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3 votes
Accepted

(Group)Homomorphism and Subgroups

The Isomorphism Theorems establish that there is a bijection between the subgroups of $G/H$ and the subgroups of $G$ that contain $H$. the bijection is given by sending the subgroup $Y\leq G/H$ to $\...
Arturo Magidin's user avatar
3 votes

Given normal subgroup $H\triangleleft G$ with prime index $p$, then $G\cong H\ \rtimes\langle\alpha\rangle.$

Upon having read through some of the above comments, in fact the proof I presented was not true, as well a counter-example exists to disprove the statement. Problem with proof. Since $\beta^{p}\in H$ ...
JAG131's user avatar
  • 421
1 vote

Can a non-Lie quotient of $GL_3(\mathbb{R})$ be $O_3(\mathbb{R}$)?

Let $C$ denote the center of $G=GL(3, \mathbb R)$, it is the group of scalar $3\times 3$ real matrices and, hence, is isomorphic to $\mathbb R^\times \cong (\mathbb R, +)\times \mathbb Z_2$. Let $N$ ...
Moishe Kohan's user avatar
  • 99.6k
1 vote

Can a non-Lie quotient of $GL_3(\mathbb{R})$ be $O_3(\mathbb{R}$)?

Consider the matrix of a transvection $M=Id+E_{1,2}$, for instance, where $E_{1,2} e_2=e_1, E_{1,2} e_1=E_{1,2}e_3=0$ : $M$ is of infinite order and conjugate to any of is power $M^n$. But a rotation ...
Thomas's user avatar
  • 7,535
1 vote

A normal subgroup is the union of conjugacy classes.

The definition of a normal subgroup is that $gN=Ng$ for all $g\in G$, or equivalently $gNg^{-1}=N$ for all $g\in G$. Note the following pitfall: for a fixed $g\in G$, it is not true that $gng^{-1}\in ...
David A. Craven's user avatar
1 vote

A normal subgroup is the union of conjugacy classes.

The reverse direction is pretty straight-forward: If $N_0\triangleleft G$, then $x\in N_0$ if and only if $[x]\subset N_0$. Thus, $N_0$ must be a union of conjugacy classes. Similarly so, the forward ...
JAG131's user avatar
  • 421

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