Identify the Quotient Group $G/H$ Using the First Isomorphism Theorem

Question

I'm trying to go through Harvard's Abstract Algebra lectures on my own, and would like a little help with one of the homeworks. The problem asks:

Let $G$ be the group of invertible real upper $2 \times 2$ matrices. Determine whether or not the following conditions describe normal subgroups $H$ of $G$. If they do, use the First Isomorphism Theorem to identify the quotient group $G/H$.
(a) $\quad a_{11}$ = 1
(b) $\quad a_{12}$ = 0
(c) $\quad a_{11}$ = $a_{22}$
(d) $\quad a_{11}$ = $a_{22}$ = 1

To do this, we'll need to go one by one and determine whether the subgroup $H$ described is normal or not. If it's normal then it must be kernel of a surjective homomorphism. It can easily be shown that

$$\det: G \rightarrow R^{*}$$

is a surjective homomorphism. Then, $G/H$ must be isomorphic to $R^{*}$ by the First Isomorphism Theorem. So, we'll go one by one and see if they're normal.

NOTE: I have put *'s in places where computation would be too long simply to indicate the presence of some value determined through multiplying through.

(a) $$H = \begin{bmatrix} 1 & b \\ 0 & d \end{bmatrix},$$

$$aha^{-1} \;=\; \begin{bmatrix} a & b\\ 0 & d \end{bmatrix} \begin{bmatrix} 1 & b'\\ 0 & d' \end{bmatrix} \begin{bmatrix} \frac{1}{a} & \frac{-b}{ad} \\ 0 & \frac{1}{d} \end{bmatrix} \;=\; \begin{bmatrix} 1 & * \\ 0 & d' \end{bmatrix}.$$

This is a normal subgroup and therefore $G/H \simeq R^{*}.$

(b)

$$H = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix},$$

$$aha^{-1} \;=\; \begin{bmatrix} a & b\\ 0 & d \end{bmatrix} \begin{bmatrix} a' & 0\\ 0 & d' \end{bmatrix} \begin{bmatrix} \frac{1}{a} & \frac{-b}{ad} \\ 0 & \frac{1}{d} \end{bmatrix} \;=\; \begin{bmatrix} a' & * \\ 0 & d' \end{bmatrix}.$$

This is NOT a normal subgroup.

(c)

$$H = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix},$$

$$aha^{-1} \;=\; \begin{bmatrix} a & b\\ 0 & d \end{bmatrix} \begin{bmatrix} a' & b'\\ 0 & a' \end{bmatrix} \begin{bmatrix} \frac{1}{a} & \frac{-b}{ad} \\ 0 & \frac{1}{d} \end{bmatrix} \;=\; \begin{bmatrix} a' & * \\ 0 & a' \end{bmatrix}.$$

This is a normal subgroup and therefore $G/H \simeq R^{*}.$

(d) This is an instance of (c), and therefore it follows trivially that it is a normal subgroup.

Is this correct? Any and all help is greatly appreciated. Thanks.

Answer 1

Part (a)

Let me state the definitions in simple terms. You have a group $$G=\left\{\begin{pmatrix}a & b\\ 0 &c\end{pmatrix}:\ a,b,c\in\mathbb R, ac\neq 0\right\}$$

And its normal subgroup $$H=\left\{\begin{pmatrix}1 & b\\ 0 &c\end{pmatrix}:\ b,c\in\mathbb R, c\neq 0\right\}$$

To identify the group $G/H$ using the first isomorphism theorem, I will construct a homomorphism $f:G\to \mathbb R^*$ such that $\DeclareMathOperator{\ker}{ker}\ker f=H$.

$\det :G\to \mathbb R^*$ is not appropriate for this purpose because $\ker(\det)\neq H$ so we cannot make any conclusion about $G/H$. In fact:$\displaystyle \ker(\det) = \left\{\begin{pmatrix}a & b\\ 0 &\frac{1}{a}\end{pmatrix}:\ a,b\in\mathbb R, a\neq 0\right\}\tag*{}$

Consider the mapping $f:G\to \mathbb R^*$ $$f\begin{pmatrix}a & b\\ 0 &c\end{pmatrix}= a$$ It is well defined because $a\neq 0$. It is also a homomorphism which is routine to prove. $$\underbrace{\begin{pmatrix}a_1 & b_1\\ 0 &c_1\end{pmatrix}}_{g_1\in G}\underbrace{\begin{pmatrix}a_2 & b_2\\ 0 &c_2\end{pmatrix}}_{g_2\in G}=\begin{pmatrix}a_1a_2 & a_1b_2+b_1c_2\\ 0 &c_1c_2\end{pmatrix}\\ \implies f(g_1g_2)=a_1a_2=f(g_1)f(g_2)$$

It is also surjective. By first isomorphism theorem, $G/H\cong \mathbb R^*$.

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Part (c)

Another normal subgroup $$K=\left\{\begin{pmatrix}a & b\\ 0 &a\end{pmatrix}:\ a,b\in\mathbb R,a\neq 0\right\}$$

Again, we need a homomorphism with this as kernel.

Consider the mapping $f:G\to \mathbb R^*$ $$f\begin{pmatrix}a & b\\ 0 &c\end{pmatrix}= \frac{a}{c}$$

Prove that

• It is well defined.
• It is a homomorphism.
• It is surjective.
• $\ker f= K$

$\therefore G/K\cong \mathbb R^*$

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Part (d)

Normal subgroup of $G$: $$P=\left\{\begin{pmatrix}1 & b\\ 0 &1\end{pmatrix}:\ b\in\mathbb R\right\}$$

Consider the mapping $f:G\to \mathbb R^*\times \mathbb R^*$ $$f\begin{pmatrix}a & b\\ 0 &c\end{pmatrix}= (a,c)$$

Yet again, prove that

• It is well defined.
• It is a homomorphism.
• It is surjective.
• $\ker f= P$

$\therefore G/P\cong \mathbb R^*\times \mathbb R^*$

Though not related to the questions in the post, I noticed that $P\cong (\mathbb R, +)$. It becomes obvious if you observe that $$\begin{pmatrix}1 & b_1\\ 0 &1\end{pmatrix}\begin{pmatrix}1 & b_2\\ 0 &1\end{pmatrix}=\begin{pmatrix}1 & b_1+b_2\\ 0 &1\end{pmatrix} $$

Now, this does not mean that $G\cong \mathbb R\times \mathbb R\times(\mathbb R, +)$ (you can't "multiply" both sides by $P$ and make this conclusion). :)

Answer 2

I know this thread is quite old, but I think the solution given is not correct.

Your argument is that because the determinant is a surjective homomorphism any quotient group $G/N$ (with $N$ a normal subgroup) is isomorphic to $R^*\,$?

From my understanding, it is true that only (a), (c) and (d) are normal. But you cannot apply the FIT (First Isomorphism Theorem) to the determinant, because the theorem tells you that $G/H$ is isomorphic to the image of a homomorphism where $H=\ker(f)$. Neither the normal subgroups of (a), (c) and (d) are the kernel of the determinant.

I think in (a), one must use the homomorphism $f(g)=a$. Now we can use the FIT and indeed $G/H$ is isomorphic to $R^*$.

For (c) I would use the homomorphism $f(g)=a/d$, and again $G/H$ is isomorphic to $R^*$.

And finally, for (d) one could use the homomorphism $f(g)=\text{diag}(a,d)$.