# Show $N$ is a normal subgroup of $G$ where $G$ is a subgroup of $GL_{2}(\mathbb{Q}).$

We have $$G= \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \text{with a and c in \{\pm 1\} and b in \mathbb{Z}} \right\}$$, which is given to be a subgroup of the group of intvertible $$2\times2$$ matrices with coefficients in $$\mathbb{Q}$$.

Now I need to show two things.

First that $$N= \left\{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \text{in G with b even} \right\}$$ is a normal subgroup of $$G$$ and that $$G/N$$ is isomorphic to $$\{\pm 1\} \times \{ \pm 1\} \times \mathbb{Z}/2\mathbb{Z}$$, with mulitplication in the first two positions and addition in the third. As a hint: consider the map $$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \rightarrow (a,c,\bar{b})$$.

Im pretty puzzeld by this first question. As far as I know, $$N$$ is a subgroup of $$G$$ if for all $$g \in G$$ we have $$gN=Ng$$. The two $$b$$'s in the different matrices are confusing so I let the $$b$$ in $$N$$ be $$d$$. Then I get for $$gN = \begin{pmatrix} a & ad+b \\ 0 & c \end{pmatrix}$$ but for $$Ng= \begin{pmatrix} a & cd+b \\ 0 & c \end{pmatrix}$$. But this means $$ad+b = cd + b \Rightarrow ad = cd$$, which is not (always) true. Where do I go wrong? I don't even know where to start with the isomorphism part.

Secondly I have to show $$N=[G,G]$$.

Also no idea where to start.

As you can tell, I'm not very close so all help is appreciated.

• Try proving $gng^{-1}\in N$, where $g\in G$ and $n\in N$. – Thomas Shelby May 23 at 8:56
• This gives me $gng^{-1} = \begin{pmatrix} 1 & -\frac{b}{c}+acd+bc \\ 0 & 1 \end{pmatrix}$, hence $-\frac{b}{c} + acd+ bc$ must be even to be in $N$. $acd$ is obviously even, and since $c \in \{ \pm 1 \}$, the other two cancel out. That indeed shows $gng^{-1} \in N$. Thanks! – Mathbeginner May 23 at 9:05
• Can you recheck your calculation? I'm getting a different answer. – Thomas Shelby May 23 at 9:25
• @ThomasShelby, I redid it and now got $\begin{pmatrix} 1 & \frac{ad}{c} \\ 0 & 1 \end{pmatrix}$. Does this agree with yours? – Mathbeginner May 25 at 11:29
• @Mathbeginner Note that $a,c\in \{1,-1\}$. So $\frac ac =\pm 1$. So it agrees with mine. – Thomas Shelby May 25 at 12:29

Let $$m=\begin{pmatrix}1 & n \\0 & 1 \end{pmatrix}$$ be an element in $$N$$ and $$g=\begin{pmatrix}a & b \\0 & c \end{pmatrix}$$ be an arbitrary element in $$G$$. It is easy to verify that $$gmg^{-1}=\begin{pmatrix}1& \pm n\\0 & 1 \end{pmatrix}.$$

For the isomorphism part, you can use the first isomorphism theorem. All you need to do is to prove that the mapping given in the hint is a surjective homomorphism with kernel $$N$$.

• Thus we let $\phi: G \rightarrow \{ \pm 1 \} \times \{ \pm 1 \} \times \mathbb{Z} / 2 \mathbb{Z}$, where $\begin{pmatrix} a & c \\ 0 & c \end{pmatrix} \rightarrow (a,c,\bar{b})$. Then I show this is a homomorphism, which is easy, but then how do I show it is surjective? – Mathbeginner May 25 at 11:42
• $a$ and $c$ are in $\{1,-1\}$ and $b\in \Bbb Z$. So given any element of $\{\pm1\}×\{\pm1\}×\Bbb Z_2$, you can easily find a preimage in $G$. (Note that $\Bbb Z_2=\{0,1\}$.) – Thomas Shelby May 25 at 12:38

$$gN=Ng$$ doesn't mean that $$g$$ commutes with all elements of $$N$$. It just means that the sets $$gN$$ and $$Ng$$ are the same set. Anyway, there is an equivalent definition of a normal subgroup: $$N\trianglelefteq G$$ if and only if $$gng^{-1}\in N$$ for all $$g\in G,n\in N$$. So take a general element $$g\in G$$ and a general element in $$n\in N$$ and show that $$gng^{-1}$$ must be in $$N$$.

As for the second question note that since $$G/N$$ is abelian we must have $$[G,G]\leq N$$. So all we need to show is the other direction. Try to show that:

$$\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}=\left[ \begin{pmatrix} 1 & \frac{b}{2}\\ 0 & 1 \end{pmatrix},\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right]$$

When $$b$$ is an even integer.

• Computing the commutator of the matrices you provided, I get $\begin{pmatrix} 1 & -b \\ 0 & 1 \end{pmatrix} \neq \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix},$ thus this does not work entirely I guess... How did you come up with $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$ though? – Mathbeginner May 25 at 13:34
• The commutator is equal to $\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$, this is what I got. You ask how did I come up with it? I just wrote general matrices in $G$ (with $2\times 2$ matrices it is possible) and saw how the commutator looks like. Then it was clear what specific matrices should I take for the commutator to be equal to $\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$. – Mark May 25 at 13:41
• Thank you. So initially it is just a general form and since you know what you are looking for, you can fill in the general inputs as actual numbers. – Mathbeginner May 25 at 13:57
• Yes. Of course it would be harder to do if the matrices were of bigger size. – Mark May 25 at 14:04