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0 votes
1 answer
47 views

For $k$-algebras $B_1, \dots, B_n$, $\# \operatorname{Hom}_k( \prod_{i=1}^nB_i, \Omega) = \Sigma_{i=1}^n \# \operatorname{Hom}_k(B_i, \Omega)$?

Let $k$ be a field with $ k \subseteq \Omega$ a algebraically closed field. Let $B_1 , \dots, B_n$ be ( possibly finite local ) $k$-algebras. Then next equality of cardinals holds $$ \# \operatorname{...
Plantation's user avatar
  • 2,656
1 vote
0 answers
44 views

Can a countable union of subgroups of uncountable index in G be equal to G? [closed]

Let G be a group and $\{H_i\}_{i<\omega}$ be a countable family of subgroups of $G$, each of them of uncountable index. Can $G=\bigcup_{i<\omega} H_i$?
Moreno Invitti's user avatar
0 votes
1 answer
122 views

Group action & cardinality of a set.

You can find here more details and explanation on this question. Question: Let $n$ be a non-negative integer. For any family $ (i_1, \ldots, i_r) $ of non-negative integers such that $ i_1 + \ldots + ...
OffHakhol's user avatar
  • 719
2 votes
2 answers
106 views

Torsion-free commutative groups of a given cardinality

If an abelian group $G$ is torsion-free, it has at least one subgroup isomorphic to $\Bbb{Z}$, given by $⟨g⟩ := \{ g^n | n \in \Bbb{Z} \}$, for any non-identity element $g$. This subgroup is obviously ...
Rivers McForge's user avatar
3 votes
1 answer
96 views

Theorem 5, Section 1.4 of Hungerford’s Abstract Algebra

If $K,H,G$ are groups with $K\lt H \lt G$, then $[G:K]=[G:H][H:K]$. If any two of these indices are finite, then so is the third. Proof: By Corollary 4.3 $G= \bigcup_{i\in I}Ha_i$ with $a_i \in G$, $|...
user264745's user avatar
  • 4,249
1 vote
0 answers
47 views

Uncountability of the set of functions from the set of natural numbers to itself

I'm new to this website so forgive me for any formatting mistakes or other errors you encounter. I would appreciate it if you pointed them out. For anyone interested, the problem is from Algebra by ...
Oliver Golde's user avatar
0 votes
0 answers
45 views

The cardinality of the maximal linearly independent susbet of a free module over a commutative ring with the identity

Let $R$ be a commutative ring with the identity and $F$ be a free-module over $R$. I assume $X$ is the basis of $F$.I know any two bases of $F$ have the same cardinality. My question is that can we ...
fusheng's user avatar
  • 1,159
2 votes
2 answers
136 views

Do hypercontinuous fields exist?

"Hypercontinuity" is a cardinality of a continuous set's power set (set of all subsets). When talking about fields, I mean the cardinality of field's set. At first glance there is nothing ...
nbvehbectw's user avatar
6 votes
2 answers
275 views

Degree of a field extension by a transcendental element

Let $F$ be a field, and let $F(x)$ be the field of fractions of the polynomial ring $F[x]$. I'm interested in the degree of the field extension $[F(x) : F]$. Obviously it is infinite, but what exactly ...
Sambo's user avatar
  • 6,610
-1 votes
1 answer
163 views

Cardinality of double cosets

We show that the sets of double cosets $K\backslash G/H$ is in bijection with $K\backslash(G/H)$. So $\vert K\backslash(G/H) \vert = \vert G/H \vert /\vert K \vert = (\vert G \vert / \vert H \vert)/\...
roi_saumon's user avatar
  • 4,256
0 votes
1 answer
56 views

Ratio of two infinite cardinal numbers

Suppose $G$ is the group of all functions between $[0,1]\to\mathbb{Z}$. Let $H$ be the subgroup defined as $H=\{f\in G: f(0)=0\}$. Then, what can be said about the cardinality of $H$ and its index in $...
vidyarthi's user avatar
  • 7,085
3 votes
2 answers
792 views

Number of elements in a Group such that $x^7=e$

Given that $G$ is a finite Group. Prove that number of elements in G such that $x^7=e$ where $x \in G$ is always Odd. My attempt: First possibility is a Trivial group since $e^7=e$. Trivial Group ...
Umesh shankar's user avatar
15 votes
1 answer
382 views

Must large (infinite) groups have large automorphism groups?

For every cardinal $\kappa$, is there a cardinal $\lambda$ such that for all groups $G$ with $|G| > \lambda$, we have $|\mathrm{Aut}(G)| > \kappa$? I believe a similar result holds for finite ...
Beren Gunsolus's user avatar
3 votes
1 answer
184 views

How to decide the cardinality of $\{\text{all group isomorphisms from }(\mathbb{R},+)\text{ onto }(\mathbb{R}^+,\cdot)\}$?

The additive group of reals $(\mathbb{R},+)$ and the multipilicative group of positive reals $(\mathbb{R}^+,\cdot)$ are isomorphic, and $x \mapsto \exp(x)$ is one isomorphism from $(\mathbb{R},+)$ ...
user avatar
1 vote
1 answer
660 views

Let $S$ be the set of all possible functions mapping $\{\sqrt 2, \sqrt 3, \sqrt 5, \sqrt 7 \}$ to $\Bbb Q$, find the cardinality of $S$.

Let $S$ be the set of all possible functions mapping $\{\sqrt 2, \sqrt 3, \sqrt 5, \sqrt 7 \}$ to $\Bbb Q$, find the cardinality of $S$. At first I wanted to use the theorem that for any sets $A$ and ...
WaterBro's user avatar
  • 221

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