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0

It depends on the field $F$. For example, if $F$ is finite then $\operatorname{PSL}(2,F)$ is certainly co-hopfian, since it's finite. But if $F$ contains a proper subfield $F'$ isomorphic to $F$ then $\operatorname{PSL}(2,F')<\operatorname{PSL}(2,F)$, so it's not co-hopfian: for example, if $\mathbb{F}_p(x)$ is the field of rational functions, ...


0

Considering S3 the permutation group. Here O((1,2))=2 & O((1,2,3))=3 & NO element of order 6=lcm(2,3). So the quetion is wrong. `


1

The claim is wrong. Consider the Heisenberg group $\begin{pmatrix} 1 & \mathbb{Z}/3 & \mathbb{Z}/3 \\ 0 & 1 & \mathbb{Z}/3 \\ 0 & 0 & 1 \end{pmatrix}$. It is a non-abelian group of order $27$ and satisfies $g^3=1$ for all $g$.


1

If $m\mid n$ or $n\mid m$, there is nothing to be shown. Let $a,b$ have orders $m,n$ and let $d=\gcd(m,n)$. Then consider $x=a+b$. Then $mx=mb\ne 0$, $nx=na\ne0$ and $\frac{nm}dx = \frac nd mx =\frac nd mb=\frac mb nb=0$ we conclude that the order of $x$ divides $\operatorname{lcm}(m,n)$, but neither $m$ nor $n$.


0

is it true that $Z(G/Z(G)) \ne \{1\}$ iff $\exists a \in G - Z(G)$. such that the generated group $ \langle Z(G),a \rangle $ is a normal subgroup (not necessarily proper) of $G$?


2

I suspect yours is an incomplete quote of Grün's Lemma: If $G$ is a perfect group, then $Z(G/Z(G)) = \{ 1 \}$. Here $G$ is a perfect group if $G = G'$. The Wikipedia link above gives the short proof, which is based on the Three Subgroups Lemma - in every group one has that the second center commutes with the derived subgroup. (There's a more general ...


1

I don't believe that this is true. In fact I don't even believe that there is always a class $2$ normal nilpotent subgroup. Let $N = {\mathbb Z}^n$ be free abelian with generators $x_1,\ldots,x_n$, let $y$ act on $N$ by $x_i^y = x_ix_{i+1}$ ($i<n$), $x_n^y=x_n$, and let $G = N \rtimes \langle y\rangle$ be the semidirect product. Then $G$ is nilpotent of ...


2

To start with the obvious, to verify that $\langle X \rangle = G$, take a finite generating set $Y$ of $G$, and systematically try to prove that every element of $Y$ is a product of the elements of $X$ and their inverses. That's naive, but it is worth trying before you do anything else, especially if you have a computer to do the matrix multiplications. It ...


2

This is false: $Q_8/\langle -1\rangle\cong V_4$, which is abelian and thus has a nontrivial center.


2

By exhibiting a nonidentity mapping. There is a certain mapping, defined on any group, that you already know of (it's inherent in one of the axioms of a group). Another exercise in group theory asks a student to show this mapping is a group homomorphism iff the group is abelian. Can you guess which map I'm talking about?


1

$G$ is cyclic, say of order $n>2$. Then $\varphi(n)>1$. So...? Spoiler Just for the sake of addressing Gina's comment: if $G$ is abelian and not $2$-elementary, $x\mapsto x^{-1}$ is a group automorphism that is not the identity. If $G$ is non abelian and $g$ is a noncentral element, then $x\mapsto gxg^{-1}$ is a nonidentity group automorphism. ...


2

Indeed since $N\lhd G$ then for all $g\in G$ we have $N^g\subseteq N$. Now do this for $g^{-1}\in G$ so we see that $N^{g^{-1}}\subseteq N$ and therefore $$(N^{g^{-1}})^g\subseteq N^g$$ But what is $(N^{g^{-1}})^g$ then?


1

Note that I did what you were looking for in light of Derek's neat hint You see that $[G:\langle y\rangle]=8$ so the possibilities for $|G|$ could be $24$.


2

The orthogonal group $O(3)$ has a nontrivial subgroup consisting of the identity matrix and its negative. These lie in the center of $O(3)$, so the subgroup is normal. To show that $SO(3)$ is normal, you might want to use the fact that $SO(3)$ is precisely the set of matrices in $O(3)$ with determinant $1$. This allows you to characterize cosets of $SO(3)$ ...


3

Let $H$ be a subgroup of $G$ with order $p$ and $K$ be a subgroup of order $q$, It is easy to see that $H$ must be normal in $G(*)$ Then $K$ act on $H$ by conjugation so we have $\phi:K\to Aut(H)\cong \mathbb Z_{p-1}$. a) If $q\nmid p-1$ then $\phi$ is trivial homomorphism which implies $G=HK\cong H\times K$ which conclude that $G$ is cylic. b) if $q\mid ...


1

Let $\{g_1,g_2,\dots,g_k\}$ be a maximal linearly independent subset of $G$ and let $H=\langle g_1,g_2,\dots,g_k\rangle$ be the subgroup of $G$ generated by those elements. The quotient $G/H$ is torsion, because of maximality, so from the exact sequence $$ 0\to H\to G\to G/H\to 0 $$ we get, tensoring by $\mathbb{Q}$, that $H\otimes\mathbb{Q}$ is isomorphic ...


3

If we do not require that the class of a subgroup generated by two elements is bounded by some fixed constant, here is one example. Consider the infinite direct sum $G = G_1 \oplus G_2 \oplus G_3 \oplus \cdots$ where $G_i$ is nilpotent of class $i$. Then $G$ is not nilpotent, since it has nilpotent subgroups of arbitrarily large class. But any finitely ...


5

When $G$ is finite the answer is "yes", since any two elements of coprime order commute, so a Sylow $p$-subgroup is normal for each prime divisor $p$ of the group order. When $G$ is infinite, I don't know, but others might. Each element of $G$ is certainly an Engel element, but there are non-nilpotent groups in which every element is an Engel element, ...


1

The last statement in your question (which is not a mere question) also suggests a solution of the more general problem, namely the $n$-variable analog of the Hall-Witt formula for any $n\geq 2$. Let $x_1$, $\ldots$, $x_n$ be the variables. If $w=w(x_1,x_2,\ldots,x_n)$ is any word in $x_i$'s and their inverses, define the word $\gamma w$ by $(\gamma ...


3

This is not true, $\operatorname{Aut}(X)\neq A\ast B$. First, note that the tree you construct is a biregular tree, where $A$ acts by fixing a vertex and permuting its edges (and the trees extending from these edges). The group $B$ acts similarly, and this extends to a faithful action of $A\ast B$. Theorem: $\operatorname{Aut}(X)\neq A\ast B$ To see ...


2

Assuming that you know that $f$ fixes every $3$-cycle, it follows that $f$ also fixes every even permutation (because $A_4$ is generated by $3$-cycles). Now $A_n$ is the commutator subgroup of $S_n$ when $n \geq 3$ (proof: write a $3$-cycle as the commutator of two transpositions). So the parity of an permutation is invariant under an automorphism of $S_n$. ...


1

The orbits of $\lambda_g$ are exactly the right cosets $\langle g\rangle/G$. On each orbit, $\lambda_g$ is cyclic of odd length thus has even sign. Also $\lambda_g$ decomposes into $|\langle g\rangle/G|$ number of disjoint cycles. Therefore the sign of $\lambda_g$ is the sum of signs of the disjoint cycles, which is even. For the converse, one can use ...


4

This is Theorem 17.2 in Fuchs' "Infinite Abelian Groups" (vol.1), the result is originally due to Prüfer and Baer: Theorem. A bounded Abelian group is a direct sum of cyclic groups. As a corollary you obtain that any indecomposable bounded Abelian group is cyclic.


1

The answer would seem to be no. A finitely generated Abelian group is a direct product of finitely many cyclic groups (some of which may be infinite). In your situation, the cyclic factors would be of bounded order, so if your group were finitely generated, it would be finite (and even cyclic by the indecomposability condition you impose). Later edit: In ...


1

Say $H\neq K$. Then there exists an element in $K$ which is not there in $H$. Let that element be $k$. Then $kH\in G/H$ is not equal to identity. As $G/H$ and $G/K$ are finite, if we can prove the existence of an injective mapping that is not surjective between the two, we can prove that they are not of the same order. If you put these facts together, ...


1

For your first question, see my answer to this question. It says that if the group $G$ is not noetherian (in the sense that every subgroup is f.g.) then $K[G]$ is not noetherian for any field $K$. Now pick, for instance, any non-virtually-polycyclic f.g. solvable group, such as solvable Baumslag-Solitar groups $BS(1,n)$ for $|n|\ge 2$, or a lamplighter ...


1

Cardinality is the only invariant of the underlying set of a group that matters. If $(G, \circ)$ is a group and you have a bijection of sets $f: G \rightarrow T$, then you can define a group operation $*$ on $T$ so that $f$ is an isomorphism. In fact the operation $*$ is determined by the requirement that $f$ should be an isomorphism. If $f$ is an ...


4

Of course! You could just take some bijection between $\mathbb{Q}\setminus\{0\}$ and $\mathbb{Z}$ and artificially make it a homomorphism. You then get the group structure for free, because you made sure it was a homomorphism and because it is a bijection. Bijections are just re-labellings, so that is all that is going on. If I relabel $3/5$ as "$7$", $2/3$ ...


1

This is a common exercise in basic complex analysis so I will help you by outlining the steps of the proof: 1) Show that the class of functions of the form $\frac{az+b}{cz+d}$ with $ad-bc \neq 0$ send lines to lines and circles to circles. 2) Show that every bijective map $T(z): \mathbb{C} \to \mathbb{C}$ has the form $az+b$ with $a \neq 0$. 3) Extend (2) ...


1

A more abstract line: For any $\sigma \in \text{Aut}(G)$, one can (easily) show that $\sigma(C(U)) = C(\sigma(U))$. The result then follows by taking $\sigma$ to be conjugation by $x$.


0

the crucial fact is that the centralizer is defined entirely in terms of constraints of the form $$ ha = ah $$ which are trivially preserved under any automorphism. in particular if $h^x = xhx^{-1}$ then we have: $$ h^xa^x = a^xh^x $$ this is to say that if $h$ commutes with $a$ iff $\forall Ax\in G$ $xhx^{-1}$ commutes with $xax^{-1}$


0

Your way is correct, so far. To complete it, we show that $u$$x^{-1}zxu^{-1}$ is in $C_G(U)$ Let $u_0 \in U$, we conclude: $$ux^{-1}zxu^{-1}u_0=u_0ux^{-1}zxu^{-1} $$ $\iff$ $$ux^{-1}zxu^{-1}x^{-1}xu_0=u_0x^{-1}xux^{-1}zxu^{-1}$$ $\iff$ $$ux^{-1}xu^{-1}x^{-1}zxu_0=u_0x^{-1}zxux^{-1}xu^{-1} $$ $\iff$ $$x^{-1}zxu_0=u_0x^{-1}zx $$ $\iff$ ...


0

Suppose $z\in C_G(xUx^{-1})$. We want to show that $z\in xC_G(U)x^{-1}$, i.e., that $z=xyx^{-1}$ for some $y\in C_G(U)$, i.e., that $y=x^{-1}zx\in C_G(U)$. Thus, we have, for arbitrary $u\in U$: $yu=x^{-1}zxu=x^{-1}zxux^{-1}x=x^{-1}xux^{-1}zx=ux^{-1}zx=uy$, so it looks like y is in that centralizer. Does that explain the forward direction well enough?


3

It looks like the group you want is the wreath product $C_{m} \wr S_{n},$ where $C_{m}$ is the cyclic group of order $m.$ This group may be visualized in several ways. It is isomorphic to the group of all monomial matrices whose non-zero entries are complex $m$-th roots of unity. (A monomial matrix is one which has one non-zero entry in each row and one ...


0

Let $K$ be the kernel of $f.$ We can say that $N$ is isomorphic to a subgroup of $G/K,$ since your condition guarantees that $f$ embeds $N$ isomorphically as a subgroup of ${\rm im} f.$


3

$kerf\cap N=e$ so $Nkerf\cong N\times kerf$ and $Nkerf$ is normal in $G$. if $|G:N|=n$ then $f(g)^n=g^n$ for all $g\in G$. $\bar f:G/N\to G/N$ by $gN\mapsto f(g)N$ is an homomorphism. proof: obvious. since $g^n\in N$ and $f(x)=x for \ \ x\in N$ the result follows. since $f(gx)=f(g)x$ for $x\in N$ the result follows.


2

The group $G$ is supposed to be abelian. The subgroup generated by the persistent element must be simple. The group can't have any other simple subgroups, because the sum of simple subgroups is a direct sum of simple groups. Also the group must be indecomposable, because the persistent element can't belong to both a direct summand and its complement. If $H$ ...


2

It is true that $(ab)^n = e = (ba)^n$ but that doesn't mean that $n$ is the order of $(ba)$. We can only say that the order of $(ba)$ divides n. Since we don't know if $n$ is the smaller power $k$ such that $(ba)^k = e$. To prove that $n$ is also the order of $ba$ you have to complete the proof. Let $k$ be the order of $(ba)$ then: ...


2

I think it's because you have shown that if the order of $ab$ is $n$, then $(ba)^n=e$. This is not the same as concluding that the order of $ba$ is $n$. For that to be the case, it must also be shown that no smaller power will also do the job. I think you can only conclude that the order of $ba$ divides $n$. After all, it is also true that $(ab)^{7n}=e$, ...


3

Cauchy's theorem states that there exists an element of order $3$, not that every element has order $3$. (You might be confusing this with Lagrange's theorem, which says that every element has order (dividing) $|G|$ for $G$ finite.) It also looks like you're assuming $G$ is cyclic. It's just abelian; you could have, for example, $G = \mathbb{Z}_3 \oplus ...


4

Let's consider what is the condition such that $G$ would have such element $a$. For simplicity, only finite $G$ are examined. First, $|G|$ must be a prime power, otherwise it is false by Cauchy's. Let $|G|=p^{n}$ ($p$ prime, $n\geq 1$). Then any abelian subgroup of $G$ must be cyclic, by the proof using cyclic decomposition as outline in the comment: for ...


2

As you may know, a vector space is a set $V$ together with operations $+:V \times V \to V$ and $\cdot:K \times V \to V$ that satisfy certain conditions, where $K$ is a field (take $K = \mathbb{R}$ for instance). Turns out that these conditions makes $(V,+)$ into an abelian group, a fancy term for a commutative group. This means that if you take $V$ and ...


1

Since $G$ is finitly generatet abelian group; $$G\cong \mathbb Z^n\oplus \mathbb Z_{n_1}\oplus...\mathbb Z_{n_k}$$ From the question; we have $G\cong \mathbb Z^n\oplus \mathbb Z_{n_i}$ where $n_i=p^n$. since intersection of components trivial in the direct sum, $G\cong \mathbb Z$ or $G\cong \mathbb Z_{p^n}$. But in $\mathbb Z$, there is no integer which ...


0

Defining scalar products in vector spaces allows you to introduce notions of angle between two vectors, and also gives a definition of the length of a vector.


2

Since $G/N$ is a group by N normal, consider $g\in G/N$. $\vert G/N\vert$=d. So by lagrange theorem, any element of $G/N$'s order($\langle g\rangle$ is a subgroup of $G/N$ and consider its left or right cosets) divides d. So $g^{d}$=e as the order of g divides d.


2

the order $(xN)$ divide $d$ then $(xN)^d=e= N$, and $x^d\in N$.


1

This is proved in section 1.7 of Dummit and Foote's Abstract Algebra. The permutation representation is a homomorphism, it is not an isomorphism in general. For example, the trivial group $G = \{e\}$ acts on $\{1, 2\}$ by the identity (i.e. $e\cdot 1 = 1$ and $e\cdot 2 = 2$), so the associated permutation representation is a group homomorphism $G \to S_2$, ...


2

Virtually all questions involving groups defined by finite presentations are theoretically undecidable in general, so you cannot hope to find a general procedure for solving these sorts of problems. The answer to your specific question is no. I adjoined the generators of $A$ to the relations of $G$ to get the quotient of $G$ by the normal closure of $A$, ...


0

I'm not so sure what you mean. There seem to be a great number of questions in your post. Here's what I'll answer: Suppose I have a group G, given as finitely many generators and finitely many relations, say (What?). Furthermore, suppose I have two finite sets A,B of elements of the group. I would like to know if the subgroups generated by A and B are ...


3

Answer in the form of a keyword which will answer your question and its generalization to $\mathbb Z^n$: Smith normal form.



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