Isomorphism between $SL(2,\mathbb{Z}) \times \mathbb{Z_2}$ and $GL(2,\mathbb{Z})$

Question

Since $SL(2,\mathbb{Z})=\{A\in M_{(2,2)}(\mathbb{Z})|\det(A)=1\}$ and $GL(2,\mathbb{Z})=\{A\in M_{(2,2)}(\mathbb{Z})|\det(A)=\pm1\}$, one can naturally guess there may exist an isomorphism between $SL(2,\mathbb{Z}) \times \mathbb{Z_2}$ and $GL(2,\mathbb{Z})$.

In my text book, the author shows that $SL(2,\mathbb{Z})\cong \mathbb{Z_4}*_{\mathbb{Z_2}}\mathbb{Z_6} $ and $GL(2,\mathbb{Z})\cong (\mathbb{Z_4}*_{\mathbb{Z_2}}\mathbb{Z_6})\times \mathbb{Z_2}$.

Hence there should be such an isomorphism.

But I failed to construct a conscise isomorphism directly from $SL(2,\mathbb{Z}) \times \mathbb{Z_2}$ to $GL(2,\mathbb{Z})$.

Is there a '(possibly) simple' isomorphism between them?

Any help will be appreciated.

Answer 1

There is no such isomorphism. The groups on the LHS and RHS have different centers: the center of $SL_2(\mathbb{Z}) \times \mathbb{Z}_2$ is $\mathbb{Z}_2 \times \mathbb{Z}_2$ but the center of $GL_2(\mathbb{Z})$ is $\mathbb{Z}_2$.

The correct statement is that the short exact sequence

$$1 \to SL_2(\mathbb{Z}) \to GL_2(\mathbb{Z}) \xrightarrow{\det} \mathbb{Z}_2 \to 1$$

splits, so $GL_2(\mathbb{Z})$ can be written as a nontrivial semidirect product $SL_2(\mathbb{Z}) \rtimes \mathbb{Z}_2$.

Answer 2

If there were such an isomorphism, then there would exist a normal subgroup $N$ of order $2$ of $\text{GL}_{2}(\mathbb{Z})$. I will show that $N$ does not exist.

Suppose contrary that $N$ exists and is generated by $n:=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. With $x:=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, we require that $xN=Nx$, or $xn=nx$. Hence, $a=d$ and $b=c$. Now, it must hold that $\det(n)=-1$, whence $a^2-b^2=ad-bc=\det(n)=-1$. That is, $a=0$ and $b=\pm 1$. It is easily seen that both $(a,b,c,d)=(0,1,1,0)$ and $(a,b,c,d)=(0,-1,-1,0)$ do not lead to $N$ being normal in $\text{GL}_2(\mathbb{Z})$.

However, it can be shown that $\text{SL}_2(\mathbb{Z}) \rtimes \mathbb{Z}_2\cong \text{GL}_2(\mathbb{Z})$. Just look at the subgroup $H\cong \mathbb{Z}_2$ generated by $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ of $\text{GL}_2(\mathbb{Z})$. We have $\text{SL}_2(\mathbb{Z})\trianglelefteq \text{GL}_2(\mathbb{Z})$, $\text{SL}_2(\mathbb{Z})\cap H$ is trivial, and $\text{SL}_2(\mathbb{Z})\cdot H=\text{GL}_2(\mathbb{Z})$. That is, $\text{GL}_2(\mathbb{Z})=\text{SL}_2(\mathbb{Z})\cdot H\cong \text{SL}_2(\mathbb{Z})\rtimes \mathbb{Z}_2$.