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Let G be a finite group. Let T be an element of Aut(G) such that
$T^{2} = I$ , and $x T = x \Leftrightarrow x = e$ .
Then, G is abelian.

Breanna Mcclure 2022-04-14

Is there a particular characterization of $A u t (Q^{3})$ ?

Bradley Barron 2022-04-14

Find the isomorphic ring with
$Z_{6} \frac{x}{⟨ x^{3} - x ⟩}$
(Chinese remainder thm)

Rubi Riggs 2022-04-13

Let G be a finite group with $c a r d (G) = p^{2} q$ 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
$\underset{S \in S y l_{q} (G)}{\cup} S ∖ {1} = {\dot{\cup}}_{S \in S y l_{q} (G)} S ∖ {1}$
i.e. that for $S, T \in S y l_{q} (G)$ with $S \neq T$ we that $S ∖ {1} \cap T ∖ {1} = \emptyset$

Ansley Sparks 2022-04-13

Jacobson radical of the ring of lower triangular $n \times n$ matrices over $Z$

Rosa Townsend 2022-04-13

Isomorphism between quotient of ring and principal ideals
Let R be a principal ideal domain and suppose that $u, u^{'} \in R$ are such that $\frac{R}{(u)} \tilde{=} \frac{R}{(u^{'})}$ as R-modules, where (u) denotes the ideal generated by u. Is it true that $u = α u^{'}$ , where $α \in R$ is invertible?

Ashleigh Mitchell 2022-04-13

Is there any relation between ${Sym}_{n}$ and ${Sym}_{k}$ where $k < n$ ? Is ${Sym}_{k} \subset {Sym}_{n}$ ? Since $k! ∣ n! \forall k < n$ they satisfy Lagrange's theorem, but of course that doesnt guarantee the existence of a subgroup

Devin Dougherty 2022-04-12

How do I prove that this ideal is not a ' ideal?
Let K be a field and $R = \frac{K [X, Y]}{(X Y)} . F or P \in K [X, Y]$ we denote $[P]$ its class in R. Show that the Ideal (XY) is not a ' ideal.

Suppose that R is a commutative ring and |R| = 30. If I is an ideal of R and |I| = 10, prove that I is a maximal ideal

Albert Byrd 2022-04-10

Use the isomorphism theorem to determine the group $G L_{2} \frac{R}{S} L_{2} (R)$ . Here $G L_{2} (R)$ is the group of $2 \times 2$ matrices with determinant not equal to 0, and $S L_{2} (R)$ is the group of $2 \times 2$ matrices with determinant 1. In the first part of the problem, I proved that $S L_{2} (R)$ is a normal subgroup of $G L_{2} (R)$ . Now it wants me to use the isomorphism theorem. I tried using
$| G L_{2} \frac{R}{S} L_{2} (R) | = \frac{| G L_{2} (R) |}{| S L_{2} (R) |},$
but since both groups have infinite order, I don't think I can use this here.

Ramiro Grant 2022-04-09

Two sided ideal in $M_{2 \times 2} (C)$

Deangelo Hardy 2022-04-08

I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I'm very sorry if it's extremely straightforward, I just think I need some time to get used to the way of thinking that's required to solve these questions.
Let $R_{1} and R_{2}$ be commutative rings with identities and let $R = R_{1} \times R_{2}$ . The question asks to show that every ideal I of R is of the form $I = I_{1} \times I_{2}$ with $I_{1}$ an ideal of $R_{1}$ and $I_{2}$ an ideal of $R_{2}$ .

Breanna Fisher 2022-04-08

Understanding a proof of: In a finite group, the number of elements of ' order p is divisible by $p - 1$ .

Amya Horn 2022-04-05

Fields with involution whose fixed field is ordered?
Let K be a field with an involution $$ , meaning $\cdot : K \to K$ is an automorphism and $(x ) * = x$ for all $ξ n K$ . Suppose further that the fixed field of $*$ is ordered (i.e., it can be given an ordering that is compatible with the field structure).

Makayla Stevens 2022-04-04

Euler's remarkable '-producing polynomial and quadratic UFDs
Example of a polynomial which produces a finite number of 's is:
$x^{2} + x + 41$
which produces 's for every integer $0 \leq x \leq 39$ .

Janiyah Hays 2022-04-03

Examples for when the quotient ring is necessarily / not necessarily an extension of the residue field.
Let R be an integral domain with unique maximal ideal $m$ . Let $F : = Frac (R)$ be the field of fraction of R.

burubukuamaw 2022-04-02

Show if M is free of rank n as R-module, then $\frac{M}{I M}$ is free of rank n as $\frac{R}{I}$ module:
Let R be a ring and $I \subset R$ a two-sided ideal and M an R-module with
$I M = {\sum r_{i} x_{i} ∣ r_{i} \in I, x_{i} \in M} .$

Oxinailelpels3t14 2022-04-02

Residue of x in polynomial ring $Z \frac{x}{f}$
When we say that α is the residue of x in $Z \frac{x}{f}$ , and $f = x^{4} + x^{3} + x^{2} + x$ , wouldn't $α$ just be x? Because if we divide with the remainder, we would get $x = 0 f + x$ ?

Miley Caldwell 2022-04-01

How to solve a cyclic quintic in radicals?
Galois theory tells us that
$\frac{z^{11} - 1}{z - 1} = z^{10} + z^{9} + z^{8} + z^{7} + z^{6} + z^{5} + z^{4} + z^{3} + z^{2} + z + 1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be $ζ^{1}, ζ^{2}, \dots, ζ^{10}$ , following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
$\begin{aligned} A_{0} & = x_{1} + x_{2} + x_{3} + x_{4} + x_{5} \\ A_{1} & = x_{1} + ζ x_{2} + ζ^{2} x_{3} + ζ^{3} x_{4} + ζ^{4} x_{5} \\ A_{2} & = x_{1} + ζ^{2} x_{2} + ζ^{4} x_{3} + ζ x_{4} + ζ^{3} x_{5} \\ A_{3} & = x_{1} + ζ^{3} x_{2} + ζ x_{3} + ζ^{4} x_{4} + ζ^{2} x_{5} \\ A_{4} & = x_{1} + ζ^{4} x_{2} + ζ^{3} x_{3} + ζ^{2} x_{4} + ζ x_{5} \end{aligned}$
Once one has $A_{0}, \dots, A_{4}$ one easily gets $x_{1}, \dots, x_{5}$ . It's easy to find $A_{0}$ . The point is that $τ$ takes $A_{j}$ to $ζ^{- j} A_{j}$ and so takes $A_{j}^{5}$ to $A_{j}^{5}$ . Thus $A_{j}^{5}$ can be written down in terms of rationals (if that's your starting field) and powers of $ζ$ . Alas, here is where the algebra becomes difficult. The coefficients of powers of $ζ$ in $A_{1}^{5}$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have $A_{1}$ as a fifth root of a certain explicit complex number. Then one can express the other $A_{j}$ in terms of $A_{1}$ . The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

Octavio Chen 2022-03-31

${End}_{ℝ [x]} (M)$ where $M = \frac{R [x]}{(x^{2} + 1)}$ is a module over the ring $R [x]$

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