Malachi Novak
2022-04-24
Answered

How can I show these polynomials are not co'?

$x,\text{}x-1$ and $x+1$ in the ring ${\mathbb{Z}}_{6}\left[x\right]$

You can still ask an expert for help

veceritzpzg

Answered 2022-04-25
Author has **16** answers

Step 1

The ideal$(x-1,\text{}x+1)$ equals the ideal

$(2,\text{}x-1)$ and this is not the whole ring, otherwise

$1\in (2,\text{}x-1)$ , so $1=2f\left(x\right)+(x-1)g\left(x\right)$ .

Now send x to 1 and get a contradiction.

The ideal

Now send x to 1 and get a contradiction.

asked 2020-11-20

Prove that in any group, an element and its inverse have the same order.

asked 2022-04-10

Use the isomorphism theorem to determine the group $G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)$ . Here $G{L}_{2}\left(\mathbb{R}\right)$ is the group of $2\times 2$ matrices with determinant not equal to 0, and $S{L}_{2}\left(\mathbb{R}\right)$ is the group of $2\times 2$ matrices with determinant 1. In the first part of the problem, I proved that $S{L}_{2}\left(\mathbb{R}\right)$ is a normal subgroup of $G{L}_{2}\left(\mathbb{R}\right)$ . Now it wants me to use the isomorphism theorem. I tried using

$\left|G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)\right|=\frac{\left|G{L}_{2}\left(\mathbb{R}\right)\right|}{\left|S{L}_{2}\left(\mathbb{R}\right)\right|},$

but since both groups have infinite order, I don't think I can use this here.

but since both groups have infinite order, I don't think I can use this here.

asked 2022-01-12

Is it true that for a Group G with Normal Group $N:\frac{G}{N}=\frac{GN}{N}?$

I think the statement is correct. But why do we have to write:$[G,G]\frac{N}{N}$ here instead of just $\frac{G,G}{N}?$

I think the statement is correct. But why do we have to write:

asked 2022-04-13

Let G be a finite group with $card\left(G\right)={p}^{2}q$ with $p<q$ two ' numbers. We denote $s}_{q$ the number of q-Sylow subgroups of G and similarly for p. I have just shown that ${s}_{q}\in \{1,{p}^{2}\}$. Now I want to show that

$\underset{S\in Sy{l}_{q}\left(G\right)}{\cup}S\setminus \left\{1\right\}={\dot{\cup}}_{S\in Sy{l}_{q}\left(G\right)}S\setminus \left\{1\right\}$

i.e. that for $S,\text{}T\in Sy{l}_{q}\left(G\right)$ with $S\ne T$ we that $\mathrm{S}\setminus \left\{1\right\}\cap \mathrm{T}\setminus \left\{1\right\}=\varnothing $

asked 2022-04-09

Two sided ideal in ${M}_{2\times 2}\left(\mathbb{C}\right)$

asked 2022-02-24

There are 2 photos inserted the one is the part 1 of the problem and the other is the part 2 . The picture with the number 1 is the part 1 and the x and y axis graph pic is part 2.

asked 2022-01-12

Let H be a subgroup of G and $x,\text{}y\in G$ . Show that

$x\left(Hy\right)=\left(xH\right)y$