How to show that we deal with a normal subgroup.

Question

In $S_{\mathbb{Z}/n\mathbb{Z}}$, the group consisting of all permutations of the set $\mathbb{Z}/n\mathbb{Z}$, we consider the subgroup G given by

G:= {${f_{a,b}: x \rightarrow ax + b | a \in (\mathbb{Z}/n\mathbb{Z})^*, b \in \mathbb{Z}/n\mathbb{Z} }$}

In G we have the following two subgroups:

• H = {$f_{a,0} \in G | a \in (\mathbb{Z}/n\mathbb{Z})^* $}

• N = {$f_{1,b} \in G | b \in \mathbb{Z}/n\mathbb{Z} $}

I have to answer the following 3 questions.

1) Explain why N is a subgroup of G, and why this subgroup is a normal subgroup.

2) Show that HN equals G

3) Show that G/N $\cong$ $(\mathbb{Z}/n\mathbb{Z})^*$

Any help would be grateful, since we did not start the course yet and because I have no background in Group Theory.

Answer 1

For 1. The inverse of an element $f_{a,b}\in G$ is given by $$f_{a,b}^{-1}(x)=a^{-1}(x-b).$$ Let $f_{1,c}\in N$. We have $$f_{a,b}f_{1,c}f_{a,b}^{-1}(x)=f_{a,b}(a^{-1}(x-b)+c)=x+ac,$$ so $f_{a,b}f_{1,c}f_{a,b}^{-1}\in N$. Hence $N$ is normal in $G$.

For 2. We have

• $N\cap H=\{\mathrm{id}\}$ (because $f_{a,b}\in N\cap H\Rightarrow a=1,b=0$)
• $$f_{a,0}f_{1,b}(x)=a(x+b)=ax+ab,$$ so every element $f_{a,b}\in G$ can be written as $$f_{a,b}=f_{a,0}f_{1,a^{-1}b}$$
• $N$ is normal in $G$.

It follows that $G$ is the internal product $HN$.

For 3. $G/N\simeq H\simeq (\mathbb{Z}/n\mathbb{Z})^*$. You can also prove it directly by noticing $$f_{a,b}f_{c,d}^{-1}\in N \Leftrightarrow a=c,$$ so you have a homomorphism $G/N\to (\mathbb{Z}/n\mathbb{Z})^*,\; f_{a,b}N\mapsto a$, which is obviously a bijection.