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- Apr 14, 2013

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Let $p<q$ different prime numbers.

a) If $q\neq 1\pmod p$ show that each group of order $pq$ is abelian.

b) If $q\neq \pm 1\pmod p$ show that each group of order $pq^2$ is abelian.

I have done the following :

a) Let $H$ be a $p$-Sylow subgroup of $G$ and let $K$ be a $q$-Sylow subgroup of $G$.

From Sylow theorem we have that the number of $p$-Sylow subgroups of $G$ is $\equiv 1\pmod p$, i.e. of the form $1+kp$ and divides the order of the group, $pg$.

The $1+kp$ doesn't divide $p$.

The $p$ and $q$ are prime numbers, $1+kp$ doesn't divide $p$ so it must divide $q$.

The $q$ is prime, so its divisors are $1$ and $q$. So $1+kp=1$ or $1+kp=q$.

From the second case we have $1+kp=q \Rightarrow kp=q-1$, a contradiction since $q\neq 1\pmod p$.

Therefore it must be $1+kp=1$, i.e. there is only one $p$-Sylow subgroup of $G$.

Similarily, there is only one $q$-Sylow subgroup of $G$.

In general : A group $G$ has exactly one $p$-Sylow group $P$ iff $P$ is normal.

Therefore $H$ and $K$ are normal subgroupsof $G$.

Since $p$ and $q$ are prime, $H$ and $K$ are cyclic. Let $H=\langle x\rangle$ and $K=\langle y\rangle$.

To show that $G$ is abelian, we show that it is cyclic.

To show that $G$ is cyclic, we have to show that $xy=yx$, since then $\text{ord}(xy)=\text{ord}(x)\cdot \text{ord}(y)=p\cdot q$.

Since $H$ and $K$ are normal, we have that \begin{align*}&xyx^{-1}y^{-1}=(xyx^{-1})y^{-1}\in Ky^{-1}=K \\ &xyx^{-1}y^{-1}=x(yx^{-1}y^{-1})\in xH=H\end{align*}

Let $c\in H\cap K$. Then it mus be $\text{ord}(c)\mid p$ and $\text{ord}(c)\mid q$, since $(p,q)=1$ it follows that $\text{ord}(c)\mid 1 \Rightarrow c=e$.

Therefore $xyx^{-1}y^{-1}=e \Rightarrow xy=yx$.

b) Let $G$ be a group of order $pq^2$.

Let $n_p$ the number of $p$-Sylow subgroups of $G$ and $n_q$ the number of $q$-Sylow subgroups of $G$.

Since $G$ has finite order and $q$ is prime with $q\mid\text{ord}(G) \Rightarrow q\mid pq^2$ and $q\nmid p$ (prime to each other) we have from Sylow Theorem that \begin{align*}&n_q\equiv 1\pmod q \ \ \ \ \ \ \ \ (\star) \\ &n_q\mid p\end{align*}

Since $n_q\mid p$ and since $p$ is prime we have that $n_q=1$ or $n_q=p$.

If $n_q=p$ then from $(\star)$ we have that $p\equiv 1\pmod q \Rightarrow q\mid p-1$, a contradiction since $q>p$.

So it must be $n_q=1$, i.e. there is only one $q$-Sylow subgroup of $G$, say $G_q$.

From Lagrange Theorem the possible orders of a subgroup of $G$ are $1, q, q^2, p, pq, pq^2$.

Therefore $G_q$ has order $q^2$.

Since $G$ has a unique subgroup of order $q^2$, since $n_q=1$,it follows that $G_q$ is normal in $G$.

So $$\{0\}\leq G_q\leq G$$ We have that $\left |\frac{G_q}{\{0\}}\right |=g^2$.

In general : Each group of order $q^2$ is abelian.

Therefore $G_q/\{0\}$ is abelian.

We also have that $\left |\frac{G}{G_q}\right |=p$.

The $p$ is prime and so $|G/G_q|$ is syclic, and so abelian.

Is everything correct and complete?