Proving that $G/N$ is an abelian group

Let $G$ be the group of all $2 \times 2$ matrices of the form $\begin{pmatrix} a & b \\ 0 & d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} \right\}$ be a subset of the group $G$. Prove that $N$ is a normal subgroup of $G$ and prove that $G/N$ is abelian group.

Here is my attempt!

To prove $N$ is normal I consider the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$ is normal, it follows that $N$ is also normal. Is this proof helpful here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets.

Thank you.

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The formatting on this question is horrendous. The punctuation is sometimes misplaced. All of this could be chalked up to a new user, unfamiliar with the format here, or even the English language. But "thank u"? Unforgivable. – user641 Dec 26 '11 at 7:12
I am joking of course, but in all seriousness the formatting here makes the question unreadable, and you should probably stop asking questions temporarily, until you are able to format them legibly. – user641 Dec 26 '11 at 7:14
@SteveD I think it's better now (hopefully I didn't introduce any errors), but I agree that the OP should try to learn TeX. It isn't so bad! – Dylan Moreland Dec 26 '11 at 7:17
@JuniorII Just because the image of a subgroup is normal doesn't mean the subgroup itself is normal. If this were true, then since every group maps to the trivial group hence has a map that sends its subgroups to a normal subgroup of the trivial group (the trivial group itself), all subgroups would be normal. Chances are you are confusing this with the statement that the image of a normal subgroup is normal, which is true. – Alex Becker Dec 26 '11 at 7:23
– Martin Sleziak Oct 16 '15 at 13:28

One way is using first isomorphism theorem.

To do this you should find a group homomorphism such that $\operatorname{Ker} \varphi=N$.

Let us try $\varphi: G\to \mathbb R^*\times \mathbb R^*$ given by $$\begin{pmatrix} a & b \\ 0 & d\end{pmatrix} \mapsto (a,d).$$ (By $\mathbb R^*$ I denote the group $\mathbb R^*=\mathbb R\setminus\{0\}$ with multiplication. By $G\times H$ I denote the direct product of two groups, maybe your book uses notation $G\oplus H$ for this.)

It is relatively easy to verify that $\varphi$ is a surjective homomorphism. It is clear that $\operatorname{Ker} \varphi=N$. Hence, by the first isomorphism theorem, $$G/N \cong \mathbb R^*\times\mathbb R^*$$ This is a commutative group.

If you prefer, for any reason, not using the first isomorphism theorem, you could also try to verify one of equivalent definitions of normal subgroup and then describe the cosets and their multiplication.

In this case you have $$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \begin{pmatrix} 1 & b' \\ 0 & 1 \end{pmatrix} \frac1{ad} \begin{pmatrix} d & -b \\ 0 & a \end{pmatrix}= \begin{pmatrix} 1 & \frac{ab'}d \\ 0 & 1 \end{pmatrix}$$ (I have omitted the computations), which shows that $xNx^{-1}\subseteq N$ for any $x\in G$.

You can find out easily that cosets are the sets of the form $$\{\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}; y\in\mathbb R\}$$ for $x,z\in\mathbb R\setminus\{0\}$ and that the multiplication of cosets representatives $\begin{pmatrix} x & 0 \\ 0 & z \end{pmatrix}$ is coordinate-wise.

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Great explanation of the problem. +1, you took nothing for granted and that is good, since we usually don't know the level of understanding of the question of the OP very well in this website context. – Patrick Da Silva Dec 26 '11 at 11:30
Isn't the first part the same as saying that $\begin{pmatrix} a & b \\ 0 & d\end{pmatrix} =\begin{pmatrix} a & 0 \\ 0 & d\end{pmatrix} \begin{pmatrix} 1 & b/a \\ 0 & 1\end{pmatrix}$? – N. S. Dec 26 '11 at 17:34
@N.S. I am not sure what you mean. By the first part you mean the solution using isomorphism theorem or the first part of this solution? – Martin Sleziak Dec 26 '11 at 17:42
Very clear (+1) – Karl Mar 12 at 11:14

I would like to add one more way to solve the problem. If $A= \begin{pmatrix} a && b \\ 0 &&c \end{pmatrix}$ and $B= \begin{pmatrix} d && e \\ 0 &&f \end{pmatrix}$ then $[A,B]=ABA^{-1}B^{-1}=\begin{pmatrix} 1 & -\frac{e}{f}+\frac{\frac{-db}{c}+\frac{ae+bf}{c}}{f} \\ 0 & 1 \end{pmatrix}$. Substituting $e=0$, $f=1$, $c=1$, $b=1$ and $d=1-k$ for any given k we see that $N=G'$ is in fact the (first) derived subgroup of $G$ and therefore $G/N$ is necessarily Abelian (even the largest Abelian factor group of $G$).

Of couse this proof presumes a little more knowledge on the part of the reader than Martin's excellent solution.

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We can apply the property of the determinant to show that the normal subgroup test is satisfied. For the second part, we can use the fact that G is abelian since it forms a diagonal matrix. Then we show that elements in the factor group commute.

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This can another solution. Consider the relation of equivalence: $$a\equiv b\pmod H \iff a^{-1}b\in H$$ where $a$ and $b$ are elements of a group $G$ while $H$ is a subgroup of $G$ Note that the equivalent class $[a]=aH$ the left coset of $a$ (generic element of this group). Note that every triangular upper matrix is congruent to diagonal matrix. Indeed if $$A=\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$$ then $A$ is congruent to: $$\begin{pmatrix} a & b' \\ 0 & d \end{pmatrix}$$ in the preceding equivalence relation. Therefore every element of $G$ stands for a diagonal matrix in the equivalence class : $$Nab=Nba$$ because $a$ and $b$ are diagonal matrix but diagonal matrix commute

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