\$.
Similarly, \$\$kh=h(h^{-1}kh)in HK\$\$, proving the reverse.
We know that if \$\$H\$\$ and \$\$K\$\$ are subgroups of a group then \$\$HK\$\$ is a subgroup if and if only \$\$HK=KH\$\$.
This proves the result.

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

• If \$\$pmid(q-1)\$\$, there exists a non-abelian group of order \$\$pq\$\$.
• Any two non-abelian groups of order \$\$pq\$\$ 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 \$\$mid N_{S_p}(P) mid = 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, \$\$mid N_{S_q}(Q) mid =q(q-1)\$\$.
Since \$\$p mid 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 \$\$PQ\$\$ is a group of order \$\$pq\$\$.
Since \$\$C_{S_q}(Q)=Q\$\$ so, \$\$PQ\$\$ is non-abelian.
This proves the first result.

Let \$\$G\$\$ be any group of order \$\$pq\$\$, let \$\$Pin Syl_p(G)\$\$ and let \$\$Qin Syl_q(G)\$\$.
We have \$\$pmid q-1\$\$ and let \$\$p=<y>\$\$.
Since \$\$Aut(Q)\$\$ is cyclic, it contains a unique subgroup of order \$\$p\$\$, say \$\$<gamma>\$\$, and any homomorphism \$\$phi : P rightarrow Aut(Q)\$\$ must map \$\$y\$\$ to a power of \$\$gamma\$\$.
There are therefore \$\$p\$\$ homomorphisms \$\$phi_i:P rightarrow Aut(Q)\$\$ given by \$\$phi_i(y)=gamma^i, 0leq i leq p-1\$\$.
Now each \$\$phi_i\$\$ for \$\$i\$\$ not equal to \$\$0\$\$ give rise to a non-abelian group \$\$G_i\$\$, of order \$\$pq\$\$.
It is straightforward to check that these groups are all isomorphic because for each \$\$phi_i,~i>0\$\$, there is some generator \$\$y_i\$\$ of \$\$P\$\$ such that \$\$phi_i(y_i)=gamma\$\$.
Thus up to a choice for the arbitary generator of \$\$P\$\$, these semi-direct products are all the same.