All questions with tag [math: group-theory]


If $H$ is a subgroup of $\mathbb Q$ then $\mathbb Q/H$ is infinite

I'm trying to work out this question: Prove that if $H$ is a proper subgroup of $\mathbb{Q}$ then $\mathbb{Q}/H$ is infinite, but each of its elements have finite order. I thought, for the first part, that I could assume for contradiction that $\mathbb{Q}/H$ is finite of order $n$, then for all $\dfrac{a}{b}\in\mathbb{Q}$, $\dfrac{a^n}{b^n}$ is ...
2022-07-25 17:47:17

Free abelian group $F$ has a subgroup of index $n$?

Suppose that we have a free abelian group $F$. How can it be proved that $F$ has a subgroup of index $n$ which $n≥1$? Honestly, according to the Theorems, I just know that if we take $X$ as a base for $F$, then $$ F= \bigoplus_{\alpha \in X} \mathbb Z_\alpha \ $$ in which for all $ \alpha \in X$; $\mathbb Z_\alpha \ $ is a copy of $ \mathbb Z $....
2022-07-25 17:47:13

subgroup structure of $S_4$

In the checklist of Young subgroups of $S_4,$ we locate $\langle(12)\rangle, \langle(13)\rangle, \langle(14)\rangle, \langle(23)\rangle, \langle(24)\rangle, \langle(34)\rangle,$ yet we do not locate $\langle(12)(34)\rangle, \langle(13)(24)\rangle,\langle(14)(23)\rangle,$ while they are all isomorphic to $S_2.$ I'm perplexed.
2022-07-25 17:46:51

If every element of a monoid is right invertible, then every element is invertible

Let $G$ be a set with associative binary procedure and also a device. Think that for every single $g\in G$ there exists $x \in G$ with $gx = 1$. Confirm that $xg = 1$ issues.
2022-07-25 17:46:33

how to calculate order cyclic multiplicative group of quadratic residues.

Let N = pq be one publicly known RSA modulus, in which p = 2p′ + 1, q = 2q′ + 1 are two large primes. p′ and q′ are also primes. All the quadratic residues modulo N forms a multiplicative cyclic group of order p′q′. My question is , how is this order computed? Thank you for your attention!
2022-07-25 17:46:14

Abel-Ruffini Theorem Clarification

Let $n \geq 5$. I want to show that $\exists p\in \mathbb{Q}[X]$ of degree $n$ with roots which are not possible to find via radicals and rational functions from the coefficients of $p$. I've got the following theorem (Abel-Ruffini) written down in my notes. (1) $\exists$ transcendental $\alpha_1,\dots\alpha_n\in\mathbb{C}$ s.t. $F=\mathbb{Q}(\a...
2022-07-25 17:44:21

A non-abelian group of order 12 with only one element of order 2.

I require to show that, if I have a non - abelian team G of order 12 with just one component has order 2, after that G is soluble and also the facility Z (G) is such that $Z(G)\cong \mathbb{Z}_2$ and also $\frac{G}{Z(G)}\cong S_3$ I do not recognize just how to address this trouble, and also any kind of aid would certainly be most welcome.
2022-07-25 17:43:52

Constructing Young subgroups of $S_4$

Given the symmetric group $S_4$ and a subgroup $H\subset S_4$ I want to construct a $Y\subset S_4$ such that $Y$ is minimal, meaning that there is no other young subgroup $Y'$ such that $H\subset Y'\subset Y$. My understanding of the problem is in two ways: 1) Consider a partition of the set $\{1,2,3,4\}$. A Young subgroup is the direct product...
2022-07-25 17:43:26

Subgroup generated by a set

A subgroup created by a set is specified as () : More usually, if S is a part of a team G, after that, the subgroup created by S, is the tiniest subgroup of G having every component of S, suggesting the junction over all subgroups having the components of S ; equivalently, is the subgroup of all components of G that can be shared as the limite...
2022-07-25 17:43:15

Is it true that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as abelian groups?

I think the answer is yes. Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\lambda_1},e_{\lambda_2}):\lambda_1,\lambda_2\in\Lambda\}\subset\mathbb{R}^2$ is a Hamel basis of $\mathbb{R}^2$. Consider set theoretic bijection $i:\La...
2022-07-25 17:38:59

commutator identity on planetmath

In there is an identification of commutators which I assume is incorrect. This identification is the 4th identification in Theorem 5 of this web page. I assume that the proper identification is this set, calculated by me: PlanetMath identification: $[x^z,y]=\left[x,y^{z^{-1}}\right]$ , where $x^z=z^{-1}xz$ Identity calculated by me: $[x^z,y...
2022-07-25 17:38:37

Free Group Generated By Image

Let $S$ be any set and let $f:S \rightarrow F$ denote the free group on $S$. By definition, this means that for any group $X$ and any function $g:S \rightarrow X$ there exists a unique homomorphism $h:F \rightarrow X$ such that $h \circ f = g$. How can I show that $f(S)$ generates $F$? By "generate $F$" I mean that the intersection of ...
2022-07-25 17:18:31

Showing two abelian groups have the same rank

Let $G=C_{p^{a_{1}}}\times C_{p^{a_{2}}}\times...\times C_{p^{a_{t}}}$ where $a_{1}\geq a_{2}\geq...\geq a_{t}$ and $H\subseteq G^{P^n}$ for some integer $n$. Please prove if $n>a_{k}$ for some $k\in\{1,...,t\}$ then, $G$ and $\frac{G}{H}$ have equal rank. The rank $G$ is minimal number generators of $G$.
2022-07-25 17:18:13

Rubik's Cube Not a Group?

I read online that although the 3x3x3 is a wonderful instance of a mathematical team, bigger dices aren't teams in all. Just how can that hold true? There is clearly an identification and also it is shut, to make sure that have to suggest that some actions aren't invertible. Yet this appears not likely to me.
2022-07-25 17:05:31

Nilpotent groups are solvable

I know this should be obvious but somehow I can't seem to figure it out and it annoys me! My definition of nilpotent groups is the following: A group $G$ is nilpotent if every subgroup of $G$ is subnormal in $G$, or equivalently if $U<N_G(U)$ for all $U<G$. And my definition of solvable groups is that a group $G$ is solvable if $U'...
2022-07-25 17:05:24

Finding generators for commutative encryption

The paper Information Sharing Across Private Databases presents a protocol for finding an intersection of two private sets. They use a commutative encryption that is quite similar to the Diffie–Hellman key exchange and operate in $\mathbb{Z}_p^*$ with $p$ being a safe prime. Basically they hash each set element $x$ and then encrypt it with their...
2022-07-25 16:53:18

Generators for $PSL(2,\mathbb{Z})$ with a specific property

Does there exist generators $S$ and also $T$ for the modular team $\Gamma=PSL(2,\mathbb{Z})$ with the adhering to property: $$S+S^{-1}+T+T^{-1}=0$$ Here is a candidate: $$S=\left[\begin{array}{cc} -1 & 0 \\ 1 & -1 \\ \end{array}\right], \,T=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right]$$ Just no...
2022-07-25 16:44:20

Double Coverings of the Double Torus

I'm trying to count all the double coverings of the double torus. I know that the fundamental group of the double torus is $$\pi_1(X)=\langle a,b,c,d;[a,b][c,d]\rangle $$ where $[a,b]=aba^{-1}b^{-1}$. I also know from the classification theorem for covering spaces that in fact it suffices to count subgroups of index 2. I'm not sure how to do th...
2022-07-25 16:43:03

Finite subgroup

In the following problem, the only parts I didn't understand were c) and e). The remaining I did. Please, help me! Let $A,B$ be subgroups of $G$ that normalize each other. Assume that the set $$X=\{[a,b]:a\in A,b\in B\}$$ is finite. We will show that $[A,B]$ is finite. Note first that, without loss of generality, we may assume that $G=AB$. W...
2022-07-25 16:42:08

Proof via Group Theory : $\mathrm{lcm}(a,b) \cdot \gcd(a,b) = |ab|$

Recently, I was informed that we can verify the famous formula about $\mathrm{lcm}(a,b)$ and $\gcd(a,b)$ which is $$\mathrm{lcm}(a,b)=\frac{|ab|}{\gcd(a,b)} $$ via group theory. The least common multiple of two integers $a$ and $b$, usually denoted by $\mathrm{lcm}(a,b)$, is the smallest positive integer that is a multiple of both $a$ and $b$ a...
2022-07-25 16:42:01