21 votes
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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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  • 123k
15 votes
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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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  • 78.8k
14 votes
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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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  • 15.6k
13 votes
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Abelianization of general linear group?

The derived group of ${\rm GL}(n,K)$ is ${\rm SL}(n,K)$ for all $n \ge 1$ and all fields $K$, so the abelianization of ${\rm GL}(n,K)$ is the multiplicative group of the field, $(K \setminus \{0\},\...
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  • 78.8k
13 votes

How to prove that $\mathrm{SL}_{2} (\mathbb F_{3})$ is not isomorphic to $S_{4}$?

$SL_2(\mathbb{F}_3)$ has two elements $I, -I$ in its center, but the center of $S_4$ is trivial.
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  • 6,395
12 votes
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The group $\mathbb Z_n$ is isomorphic to a subgroup of $GL_2(\mathbb R)$.

Consider the group of those matrices of the form\begin{pmatrix}\cos\left(\frac{2k\pi}n\right)&-\sin\left(\frac{2k\pi}n\right)\\\sin\left(\frac{2k\pi}n\right)&\cos\left(\frac{2k\pi}n\right)\end{...
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11 votes
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Is this group of matrices cyclic?

The group is indeed cyclic and is generated by the matrix $$ J=\left( {\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} } \right) $$ As you have for $n \in \mathbb Z$ $$ J^n ...
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11 votes
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Conjugacy classes in $SU_2$

Since the characteristic polynomial of any matrix $A \in SU(2)$ has real coefficients, its eigenvalues are complex conjugates $\lambda, \bar{\lambda}$; since $\lambda \bar{\lambda} = \det A = 1$, the ...
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11 votes

$\det(A)={\pm 1} \iff A\in\mathcal O_n(\Bbb R)$?

Not every matrix with $\det(A)=\pm 1$ is orthogonal. For instance, the matrix $$ A=\begin{bmatrix}1&1\\0&1\end{bmatrix} $$ is not orthogonal.
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10 votes
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The rotation group $SO(3)$ may be mapped to a $2$-sphere by sending a rotation matrix to its first column. How can I describe the fibres of the map?

Hint A $3 \times 3$ matrix $A$ is in $SO(3)$ iff its columns form an oriented orthonormal basis of $\mathbb{R}^3$. Remark The universal (and two-fold) cover of the space $SO(3)$ is $SU(2)$, which we ...
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10 votes

Does SO(3) preserve the cross product?

Two ways of seeing it more directly: The cross product of two vectors can be expressed in terms of the norms of the vectors and the angle between them, and those properties are preserved by rotations....
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  • 2,006
9 votes
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General Linear Group $GL_n(R)$ not Finitely Generated

Consider determinants. If $F$ is an infinite field, the determinants of a generating set of $\textrm{GL}_n(F)$ must generate the multiplicative group $F^*$. But $F^*$ contains a copy of either $\Bbb Q^...
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9 votes
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Diagonalizability of elements of finite subgroups of general linear group over an algebraically closed field

Let $K$ be a field. As you have already mentioned, a matrix $A \in \operatorname{M}_n(K)$ is diagonalizable (over $K$) if and only if there exists a polynomial $f(t) \in K[t]$ with $f(A) = 0$ such ...
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8 votes
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Isomorphism between $SL(2,\mathbb{Z}) \times \mathbb{Z_2}$ and $GL(2,\mathbb{Z})$

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(\...
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8 votes
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General Linear Group $GL_n(\mathbb{Z})$ of Integers is finitely generated

The Smith normal form algorithm shows that a matrix in $GL_n(\mathbb{Z})$ can be diagonalized by elementary row and column operations. A row operation is the same as multiplying on the left by a power ...
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  • 100k
8 votes
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Commutator Group of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$

If $x\in\mathbb R$, then$$\begin{pmatrix}1&x\\0&1\end{pmatrix}=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1&x\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{...
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8 votes
Accepted

Group of matrices form a manifold or euclidean space

Take $SO(2,\mathbb{R})$, for instance. This is the group of the $2\times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,\mathbb{R})=\left\{\begin{bmatrix}\cos\theta&-\sin\theta\\\...
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8 votes

Does $G$ being a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is a subgroup of $GL(n, \mathbb{Z})$?

Here's one thing we can say. Suppose a finite group $G$ embeds into $GL_n(\mathbb{F}_p)$ for all $p$; actually we can ask for the much weaker condition that $G$ embeds into $GL_n(\prod \mathbb{F}_p)$, ...
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7 votes

Is this group of matrices cyclic?

if we define: $$ B=\left( {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right) $$ then $$ A = I+B \\ B^2 = 0 $$ so by the binomial theorem, for $n \ge 0$: $$ A^n =(I+B)^n = ...
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  • 17.6k
7 votes

How to prove that $\mathrm{SL}_{2} (\mathbb F_{3})$ is not isomorphic to $S_{4}$?

$$A:=\begin{pmatrix}2&0\\1&2\end{pmatrix}\implies A^2=\begin{pmatrix}1&0\\1&1\end{pmatrix}\;,\;\;A^3=\begin{pmatrix}2&0\\0&2\end{pmatrix}=-I\implies A^6=I$$ and thus SL$_2(\...
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  • 206k
7 votes

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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7 votes
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Why is $SL(n, \mathbb{R})$ the kernel of $\det : GL(n, \mathbb{R}) \mapsto\Bbb R^*$?

But $\det$ is not a linear map. It is a group homomorphism. Its kernel is$$\left\{M\in GL(n,\Bbb R)\,\middle|\,\det(M)=1\right\},$$since $1$ is the identity element of $\bigl(\Bbb R\setminus\{0\},\...
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6 votes

Pairs of $2\times 2$ matrices generating free groups.

There's a 1969 paper by Lyndon and Ullman that appears to be it, plots and all. R. C. Lyndon and J. L. Ullman, Groups generated by two parabolic linear fractional transformations. Canad. J. Math. ...
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6 votes
Accepted

Does an injective homomorphism always exists from $G$ into $GL_n(R)$ where order of $G$ is $n$?

Yes, because by Cayley's theorem, $G$ can be considered as a subgroup of $S_n$ (by $G \rightarrow \rm{Sym}(G), g \mapsto (h \mapsto gh)$. But $S_n$ acts on the standard basis $(e_1, \dots, e_n)$ of $\...
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  • 33k
6 votes
Accepted

Is a nontrivial finite group of order $n$ always isomorphic to a subgroup of $GL_{n-1}(\mathbb{Z})$?

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 $...
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6 votes
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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 ...
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  • 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 ...
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  • 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 ...
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  • 5,375
6 votes
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$PSL(2,13)$ 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 ...
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  • 78.8k
6 votes
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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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