# Show that the groups are abelian

• MHB
• mathmari
In summary, a group is considered abelian if its operation is commutative, meaning that the order in which elements are multiplied does not affect the end result. To prove that a group is abelian, it must be shown that for any two elements a and b, a * b = b * a. Some properties of abelian groups include uniqueness of the identity element, existence of inverses for every element, and commutativity. Examples of abelian groups include the integers, real numbers, and complex numbers. Showing that a group is abelian is important for easier mathematical manipulation and identification of patterns and relationships within the group.

#### mathmari

Gold Member
MHB
Hey! :giggle:

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? :unsure:

Hi there! Your response looks great overall. I just have a few suggestions for improvement:

1. In part (a), you could add a bit more explanation for why $H$ and $K$ must be normal subgroups. For example, you could mention that $H$ and $K$ are the only $p$-Sylow and $q$-Sylow subgroups respectively, and therefore must be normal by a property of Sylow subgroups.

2. In part (b), when you say "the possible orders of a subgroup of $G$ are $1, q, q^2, p, pq, pq^2$", you could explain why this is the case. For example, you could mention that this follows from the fact that the order of a subgroup must divide the order of the group.

3. In both parts, you could also mention that since $p$ and $q$ are prime, the orders of $H$ and $K$ must be relatively prime, and therefore $H$ and $K$ must have trivial intersection. This would help to explain why $xy=yx$ in part (a) and why $G_q/\{0\}$ is abelian in part (b).

Overall, your response is correct and complete. Great job! :)

## 1. What does it mean for a group to be abelian?

For a group to be abelian, it means that its group operation is commutative. This means that the order in which elements are multiplied does not affect the result.

## 2. How can you prove that a group is abelian?

To prove that a group is abelian, you must show that for all elements a and b in the group, the operation ab is equal to the operation ba. This can be done by using the group's defining properties and properties of commutativity.

## 3. Are all groups abelian?

No, not all groups are abelian. There are many non-abelian groups, such as the symmetric group and the dihedral group. These groups have operations that are not commutative.

## 4. What are some examples of abelian groups?

Some examples of abelian groups include the integers under addition, the real numbers under addition, and the group of 2x2 invertible matrices with real entries under matrix multiplication.

## 5. Why is it important to show that a group is abelian?

Showing that a group is abelian can provide useful information about the group's structure and properties. It can also simplify calculations and proofs involving the group, as commutativity often makes operations easier to work with.