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Let $G = \left\{\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \textrm{ with } a=\pm 1, b \in \mathbb{Z}, c= \pm 1 \right\} \subset GL_2(\mathbb{R})$

and $H = \left\{\begin{pmatrix} a & b \\ 0 & 1\end{pmatrix} \in G \right\}$.

Show or disprove that $G$ is isomorphic to $H \times \{\pm 1\}$.

I tried constructing the obvious mappings $G \to H \times \{\pm 1\}$ but they turned out to be non-homomorphic. I then tried to disprove the conjecture by looking at the orders of elements in $G$ and $H \times \{\pm 1\}$, but they both turned out to posses only countable infinite elements of order 2 and $\infty$. Furthermore both $G$ and $H \times \{\pm 1\}$ are Abelian.

Now I'm stuck. In the preceding exercise I have proven $G/[G,G] \simeq (\mathbb{Z}/2\mathbb{Z})^3$ so maybe this can be used?

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You tried $\pmatrix{a&b\cr0&c\cr}$ goes to $(\pmatrix{a&b\cr0&1\cr},c)$? – Gerry Myerson Apr 20 '11 at 0:55
Yes, but it's no homomorphism. – ZulfiqarIII Apr 20 '11 at 0:59
By the way, you say $G$ is abelian, but that's not consistent with your result on the commutator. – Gerry Myerson Apr 20 '11 at 1:49
Gerry: Good point. – ZulfiqarIII Apr 20 '11 at 2:12
up vote 3 down vote accepted

In H × {±1}, the ±1 part commutes with everything. The matrix with (a=1,b=0,c=−1) does not commute with the matrix with (a=1,b=1,c=1).

Can you find some non-identity element of your group G that does commute with everything?

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In $G$ I find $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ to be commuting with everything; in $H\times \{ \pm 1 \}$ there are also two: $\left(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \pm 1 \right)$? – ZulfiqarIII Apr 20 '11 at 2:18
Exactly. So you want to send [-1,0;0,-1] to ([1,0;0,1],-1). Figure out where to send [a,b;0,1] (easy) and then you are done. – Jack Schmidt Apr 20 '11 at 2:22
Thank you, i haz fixed. – ZulfiqarIII Apr 20 '11 at 3:22

There is an element of $H\times\{\pm 1\}$, and only one element, which is not the identity and commutes with everything in $H\times\{\pm 1\}$: the element that has the identity matrix in the $H$ component, and $-1$ in the $\{\pm 1\}$ component.

There is an element of $G$, and only one element, which is not the identity and commutes with everything in $G$.

If there is going to be an isomorphism from $G$ to $H$ (or going the other way), then that unique element of $G$ has to be identified with the corresponding element of $H$.

What else? Every element of $H$ is already in $G$. What's more, $H$ is normal in $G$. What's the quotient $G/H$?

Since $G$ is the disjoint union of the $H$-cosets, maybe this will tell you a way to get a map?

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