New answers tagged

2

You have proven that $h(ax)=a\,h(x)$ for all $a\in G$. With $x:=e$, we get $h(a)=a\,h(e)$ for every $a\in G$. If $b$ is the inverse of $h(e)$, then [...]. I leave the rest to you.


1

In universal algebra this concept is also known as the kernel of a morphism $f : X\to Y$. However, topological spaces are not universal algebras and calling these things "kernels" is not a good idea, because the term is reserved for a better known concept already. Incidentally, in a typical "concrete category of sets with some structure", like $\mathsf{Top}$...


0

$H=\{d^j: 1\leq j\leq $ ord$(d)\}$ $\pmod n $ where ord$(d)$ is the least $m\in \mathbb Z^+$ such that $d^m\equiv 1 \pmod n . $ And $|G/H|=|G|/|H|=\phi (n)/$ ord$(d).$......($\phi$ is the totient.)


0

Let $M=\mathbb{R}$. There exist (uncountably many) functions $f:\mathbb{R}\to\mathbb{R}$ which satisfies $f(x+y)=f(x)+f(y)$ but not linear (to prove it, we use Hamel basis, basis of $\mathbb{R}$ as $\mathbb{Q}$-vector space). Although this functions are not constructible (use AC), I think $f\circ g\neq g\circ f$ for appropriate Hamel functions $f, g:\mathbb{...


3

An abelian group is simple iff it is finite of prime order $\;p\;$ , and in that case its automorphism group is cyclic of order $\;p-1\;$ , so all its automorphisms commute with each other. If $\;f\;$ is an endomorphisms of such a cyclic group of order a prime then it is either the trivial homomorphism or the identity one (as it has no non-trivial subgroups)...


0

the notion of matrical product must respect how the matrix product is done, and also how can represents a vector in $K^n$, a matrix $M$ that has $n$ rows and $m$ colones, it is called of type$(n, m)$. Let matrix $M$ and $N$ of type respectively $(n, m)$, $(l, k)$, then the matrix product $MN$ exists iff $m = l$ and in this case the matrix is of the type$(...


2

If you consider $\mathbb{R}^n$ as the space of row vectors, that is, $1\times n$ matrices, the only reason for using the transpose is to make composition of linear maps corresponding to matrix multiplication. Define $f_A(x)=xA^T$; if $B$ is another $n\times n$ matrix (invertibility is not of a concern), you also have $f_B(x)=xB^T$. Then $$ f_A\circ f_B(x)=...


0

Let $f_A(x)=xA^t$ (where $x$ is a row vector) and $g_A(x)=Ax$ (where $x$ is a column vector). Note that $$ f_A(x)^t=(xA^t)^t=Ax^t=g_A(x^t). $$ Hence, $f_A$ and $g_A$ are essentially the same transformation, except that $f_A$ uses row vectors and $g_A$ uses column vectors. If you begin with some $x\in\mathbb R^n$, treat it as a row vector, apply $f_A$ and ...


1

Yes, your solution is correct. The number of ways to choose the three fixed points is ${8 \choose 3}$, the number of ways to choose the transposition from the remaining $5$ points is ${5 \choose 2}$, and the number of ways to form a $3$-cycle from the remaining $3$ points is $(3-1)!=2$. Thus, the number in question is ${8 \choose 3} \cdot {5 \choose 2} \...


1

We want to find the number of permutations in $S_5$ having exactly one fixed point. This is the number of permutations of shape $(a)(bcde)$ or $(a)(bc)(de)$. The number of permutations having the first shape is $5 \times 3! = 30$, and the number of permutations having the second shape is $5 \times 3 = 15$, for a total of $45$ permutations.


0

For (BN5), we know that $G=BNB = UTNTU=UNU$ (since $T \le N$), and since $U \le \tilde{B}$, we get $\tilde{G}=\tilde{B}N\tilde{B} = \tilde{B}\tilde{N}\tilde{B}$. You have not stated (BN4) correctly. You need to prove that $\tilde{B}^n \ne \tilde{B}$ for any $n \in N$ that maps onto some $s \in S$. Since $T \unlhd N$, $n$ normalizes $T$, so if if $n$ ...


0

As Frank notes for M8, $G \leq SU_{2} (\mathbb{C})$. There is a 2-to-1 group homomorphism: $$SU_{2} (\mathbb{C}) \to SO_{3} (\mathbb{R})$$ This is the universal cover of $SO_{3} (\mathbb{R}) \cong \mathbb{R}P^{3}$ (Artin covers this-I believe- in chapter 9). Since $G$ is simple, this means $G \leq SO_{3} (\mathbb{R})$ also, which means that $G$ is a finite ...


1

Hint: the center $Z(D_{12}) \neq 1$


1

There is an index $2$ subgroup $C$ which is cyclic of order $33$, generated by some element $\sigma$, so $C=\langle \sigma\rangle$. There is another element $\tau$ which has order $2$. Clearly $\tau\not\in C$, and also clearly the two cosets are $C$ and $\tau C$. Indeed, we can take $\tau$ to correspond to reflection and $\sigma$ to rotation, so we also have ...


0

Many different ways are there to answer your question, but the answer you will expect will depend on what properties of these two groups you are familiar with. It would be worth for you to prove these properties, to clarify the answer. In dihedral groups, at least half elements are of order $2$; it contains an element of order equal to half the order of ...


0

If $G$ is commutative then the problem is done.So let us first assume that $G$ is not commutative.Then according to Jackson Hsu $|G/Z(G)|=4$.So $|Z(G)|=2$.Let $K=Z(G)$.Then $H \cap K$={$e_G$} or $K$.If the intersection contains only the identity then $|HK|=8$ i.e. $HK=G$.So $G$ becomes an internal direct product of $H$ and $K$.Hence $G \cong H \times K$.But ...


1

In $S_5$, a permutation will have one of the following cycle types (these are also the conjugacy classes of $S_5$): identity $1$ transposition $(1,2)$ double transposition $(1,2)(3,4)$ $3$-cycle $(1,2,3)$ $4$-cycle $(1,2,3,4)$ $5$-cycle $(1,2,3,4,5)$ disjoint $2$-cycle and $3$-cycle $(1,2)(3,4,5)$ The only cycle types that fix exactly one number are the ...


1

$D_{33}$ is the symmetry group of the regular polygon $A_1A_2\cdots A_{33}$ with 33 sides. If $\sigma$ is the generator of rotation cyclic group of the polygon and $\tau$ a reflection of the polygon on the altitude passing through $A_1$ (passing through $A_1$ and the mid point of the opposite side), then $\sigma^{33} = e$ and $\tau^2 =e$ where $e$ is the ...


0

Every reflection is by definition self-inverse, order 2. The smallest rotation $r_1$ (other than the identity) generates all other rotations in a cycle, including reaching the identity after $33$ steps. Some of the rotations will have shorter cycles, though, in the classic pattern of cyclic groups - specifically, for a rotation generated by $r_1^k$, it will ...


2

If $a$ is a fixed point, all the other symbols must be moved. This is same as the number of derangements on four symbols and hence 9 in number. Thus the total number of permutations with exactly one fixed point is $9 \times 5 = 45$. If we want the number of permutations in $S_n$ with exactly one fixed point, then it is \begin{equation*} n! \left(1- \frac{1}...


2

SMALL number theory person here. First of all, the problem doesn't really depend on you taking the finite subsets. We may as well just call $A = \{ \sum_{n\in\mathbb{N}} \delta_n a_n : \mbox{$\delta_n\in\{0,1\}$ and all but finitely many $\delta_n$ are zero}\}$ and call $R$ the set of remainders of the elements of $A$ when divided by $q$. The problem is ...


0

Let $G$ be a group of order $8$. Then $Z(G) > 1$ since $G$ is a $2$-group. So $|Z(G)| = 2, 4,$ or $8$ and $G/Z(G)$ has order $4, 2,$ or $1$. If $|G/Z(G)|$ has order $1$ or $2$, or is cyclic of order $4$, then $G$ is abelian and so $g^2 \in G = Z(G)$ for any $g \in G$. (As noted above, one should verify the result: If $G/Z(G)$ is cyclic, then $G$ is ...


1

There are $5$ possible fixed points, but we also need that no other point is fixed, so the answer will 5 times the number of permutations of $4$ elements without fixed points. Such permutations in $S_4$ will have to be a union of cycles of length at least $2$: $(ab)(cd)$ and $(abcd)$. In the former case, we may assume $a=1$, and then there are $3$ ...


3

In general (i.e. for $N \neq 4$), the only normal subgroups of $S_N$ are $S_N$ itself, $A_N$, and $1.$ Therefore no, because most $G$ will not map to one of these. ($S_4$ has an additional normal subgroup, the Klein $4$-group hiding in it.)


13

HINT: Try to prove that for $n \ge 5$, $A_n$, the alternating group of $n$ elements is the only proper and nontrivial normal subgroup of $S_n$. UPDATE: This has to do something with the fact that $A_n$ is simple for $n \ge 5$. After proving this and checking the cases $n \le 4$ we can conclude that a normal subgroup of symmetric group has order $1, 4, \frac{...


2

The definition of "isomorphic" is an existence statement. Given two groups $G$ and $H$, to say that "$G$ and $H$ are isomorphic" means, by definition, that "There exists an isomorphism $f : G \to H$". Like any existence statement, the most direct way to prove that $G$ and $H$ are isomorphic is to show that an isomorphism $f$ exists. And to do this requires ...


4

I think you should review the definition of isomorphism (and probably homomorphism too). An isomorphism between $G$ and $H$ is a function $f$ from $G$ to $H$ (so, it has to be defined on all elements of $G$, not just the identity of $G$) such that $f$ is surjective: it hits everything in $H$. (That is, for each $h\in H$ there is some $g\in G$ such that $f(...


1

An isomorphism $f:G \to H$ is in particular a function or mapping between the underlying sets. This means that to define $f$ you need to specify an element $f(g) \in H$ for every $g \in G$, not only the identity. Of course you could always define $f(g) = \mathit{identity}_H$ for every $g$, but this so-called trivial homomorphism is never an isomorphism ...


0

Every reduced word $w=a_1a_2\cdots a_t$ can be rewrote as $$w=(\overline{e}a_1\overline{ea_1}^{-1})(\overline{a_1}a_2\overline{a_1a_2}^{-1})(\overline{a_1a_2}a_3\overline{a_1a_2a_3}^{-1})\cdots (\overline{a_1a_2\cdots a_{t-1}}a_t\overline{a_1a_2\cdots a_t}^{-1})$$


-3

so the questionis is haw many $\sigma\in S_5$ that fixe only one point $k$ in $\{1,...,5\}$, for each $k$ there are exactely $3$, and the choic of $k$ are $ 5$, then in total are $15$ such $\sigma$.


0

The inner automorphism group $\text{Inn}(G)$ of $G$ consists of group automorphisms of the form $g\mapsto xgx^{-1}$ (known as the conjugation by $x\in G$). It is easily seen that $\text{Inn}(G)\cong G/Z(G)$. As psj36 has noted, if $G/Z(G)$ is cyclic, then $G$ is abelian. However, the claim is trivial when $G$ is abelian. Hence, we only need to check when ...


2

Let $G$ acting on a set $X$ by $(g,a)\mapsto g.a$, for each $a\in X$ the stabilizer of $a$ under ( or relatively to) this action is the maximal sub-group $N$ of $G$, s.t $g.a=a, \forall g\in H$; the standard notation of this stabilizer is $St(a)$. When $G$ acts on the set of all its subgroup by conjugation the stabilizer $St(H)$ of a subgroup $H$ ...


1

In answer to your specific points, 1) The inductive assumption is not being applied to the whole expression, but only to the first bracket, which runs to some smaller number than $k$. At that point, the intent is to show that the final element from the first bracket can be shifted into the start of the second bracket. 2) As above, the square bracket ...


1

1) How is it that Eq(2) contains $a_k$ but in that section of the proof the assumption is that the theorem is true for n≤k-1. Is not this assumption being overreached by the equation running to ak and therefore invalid? It's just a proof by complete induction: they assume that the result is true for $n \leq k-1$ and show it is also true for $n=k$. If the ...


1

The answer depends on what you mean by "find". As you have noticed, $$ Aut(\mathbb Z_{276}) \cong U(276) \cong U(23) \times U(3) \times U(4) \cong \mathbb Z_{11} \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 $$ Therefore, the $2$-Sylow subgroup of $Aut(\mathbb Z_{276})$ corresponds to $0 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$. ...


1

The obvious way to do this is to let the entries in $F$ be a set of $n^2$ unknowns to be solved for. It is necessary and sufficient for the equations to be satisfied for all $g$ in a generating set $S$ of $G$. So you get a system of $|S|n^2$ linear equations in $n^2$ unknown to solve. That should work up to dimension $22$. It depends a bit on the field. If ...


2

Every group of order $p^2$ is abelian and the center is always normal. The hypothesis implies that $|Z(G)|=p$. Its quotient is therefore abelian and so $G'\subseteq Z(G)$. The other containment follows from minimality.


1

In $|S_{20}|$, the highest power of $7$ which divides $20!$ is $7^2$. So it is clear that the Sylow-$7$ subgroup of $S_{20}$ is of order $7^2$. Group of order $7^2$ is either cyclic or isomorphic to $Z_7\times Z_7$. If it is cyclic, then $S_{20}$ will have an element of order $49$, and it should be product of disjoint cycles. Check, whether this is ...


1

Hint: You just need to suppose that $G$ is endowed with a finite Haar $\mu$measure which is the case if $G$ is compact, for $x\in P,u,v\in T_xP$, take any differentiable metric $\langle,\rangle$ on $P$ and set $\langle u,v\rangle_G=\int_G\langle Dg_x(u),Dg_x(v)\rangle d\mu$


4

There is no equivalence (or 'bi-implication'), only the following implication: If $G$ is a group acting on a non-empty set $A$, then the relation $\sim$ on $A$ defined by $$a\sim b\qquad\Leftrightarrow\qquad (\exists g\in G)(a=gb),$$ is an equivalence relation. The 'iff' in your question only serves to define the relation on $A$.


1

The question of Alnitak refers to the class of finite groups known as CLT groups, where CLT stands for Converse Lagrange Theorem: $G$ is a CLT group if for each positive integer $d$ dividing $|G|$, $G$ has at least one subgroup of order $d$. These groups have been studied extensively. It turns out for example that all supersolvable groups are CLT, and all ...


2

The assumption that $G$ is finite is redundant; a $p$-group is a group where every element has order a (finite) power of $p$ or, which is the same, if for every element $x$, $x^{p^a}=1$, for some $a>0$. Let $x\in G$; then $(xH)^{p^a}=H$, for some $a$, because $G/H$ is a $p$-group. This means that $x^{p^a}\in H$, but then $(x^{p^a})^{p^b}=1$, for some $b$,...


1

Yes, (P1) implies (P2). You can find Hall subgroups of all orders by taking intersections of the largest Hall subgroups, which (P1) assumes exist. For example, if $|G|=p^nm=q^rs$ with $p$ and $q$ distinct primes, and $H$ and $K$ are subgroups of orders $m$ and $s$, then since $|HK| \le |G|$, we get $|H \cap K| \ge m/q^r = s/p^n$. But $|H \cap K|$ is not ...


1

Your solution is correct,here is a solution without using the Structure theorem. Proof: Suppose $G$ is cyclic then $G \cong \frac {\mathbb Z}{25 \mathbb Z}$. So assume that $G$ is not cyclic and $G$ is abelian (as you mentioned in your solution).Now note that we have a group action $\frac {\mathbb Z}{5 \mathbb Z} \times G \to G$ defined as $(a,g) \to g^a$ (...


4

Suppose $G$ is a nonabelian group of order $p^3$ for $p$ prime. Since $G$ is a $p$-group, $Z(G) > 1$; since $G$ is not abelian $Z(G) < G$. So $|Z(G)| = p$ or $p^2$. If $|Z(G)| = p^2$, then $G/Z(G)$ would be a cyclic group of order $p$ and thus implying $G$ is abelian. So we must have $|Z(G)| = p$; thus $G/Z(G)$ is a group of order $p^2$ and hence ...


2

if $|Z(G)|=p^3$ then $G$ is a abelian group.It is contradiction. if $|Z(G)|=p^2$ then $[G:Z(G)]=p$ thus $\frac{G}{Z(G)}$ is cyclic and $G$ is a abelian group.It is contradiction. Therefore $|Z(G)|=p$


5

Hint: (1) If $G$ is non-abelian then $G/Z(G)$ is non-cyclic. (2) Every group of order $p^2$ is abelian. You already noted that the center should be non-trivial for your group.


3

Yes; we can say more: there are at least three non-abelian groups of order $16$ with an element of order $8$. We can say something more again - there are exactly four non-abelian groups of order $16$ with an element of order $8$. To see best possible result, see this page (The Theory of Finite Groups, Hans Kurzweil, Bernd Stellmacher, p.108)


0

Consider permutation group on $4$ letters only, since you will find arguments simple. One can then easily see it for general $n$. Consider the polynomial $$\Phi(x_1,x_2,x_3,x_4)=(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4)(x_3-x_4).$$ Let $\sigma.\Phi$ denote the polynomial obtained by permuting the indices in $\Phi$ according to $\sigma$. Thus, for $\...



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