# Thread: Using Sylow to solve the following:

1. ## Using Sylow to solve the following:

Dear All,

I am stuck on a partucular question which asks the following:

Let G be a group of order pq where p and q are distinct primes. Prove that G is soluble:

I beileve that one method is to use Sylow but I am having trouble solving it. Can someone please assist.

Thanks.

2. Hello.

Recall that a group $G$ is solvable if we can find a series of groups $0=G_0\leq G_1\leq \ldots \leq G_n=G$ such that $G_i$ is normal in $G_{i+1}$ and $G_{i+1}/G_i$ is abelian.

Suppose that a group $G$ is a group of order $pq$, where $p$ and $q$ are distinct primes (and without loss of generality, suppose $p. By the first Sylow theorem, there are subgroups of $G$ of order $q$. By the third Sylow theorem, the number of of these subgroups $n_q$ divides $p$ and satisfies $n_q\equiv 1\pmod{q}$. We can conclude that $n_q=1$. Take $P$ to be that subgroup. By the second Sylow theorem, $P$ is normal in $G$. We can write the series $0\leq P\leq G$, noting that $P/0$ is abelian since it is prime order $p$ and $G/P$ is abelian since it is prime order $q$.

This proves it.

3. Alternatively, consider

$1 \unlhd Z(G) \unlhd G.$

By Lagrange's theorem, $|Z(G)| \in \{1,p,q,pq\}$. If $Z(G)=1, \, pq$, then we're done, so let $|Z(G)|=p$.
Then $|G/Z(G)|=q$. Since $q$ is prime, $G/Z(G)$ is cyclic. The case for $|Z(G)|=q$ is similar.