# Tag Info

## Hot answers tagged linear-groups

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### Does SO(3) preserve the cross product?

You may use the scalar triple product formula $r \cdot (p\times q)=\det(r,p,q)$ to prove that $$gr \cdot (gp\times gq)=gr \cdot g(p\times q)\tag{1}$$ ($=\det(r,p,q)$) for any vector $r$. Since $g$ ...
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### Does $G$ being a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is a subgroup of $GL(n, \mathbb{Z})$?

No, the quaternion group $Q_8$ is a subgroup of ${\rm GL}(2,p)$ for all odd primes $p$, but not of ${\rm GL}(2,{\mathbb Z})$.
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### Finite-order elements of $\text{GL}_4(\mathbb{Q})$

Let $A$ be a matrix of finite order $n$. Consider its minimal polynomial $m(x)$. More generally,
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### Diagonalizability of elements of finite subgroups of general linear group over an algebraically closed field

Let $A\in G$ be a matrix. After changing bases, $A$ is in JNF. Let us write $A=D+N$ with $D$ diagonal and $N$ the nilpotent part. Now since $G$ is finite, there is some $m\in\Bbb N$ with $A^m=I$, the ...
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You found an injection from $G$ to $GL_n(\mathbb Z)$. Notice that every element in the image preserves the subgroup $H$ of $\mathbb Z^n$ generated by $(1,\dots,1)$, so in this way we get an action of $... 6 votes Accepted ### Free subgroup of linear groups Yes. Note first that it suffices to answer the question in$\textit{SL}(2,\mathbb{C})$. In particular, if$A,B\in\textit{GL}(2,\mathbb{C})$, let$\hat{A},\hat{B}$be scalar multiples of$A,B$that ... • 46.5k 6 votes ### Suppose A is diagonalizable. Can this diagonalization necessarily be done by a matrix P in the special linear group? Just divide one column of$P$by its determinant. This argument shows a more general fact:$SL_n$acts transitively on each$GL_n$-orbit (if the field is algebraically closed as pointed out in the ... • 29.8k 6 votes ### Does SO(3) preserve the cross product? Since both expressions$g(p\times q)$and$g(p)\times g(q)$are linear in each of the variables$p,q$, it suffices to check the equality for the nine cases$p,q=e_1,e_2,e_3$, i.e. when$p,q$are basis ... • 5,375 6 votes Accepted ###$PSL(2,13)$has no subgroup of prime index. Suppose$G = {\rm PSL}(2,13)$embeds in$A_{13}$. Then clearly$G$acts transitively on the$13$points. Now a Sylow$7$-subgroup$P$of$G \le A_{13}$would be generated by a single$7$-cycle, and ... • 78.8k 6 votes Accepted ### Explicit isomorphism between${\rm SL}(2,{\Bbb R})$and${\rm SU}(1,1)$The elements of$SU(1,1)$are the matrices of the form$$\begin{bmatrix}\alpha+\beta i&\gamma+\delta i\\\gamma-\delta i&\alpha-\beta i\end{bmatrix}$$with$\alpha,\beta,\gamma,\delta\in\Bbb R\$ ...

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