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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.

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  • $\begingroup$ Try proving $gng^{-1}\in N$, where $g\in G$ and $n\in N$. $\endgroup$ – Thomas Shelby May 23 at 8:56
  • $\begingroup$ 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! $\endgroup$ – Mathbeginner May 23 at 9:05
  • $\begingroup$ Can you recheck your calculation? I'm getting a different answer. $\endgroup$ – Thomas Shelby May 23 at 9:25
  • $\begingroup$ @ThomasShelby, I redid it and now got $\begin{pmatrix} 1 & \frac{ad}{c} \\ 0 & 1 \end{pmatrix}$. Does this agree with yours? $\endgroup$ – Mathbeginner May 25 at 11:29
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    $\begingroup$ @Mathbeginner Note that $a,c\in \{1,-1\} $. So $\frac ac =\pm 1$. So it agrees with mine. $\endgroup$ – Thomas Shelby May 25 at 12:29
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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 $N$.

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  • $\begingroup$ 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? $\endgroup$ – Mathbeginner May 25 at 11:42
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    $\begingroup$ $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\} $.) $\endgroup$ – Thomas Shelby May 25 at 12:38
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$gN=Ng$ 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.

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  • $\begingroup$ 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? $\endgroup$ – Mathbeginner May 25 at 13:34
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    $\begingroup$ 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 $\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$. $\endgroup$ – Mark May 25 at 13:41
  • $\begingroup$ 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. $\endgroup$ – Mathbeginner May 25 at 13:57
  • $\begingroup$ Yes. Of course it would be harder to do if the matrices were of bigger size. $\endgroup$ – Mark May 25 at 14:04

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