 Recent questions in Abstract algebra

Recent questions in Abstract algebra idiopatia0f 2022-01-15 Answered

How can one show (hopefully in an elementary manner) that there exist irreducible polynomials of arbitrary degree and number of variables over arbitrary field? Does $$\displaystyle\forall{n},{d}\in{\mathbb{{{N}}}}\forall$$ field $$\displaystyle{\mathbb{{{F}}}}$$ exist an irreducible $$\displaystyle{f}\in{\mathbb{{{F}}}}{\left[{x}_{{{1}}},\cdots,{x}_{{{n}}}\right]}$$ of degree d? Judith McQueen 2022-01-15 Answered

Group theory exercise from Judson text $$\displaystyle{S}={\frac{{{R}}}{{-{1}}}}$$ and define a binary operation on S by $$\displaystyle{a}\times{b}={a}+{b}+{a}{b}$$. Prove that $$\displaystyle{\left({S},\times\right)}$$ is an abelian group. Lennie Davis 2022-01-15 Answered

Let $$\displaystyle{f}:{\left({\mathbb{{{Z}}}}_{{{28}}},+\right)}\rightarrow{\left({\mathbb{{{Z}}}}_{{{16}}},+\right)}$$ be a group homomorphism such that $$\displaystyle{f{{\left({1}\right)}}}={12}$$ Find $$\displaystyle{k}{e}{r}{\left({f}\right)}$$ Jason Yuhas 2022-01-15 Answered

Difficulty showing that a group G applied on $$\displaystyle{X},{G}_{{{x}}}{\left({x}\in{X}\right)}\ \text{and}\ {G}_{{{y}}},\frac{{w}}{{y}}\in{G}{\left({x}\right)}$$ are same iff Gx is a normal subgroup of G jubateee 2022-01-15 Answered

If S is a set of functions from X to Y then I can consider the action of a group G on S via its action on X and Y by the formula $$\displaystyle{\left({g}\cdot{f}\right)}{\left({x}\right)}={g}\cdot{f{{\left({g}^{{-{1}}}\cdot{x}\right)}}}$$, So we are considering left actions both on X and Y. Then, the definition of equivariant map pops out but I don't really understand how it is related to previous statement. An function $$\displaystyle{f}:{X}\rightarrow{Y}$$ is equivariant if it satisfies $$\displaystyle{f{{\left({g}\cdot{x}\right)}}}={g}\cdot{f{{\left({x}\right)}}}\forall{g}\in{G}$$. What's happening here? Are we assuming that the group G acts trivially on Y? How to get this definition from the previous statement? stop2dance3l 2022-01-15 Answered

Let G be a group and H a subgroup of G. Show that $$\displaystyle{a}{H}={H}{a}$$ if and only if $$\displaystyle{a}{h}{a}^{{-{1}}}\in{H}$$ for every $$\displaystyle{h}\in{H}$$. Suppose that $$\displaystyle{a}{H}={H}{a}$$. Then for $$\displaystyle{h}'\in{a}{H}$$ we have that $$\displaystyle{h}'={a}{h}$$ for some $$\displaystyle{h}\in{H}$$. Right multiplying by $$\displaystyle{a}^{{-{1}}}$$ we have that $$\displaystyle{h}'{a}^{{-{1}}}={a}{h}{a}^{{-{1}}}$$. As $$\displaystyle{a}{H}={H}{a}$$ we have that $$\displaystyle{h}'={a}{h}={h}{a}$$ so $$\displaystyle{h}'{a}^{{-{1}}}=\underbrace{{{h}{a}{a}^{{-{1}}}}}={a}{h}{a}^{{-{1}}}\Rightarrow{a}{h}{a}^{{-{1}}}\in{H}$$. Conversely suppose that $$\displaystyle{a}{h}{a}^{{-{1}}}\in{H}$$. The claim that $$\displaystyle{a}{H}={H}{a}$$ so we will do both inclusions. Let $$\displaystyle{h}'\in{a}{H}$$. Then $$\displaystyle{h}'={a}{h}$$ for some $$\displaystyle{h}\in{H}$$. Now right multiplying by $$\displaystyle{a}^{{-{1}}}$$ we ahve that $$\displaystyle{h}'{a}^{{-{1}}}={a}{h}{a}^{{-{1}}}\in{H}$$. So multiplying again from right I have that $$\displaystyle{h}'{a}^{{-{1}}}{a}={a}{h}{a}^{{-{1}}}{a}\Rightarrow{h}'={a}{h}\in{H}{a}$$. For the other direction let $$\displaystyle{h}{''}\in{H}{a}$$. Then $$\displaystyle{h}{''}={h}{a}$$ for some $$\displaystyle{h}\in{H}$$. Right multiplying by $$\displaystyle{a}^{{-{1}}}$$ then left multiplying by a and again right multiplying by $$\displaystyle{a}^{{-{1}}}$$ I get that $$\displaystyle{a}{h}{''}{a}^{{-{1}}}{a}^{{-{1}}}={a}{h}{a}^{{-{1}}}\in{H}$$. Now left multiplying by a I have that $$\displaystyle{a}{a}{h}{''}{a}^{{-{1}}}{a}^{{-{1}}}={a}{a}{h}{a}^{{-{1}}}\in{a}{H}$$. Is it necessarily true that $$\displaystyle{a}{a}{h}{''}{a}^{{-{1}}}{a}^{{-{1}}}={h}{''}$$ and that $$\displaystyle{a}{a}{h}{a}^{{-{1}}}={a}{h}$$? This would seem to conclude the result, but I'm not sure if it's allowed. Also is should I do these kind of ''multiplying'' here? It seems that I am going in circles multiplying everywhere. Sallie Banks 2022-01-15 Answered

Product of cosets of normal subgroup and well-definedness Let G be any group and N be a subgroup. $$\displaystyle{\left({a}{N}\right)}{\left({b}{N}\right)}={a}{b}{N}$$ Maria Huey 2022-01-15 Answered

Determine Group structure of given $$\displaystyle{2}\times{2}$$ matrix group over $$\displaystyle{\mathbb{{{F}}}}_{{{3}}}$$ Susan Nall 2022-01-14 Answered

Number of elements in $$\displaystyle{S}_{{{13}}}$$ conjugate to both $$\displaystyle{\left({12}\right)}{\left({34}\right)}$$ and $$\displaystyle{\left({123}\right)}{\left({45}\right)}$$ then is the answer 0? Because $$\displaystyle{\left({12}\right)}{\left({34}\right)}$$ and $$\displaystyle{\left({123}\right)}{\left({45}\right)}$$ have not got the same cycle structure. sunshine022uv 2022-01-14 Answered

Find a polynomial $$\displaystyle{P}_{{{n}}}\in{C}{\left[{X}\right]}$$ such as $$\displaystyle{S}_{{{n}}}={k}{e}{r}{P}_{{{n}}}{\left({D}\right)}$$ With $$\displaystyle{S}_{{{n}}}{\left({n}\geq{1}\right)}$$ all of the functions such as $$\displaystyle{y}^{{{\left({n}\right)}}}={y}$$ (where $$\displaystyle{y}^{{{\left({n}\right)}}}$$ is the nth derivative of y) D the endomorphism sending the functions $$\displaystyle{C}^{{\infty}}$$ on their derivatives $$\displaystyle{D}:{y}\rightarrow{y}'$$ I already proved that D is an endomorphism and $$\displaystyle{S}_{{{n}}}$$ a vector space containing the functions $$\displaystyle{k}{e}^{{{x}}}$$. But my probleme is that i dont really understand the meaning of $$\displaystyle{P}_{{{n}}}{\left({D}\right)}$$, and so dont see what $$\displaystyle{S}_{{{n}}}={k}{e}{r}\ {P}_{{{n}}}{\left({D}\right)}$$ could be. It probably have to do with eigenvalues, eigenvectors...But really I dont see anything. Wanda Kane 2022-01-14 Answered

Proof attempt: Since f is irreducible and has a as root then K(a) is isomorphic to $$\displaystyle{\frac{{{K}{\left[{x}\right]}}}{{{\left({f}\right)}}}}$$ with the isomorphism $$\displaystyle{X}+{f}\rightarrow{a}$$. And we can do the same for b and we get $$\displaystyle{K}{\left({b}\right)}$$ is isomorphic to $$\displaystyle{\frac{{{K}{\left[{x}\right]}}}{{{\left({f}\right)}}}}$$. Therefore $$\displaystyle{K}{\left({a}\right)}$$ is isomorphic with $$\displaystyle{K}{\left({b}\right)}$$. a) f is irreducible and has a as root then $$\displaystyle{K}{\left({a}\right)}$$ is isomorphic to $$\displaystyle{\frac{{{K}{\left[{x}\right]}}}{{{\left({f}\right)}}}}$$ we took it for granted in the lecture but why does this hold? maduregimc 2022-01-14 Answered

Consider $$\displaystyle\overline{{{F}}}_{{{p}}}$$, the algebraic closure of $$\displaystyle{F}_{{{p}}}$$. I want to see that: for every proper subfield $$\displaystyle{K}\leq\overline{{{F}_{{{p}}}}},\frac{\overline{{{F}}}_{{{p}}}}{{K}}$$ is not a finite extension. It is known that, and can be somewhat easily shown that $$\displaystyle\overline{{{F}}}_{{{p}}}=\cup_{{{n}\geq{1}}}{F}_{{{p}^{{{n}}}}}$$. Now, if any of the proper subfields have the form $$\displaystyle{F}_{{{p}^{{{n}}}}}$$, it is easy enough to see that $$\displaystyle\overline{{{F}}}_{{{p}}}\ne{F}_{{{p}^{{{n}}}}}{\left(\alpha_{{{1}}},\cdots,\alpha_{{{m}}}\right)}$$ for some $$\displaystyle\alpha_{{{i}}}$$ by going high up enough, i.e, to some big enough m such that $$\displaystyle\alpha_{{{i}}}\in{F}_{{{p}^{{{m}}}}}\subseteq\overline{{{F}_{{{p}}}}}$$ The problem is characterizing the proper subfields. Is every subfield of $$\displaystyle\overline{{{F}_{{{p}}}}}$$ going to have this form? Can we have an infinite intermediate subfield? Roger Smith 2022-01-14 Answered

A function f is a relationship between sets, say A and B...we denote this function relation as $$\displaystyle{f}:{A}\rightarrow{B}\ldots$$ $$\displaystyle{A}\times{B}$$ denotes the set of ordered pairs of elements from A and B... An operation is a function of the form $$\displaystyle{f}:{A}\times{B}\rightarrow{C}$$. One should think of an operation as a process of bringing two objects together and creating a third operation. what does: ''An operation is a function of the form $$\displaystyle{f}:{A}\times{B}\rightarrow{C}$$. One should think of an operation as a process of bringing two objects together and creating a third operation.'' exactly mean? what would a good example look like? Annette Sabin 2022-01-14 Answered

If H and K are subgroups of ' order p, show that $$\displaystyle{H}={K}\ \text{or}\ {H}\cap{K}={1}$$ My thought would be the Lagrange Theorem. If H is a subgroup of order p of group G, then |H| divides |G|. If |G| is divided by two subgroups with the same order, then the result is the same. The condition where the subgroups are are the same makes sense (mostly), but the other condition makes no sense. I honestly have no idea how one would prove this. Edit: Prime group is cyclic. Cyclic group generated by a single element. If the groups aren't the same, $$\displaystyle{\left({H}={K}\right)}$$, then the only element they have in common is the identity element (in this case 1)? Nicontio1 2022-01-14 Answered

Determine $$K\triangleleft G,\ \frac{G}{K}\simeq H_{1}$$ and $$\displaystyle{K}\stackrel{\sim}{=}{H}_{{{2}}}$$ feminizmiki 2022-01-14 Answered

Which of the following is not correct? a) $$U(14)$$ is isomorphic to $$S3$$ b) $$\displaystyle{U}{\left({14}\right)}$$ is isomorphic to $$\displaystyle{U}{\left({7}\right)}$$ c) $$\displaystyle{U}{\left({8}\right)}$$ is isomorphic to $$\displaystyle{U}{\left({12}\right)}$$ d) none of these osteoblogda 2022-01-14 Answered

Show that an intersection of normal subgroups of a group G is again a normal subgroup of G Deragz 2022-01-14 Answered

Prove that $$\displaystyle{\left\lbrace{a}+{b}{i}:{a},\ {b}\in{R}\right\rbrace}$$ is a subring of the real Hamilton Quaternions H, which is a field but it is not contained in the center of H. lunnatican4 2022-01-14 Answered

Finding the Units in the Ring $$\displaystyle{\mathbb{{{Z}}}}{\left[{t}\right]}{\left[\sqrt{{{t}^{{{2}}}-{1}}}\right]}$$ Arthur Pratt 2022-01-14 Answered

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