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1

No, it is not enough to assume that the multiplication map is continuous. See questions here and here. However, if the topology on $G$ is nice enough, then it is enough to assume that just the multiplication map is continuous. Theorem (Ellis, 1957): Let $G$ be a group with a locally compact Hausdorff topology such that the multiplication map $G \times ... 0 The non-trivial words of length 1 are $$\mathcal{W}_1 =\left\{A,B,A^{-1},B^{-1}\right\} \text{.}$$ If$\mathcal{W}_n$is the set of non-trivial words$\omega = \omega_1\ldots\omega_n$of length$n$, then the set of non-trivial words of length$n+1$is\begin{eqnarray} \mathcal{W}_{n+1} &=& \left\{\omega A \,:\, \omega \in \mathcal{W_n}, ... 0 Expanding user133458's answer: Let's consider$\mathbb{Z}_n$, and$\bar{x} \in \mathbb{Z}_n$, so that$(x, n) = 1$. Since the order of$\bar{1}$is clearly$n$, we can find the order of$\bar{x}$as follows: In a finite cyclic group$G=<g>$, we know that$o(g^i) = \dfrac{n}{(i, n)}$. In our case, the group is$\mathbb{Z}_n = < \bar{1} >$, and ... 0 Mumble! Gripe! Once again I seem to have answered the pre-edited version of the question! Ah well, at least I can take consolation in the fact that I do not appear to be alone! I won't attempt to prove the title assertion, because it is false. I will however give a simple counterexample: Let$N_1 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} ...

2

What is to be proved is the following: $$e^{A \otimes I_b +I_a \otimes B} = e^A \otimes e^B~$$ where $I_a,A \in M_n$ , $I_b, B \in M_m$ This is true because $$A \otimes I_b~~~~\text{and}~~~~ I_a \otimes B$$ commute, which can be shown by using the so called mixed-product property of the Kronecker product. i.e. $$(A \otimes B)\cdot (C \otimes D) = ... 0 First and foremost, the result is not true as stated. It is only true of A and B commute, which is a very restrictive condition for matrices. To handle the commutative case, one can first consider the formal power series case. In the ring \Bbb Q[[X,Y]] of formal power series with rational coefficients in commuting indeterminates X,Y, one defines ... 1 If A and B are n\times n, then by Taylor expansion we have:$$e^A=\sum_{k=0}^{\infty}\frac{A^k}{k!}$$Therefore:$$e^Ae^B=\sum_{k_1=0}^{\infty}\frac{A^{k_1}}{k_1!}\sum_{k_2=0}^{\infty}\frac{B^{k_2}}{k_2!}\Rightarrow e^Ae^B=\sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\frac{A^{k_1}}{k_1!}\frac{B^{k_2}}{k_2!}\Rightarrow ...

0

A way to proceed. If $A$ and $B$ commute they are simultaneously diagonalizable (if they are diagonalizable, otherwise one must fall back to Jordan decomposition). For diagonal matrices the formula is easy, because you reduce to the property of exponential for real numbers.

3

One way to solve this problem is to use the left regular representation. $V$ acts on itself by left multiplication. First, we choose a numbering of the elements of $V$. Somewhat arbitrarily, I choose to label $1,a,b,ab$ by $1,2,3,4$, respectively. First note that $1$ must act as the identity permutation. We observe that $a$ acts by left multiplication ...

1

An easy way to do this is just to examine the permutations of order 2, since all elements of the Klein-4 group have order 2. These are just the set of all transpositions and disjoint transposition. Note that the transpositions alone do not form a normal subgroup, so we have all of the pairs of disjoint transpositions: $$\{(),(12)(34),(13)(24),(14)(23)\}.$$ ...

3

You can embed the Klein 4-group (which is not cyclic, since it contains no element of order $4$!) into $S_4$ in four ways: $$\langle (12),(34)\rangle=\{1,(12),(34),(12)(34)\}$$ $$\langle (13),(24)\rangle=\{1,(13),(24),(13)(24)\}$$ $$\langle (14),(23)\rangle=\{1,(14),(23),(14)(23)\}$$ $$\langle(12)(34),(14)(23)\rangle=\{1,(12)(34),(14)(23),(13)(24)\}$$ ...

0

my try, If G is abelian, then any irreducible character is linear which has degree 1. If G is non-abelian group. Let χ be irreducible of G, so that Kernel of χ is normal of G (not sure) but G is simple, so that kernel of χ is trivial and this lead us to a contradiction, so that kernel of χ must be the whole group G, so that χ is the trivial character. need ...

0

Numbers that are relatively prime to 6 generate $\mathbb{Z_6}$. It's a theorem.

2

$5 = -1$ who generates the same way 1 does.

5

The orthogonality condition is equivalent to the columns of $R$ being an orthonormal basis. You can pick first column $v_1$ from $S^{n-1}$, with $(n-1)$ dimensions to choose from. second column $v_2$ from $S^{n-1}\cap \{v_1\}^\perp = S^{n-2}$, with $(n-2)$ dimensions. third column $v_3$ from $S^{n-1}\cap \{v_1,v_2\}^\perp = S^{n-3}$, with $(n-3)$ ...

4

I don't want to work that hard. Take any skew symmetric square matrix, $S^T = -S.$ We get $e^S \in SO_n(\mathbb R),$ and a neighborhood of the identity is automatically covered, in bijection with a neighborhood of the $0$ matrix. Oh, $S$ and $S^T$ commute, so $e^S (e^S)^T = e^S e^{S^T} = e^{S + S^T} = e^0 = I,$ and $\det e^S = e^{\operatorname{trace} S} = ... 4 The generator is the number in the$\langle{ }\rangle$. THus the generator I'm describing above is the element$(1234)$. Why? $$(1234)^1=(1234)$$ $$(1234)^2=(1234)(1234)=(13)(24)$$ $$(1234)^3=(13)(24)(1234)=(1432)$$ $$(1234)^4=(1432)(1234)=(1)$$ As Vladhagen mentioned in his answer, this is a subgroup of order 4. 5 Isn't this just any 4-cycle? The 4-cycles, and the 4-cycles only, form cyclic groups of order 4 in$S_4$. Am I misunderstanding the question? 0 Remember that groups have operations with associative laws. You did not note that$\mathbb{R}$is the additive group of reals. Now you can take$a,b\in \mathbb{R}$with$a+b \in \mathbb{R}$since groups are closed. Then$\phi(a+b)=e^{(a+b)(2\pi i)}=e^{2\pi ai}e^{2\pi bi}=\phi(a)\phi(b)$and so$\phi$is a homomorphism. 1 Since$H$is normal,$aH = Ha$. (a) Assume there is an element$x \in aH \cap Ha^2$. Then$x = ha = ka^2 \in Ha \cap Ha^2$, for some$h,k \in H$. Therefore$a = k^{-1} h \in H$(because$H$is a subgroup), contradiction. Conclusion:$aH \cap Ha^2$is empty. (b) By the same reasoning, if$aH \cap Ha^3$is nonempty, then$ha = ka^3$for some$h, k \in H$... 2 We know that$H\cap K$has at least the element$e$. suppose that$H\cap K$has another element$g$so that$g\in H$and$g\in K$. the$|H\cap K|$divides the order of$H$and$K$but$5$and$12$don't have common divisors. So the only element in$H\cap K$is$e$. 3 Let$x\in H\cap K$then by Lagrange theorem the order of$x$divides the two coprime orders:$|H|$and$|K|$so$o(x)=1$and then$x=e$. Conclude. 3 Hint: Consider$|H\cap K|\,|\,|H|$and$|H\cap K|\,|\,|K|$. 0 Hint: Consider the determinant. 3 As vadim123 mentioned in his comment, there is always a trivial order. Suppose you have some non-trivial order that respects the group structure on your group$G$. Suppose$g\in G$and$1\leq g$. Then multiplying by$g$you will get$g\leq g^2$,$g^2\leq g^3$et.c. But since$G$is finite, eventually$g^n=1$for some$n$. Then you have$1\leq g\leq g^2\leq ...

0

The unrestricted wreath product is equivalent to the semi-direct product. The restricted wreath product is formulated using the direct sum as a base of the wreath product. See here.

0

For the sake of providing an answer. Extend the ground field to $\mathbb{C}.$ Then the representation of the group is completely reducible. Since the group has odd order, it has no complex irreducible character of degree $2.$ Hence the group is reducible, and must be conjugate to a group of diagonal matrices. Thus it is Abelian. Furthermore, since the ...

0

$g = gh^{-1}h$. Can you take it from here?

1

Note that $gh^{-1} \in \mathbf K$. It follows that $$g = (gh^{-1})h \in \mathbf Kh = h\mathbf K$$

8

The multiplicative groups ${\Bbb F}_3(X)^\times$ and $\Bbb Q^\times$ are both isomorphic to the sum of $C_2$ (corresponding to $\Bbb F_3^\times$ and $\Bbb Z^\times$) with a free abelian group of countable rank (generated by the primes of $\Bbb Z$ and the irreducible polynomials in $\Bbb F_3[X]$ respectively).

1

For the sake of brevity and by way of an incentive to learn more about this wonderful theory, here is a solution using the Polya Enumeration Theorem. All we need here is to calculate the cycle index $Z(G)$ of the symmetry group $G$ of the edges of a square, substitute our $N$ colors into the cycle index and extract coefficients. We get for the first question ...

3

Suppose $K$ is a finite group with a fixed-point-free automorphism of prime order. Then Thompson (1959) showed $K$ is nilpotent. Suppose $K$ is a finite solvable (or locally nilpotent) group with a fixed-point-free automorphism of prime order $p$. Then Higman (1957) showed $K$ is nilpotent with nilpotency class bounded above by a function, $k(p)$, of $p$. ...

2

Since determinants of invertible matrices with real entries are just nonzero real numbers, the question is whether every nonzero real number is the determinant of some such matrix. If $a$ is a real number, then it is the determinant of $\begin{bmatrix} 1 & 0 \\ 0 & a \end{bmatrix}$. So the answer is affirmative.

5

Yes. Consider the diagonal matrix with $r, 1, 1, ..., 1$ on the diagonal, where $r$ is a nonzero real. That has determinant $r$ so "det" is a surjective map from $GL(n, \mathbb R)$ to $\mathbb R^{*}$. And since $\det(AB) = \det(A) \det(B)$, it's actually a homomorphism.

1

You are right about the elements of $U(20)$ and the order of the group. Another way to name the elements (here listed in the same order as you do) is: $$U(20) = \{ 1,3,7,9,-9,-7,-3,-1 \}$$ and that might be easier to multiply (and should not confuse you when you do modulo 20 operations). Your generating set $\langle 1,3 \rangle$ does not need the element ...

2

Since you are only required to find a minimal generating set and not a set of minimum size, you can just start with the whole group and throw out elements until you can't throw out any more. This shouldn't be too hard. So to start out $$U(n) = \langle 1,3,7,9,11,13,17,19 \rangle$$ Certainly you can throw out $1$. Then $3 \cdot 3 = 9, 11 \cdot 3 = 13, 11 ... 0$|U(n)|=\varphi(n)$where$\varphi$is Euler-phi function. And it is multiplicative,i.e$\varphi(20)=\varphi(4).\varphi(5)=2.4=8U(n)\cong Aut(Z_n)$thus,$U(20)\cong Aut(Z_{20})\cong (Aut(Z_4)\times Aut(Z_5)\cong Z_2\times Z_4$Thus,it is not cyclic. Note that,I used following facts;$Z_{mn}\cong Z_m\times Z_n$if and only if$gcd(m,n)=1$and in that ... 2 Assume that$A \neq 0$. Since$pA=0$we can consider it as a vector space over$F_p$, and by taking the$G$-submodule generated by some nonzero element, we can assume that it is finite dimensional. In particular an action of$G$is just a homomorphism$\varphi:G \to GL_m(F_p)$. Suppose first that$G$is cyclic generated by$g$. Since$g^{p^m}=e$where ... 2 Since we are painting the edges of squares, we assume that these four colorings are all considered the same, and should not be counted separately: (It should be clear that painting triangular wedges is the same as painting edges, since there is a natural bijection between wedges and edges.) It's not completely clear in the question, but we should ... 4 The parity of an element$sgn(\sigma)$gives you a linear (irreducible) character (one-dimensional representation). Hence, by Frobenius orthogonality relations, the inner product with the other characters is$0$, and this is exactly your sum! (Note that conjugate permutations have the same sign) 1$\phi$is an automorphism of the$G$-set$S$, i.e.$\phi: S \to S$is a bijection with$\phi(gt) = g\phi(t)$for all$g \in G$,$t \in S$. For our particular choice of$s$in the lemma we know$\phi(s) = ns$, so$\phi(hs) = h\phi(s) = hns$(since$h \in H \subset G$). 3 There is a theorem of Ito: let$q$be a prime power,$d$a positive integer, and$p$a prime divisor of$q−1$with$d \leq p$. Fix some field$F$of order$q$and some element$\alpha$of this field of order$p$. The Frobenius complement$H$is the cyclic subgroup generated by the diagonal matrix whose$(i,i)$th entry is$\alpha^i$. The Frobenius kernel$K$... 3$Jx=Ky\implies J=Kyx^{-1}$let say$r=yx^{-1}$Thus,$Kr$is a group which is a right coset of$K$, which is only possible when$Kr=K\implies K=J$2 I will summarize the proof from the book "Theory of Groups of Finite Order" by W. Burnside. It would not be very easy to think of this if you had not seen it before! With your notation, conjugation by$a$is inducing a fixed-point-free automorphism of order$3$of$G$. You know that$x$commutes with$a^{-1}xa$and similarly with$axa^{-1}$. If you ... 1 They don't have to be... recall that you only get a matrix when you fix a basis... For example, let$p=a_3x^2+a_2x+a_1$be in$V$the vector space of quadratic polynomials and$\Phi$the action of$S_3$on$V$given by $$\Phi(p,\pi)=\pi(a_3)x^2+\pi(a_2)x+\pi(a_1).$$ This thing is a representation (unless I am acting on the wrong side... not sure) Of ... 1 All you have to do is note that if$A$and$B$are two linearly ordered sets which satisfy the following property$(*)$, then their lexicographic product satisfies the same property as well:$(*)$Every interval has a countable cofinality (i.e. there is a sequence converging to the endpoint of the interval). Clearly$\Bbb R$and$\Bbb N$both satisfy ... 3 You can't embed$\omega_1$into the$\Bbb{R}^n$or$\Bbb{N}^n$(in the sense of topological embedding.) Because if$A$is a subset of$\Bbb{R}^n$then it has only countably many isolated point, but$\omega_1$has$\aleph_1$many isolated points. For the same reason, you cannot embed$\omega_1$into$\Bbb{N}^n$, even you cannot embed it into$\Bbb{R}^\omega$... 0 Actually, I want to write it as a comment If you showed that there is an element$b$of$G$with oreder$p$not contained in$<a>$, Use induction on$|G|$,$G/<b>$satisfes the hypotesis,and$\bar{a}$has maximum possible order in$G/<b>$since$<a>\cap <b>=1$.(as$\bar{<a>}\cong <a>/(<a>\cap <b>) $) ... 7 There are infinite groups which cannot be represented by finite-dimensional matrices over any commutative ring. If a group can be represented by matrices in such a way then it is called linear (perhaps this definition really requires field not commutative ring, but everything in this answer works for the more general commutative ring definition). Not all ... 3 A representation of a group$G$is a morphism$\rho:G\rightarrow\text{GL}(V)$where$V$is a vector space over a field$K$. There are always at least the following two representations: the trivial representation, where$\rho(g)=\text{id}_V$for all$g\in G$. the regular representation, where$V$is the space of$K$-valued functions on$G\$ and ...

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