Tag Info

New answers tagged

0

I don't see an analytical solution for this. Since you write that brute force enumeration becomes computationally intractable once $n$ nears $10$, I'm assuming that you're interested in a computational solution. This can easily be done by keeping track, in each of the $n$ steps, of the number of ways the items of interest could have been permuted so far. ...


0

As you remarked, there are four non-trivial subgroups of $\mathbb Z_{18}$ given by the divisors of $18$ (others than $1$ and $18$). Now it's good to know what's the sum and the intersection of such subgroups. In general, in $\mathbb Z_n$ we have $\bar k\mathbb Z_n+\bar l\mathbb Z_n=\overline{\gcd(k,l)}\mathbb Z_n$ and $\bar k\mathbb Z_n\cap\bar l\mathbb ...


3

Try $\mathbb Z[\frac12]$, that is, the set of binary fractions. More generally, $\mathbb Z[\frac1p]$, that is, the set of fractions whose denominators are powers of the prime $p$.


-1

You have correctly shown that $\Bbb{C}^3$ is the direct sum of two smaller representations (with one minor exception: $V \cap W$ is the zero vector space, not the empty set). However, you have not given any argument that those representations are actually irreducible. It is clear that $W$ is irreducible, since it's $1$-dimensional, but you should give some ...


0

Any notation in mathematics is dependent on context, which you did not provide much of. (Where are you seeing these symbols - in your textbook? If so, which textbook? What is the surrounding sentence, or paragraph, where these notations show up?) However, given that you've said this is group theory, what seems likely to me is that you are learning about the ...


0

Another way to see that $|zw| = |z||w|$: Let $z = a+ib$, and $w = c + id$. Then: $|zw| = |(a+ib)(c+id)| = |(ac-bd) + i(ad+bc)| = \sqrt{(ac-bd)^2 + (ad+bc)^2}$ $= \sqrt{a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2}$ $= \sqrt{a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2} = \sqrt{a^2(c^2 + d^2) + b^2(c^2 + d^2)}$ $= \sqrt{(a^2 + b^2)(c^2 + d^2)} = ...


1

$\mathbb C^{\times}$ is an Abelian group. The map $z \mapsto |z|$ is a group homomorphism $\mathbb C^{\times} \to \mathbb R^{\times}$. The circle is the kernel of this map and so is a subgroup of $\mathbb C^{\times}$. The crucial point is that $|zw|=|z|\ |w|$ and this can be proved as follows: $$|zw|^2=(zw)(\overline{zw})=z w \bar z \bar w=z \bar z w \bar ...


1

Hint: if $z\in G$, there exists $\varphi\in(-\pi,\pi]$ with $z=e^{i\varphi}$. (Of course one can choose $\varphi \in [0,2\pi)$ or $\varphi \in I$ for any half closed intervall $I$ of length $2\pi$).


0

I am really not sure of this, but here's what I could do! Let $\tau$ be in the Galois group. $\tau$ is uniquely determined by what it does to $\zeta$. And $\tau_{k}(\zeta)=\zeta^{k}$. Now if $m_{1}$ and $m_{2}$ lie in the same coset of $S$, then $m_{1}m_{2}^{-1} \in S$, So $m_{1}m_{2}^{-1}=t$ for some $t$ in $S$. But we know that we can find an elememt ...


6

The simplest way to prove that is to find the solution. The crutial thing is that any element of a group has an inverse. $$xaxb=xc$$ Multiply by $x^{-1}$ on the left: $$x^{-1}xaxb=x^{-1}xc$$ Since $x^{-1}x = e$ (where $e$ is the group neutral element): $$axb=c$$ Multiply by $a^{-1}$ on the left and by $b^{-1}$ on the right: $$x=a^{-1}cb^{-1}$$ So, $x$ is ...


3

$$\begin{align} xaxb=xc & \implies x^{-1}xaxb=x^{-1}xc\\ & \implies axb=c\\ & \implies a^{-1}axbb^{-1}=a^{-1}cb^{-1}\\ & \implies x=a^{-1}cb^{-1}. \end{align}$$


-1

Given $X$, $Y$ and $Z$, composition of homomorphisms can be viewed as a map $$ Hom(Y,Z)\otimes Hom(X,Y) \to Hom(X,Z).$$ This map is surjective if $Y$ has dimension large enough (namely, has dimension at least $\min(\dim X, \dim Z)$). Thus $Hom(X,Z)$ is a quotient of $Hom(Y,Z)\otimes Hom(X,Y)$. However, since we work with vector spaces, every surjection ...


1

Consider the bijection $f\colon (1,2)\to(0,\infty)$, $t\mapsto \frac{2-t}{t-1}$. Show that $$f(x*y)=f(x)f(y).$$ Conclude that $(G,*)\cong ((0,\infty),\cdot)$.


1

I doubt there's a clever way to do this. I recommend writing out the desired final result in expanded form then doing the computation to see that you get the same thing. For finding the identity and inverses you'll need to solve equations. With a goal in mind I expect the computation will end up not being as bad as you think. (Note commutativity can be shown ...


5

While the other answers lead you to the precise order of $GL_n(\mathbf{F})$, I'll explain a simple way to see the inequality you asked about. As you note, any $n \times n$ matrix has $n^2$ entries; if these entries are coming from $\mathbf{F}$, then the set of all $n \times n$ matrices has order $q^{n^2}$, since there are $q$ choices for each of the $n^2$ ...


0

This is not a full answer, but maybe it can helps... As mt_ said, $\overline{f}$ and $\tilde{f}$ does not mean anything here. However, there is a map $H^2(G,\mathbb{Z}^\times)\rightarrow H^2(G,R^\times)$ (at least if $G$ acts trivialy on $-1\in R^\times$). So that if $f\in H^2(G,\mathbb{Z}^\times)$ there is a corresponding element in $f_R\in ...


0

I'd be careful with statements about what is "obvious", but the way I came up with trying the commutator was: We want most of the elements to remain invariant and only a few of them to be permuted. Applying the generators $x,y$ and their inverses leaves the "bulk" of the elements unchanged if we have equal numbers of $x$ and $x^{-1}$ and equal numbers of $y$ ...


1

Here are some remarks that should help you to improve your answers,first it seems that in each question you have some field you are working on, maybe it would be good to highlight this more clearly because the answers may change depending on the field... 1)(Field $\mathbb{R}$) does there always exist a rotation with an angle different from $\pi$? (think ...


1

Look what happens when you iterate the cycle. It might help to draw it out. Let $\sigma$ denote the cycle. When you apply it twice, 1 goes to 3 goes to 5 goes to 7, skipping every other one, and hence $\sigma^2=(1357)(2468)$. When you apply it three times, 1 goes to 4 goes to 7 goes to 2, etc., skipping every two, so $\sigma^3=(14725836)$. So when you do it ...


0

I am going to start where Derek finished, with the aim of giving some justifications as to why the groups are mutually non-isomorphic. To summarize, there are seven possibilities: we may choose $x, y, z$ such that $P = \langle x \rangle \cong \mathbf{Z}_5$, $Q = \langle y \rangle \cong \mathbf{Z}_{11}$, $R = \langle z \rangle \cong \mathbf{Z}_{61}$, and ...


1

I'm not sure what exactly I'm allowed to use or what Rotman had in mind, but I think this approach uses no more group theory than the (unique) factorization of permutations and the basic idea of what a homomorphism is. If $n$ is odd there exists a subset $S\subset S_n$ of pairwise commuting elements of order $2$ with $|S|=\tfrac{n-1}{2}$. For example ...


0

I think the equations asks you to prove that given an action of $G$ on $X$ ,$g\star x $, then we can find a homomorphism from $G \to S_{X}$ (the homomorphism you gave works fine!). and given a homomorphism from $G$ to $S_{X}$, you can find an action of $G$ on $X$.


1

Let $\varphi:G\rightarrow S_X$ be a homomorphism. Then we can define a group action of $G$ on $X$ by $g.x:=\varphi(g)(x)$. Conversely, given a group action of $G$ on $X$, we can define a homomorphism $\psi:G\rightarrow S_X$ by $\psi(g)(x):=g.x$ (technically, it should be checked that these define group actions and homomorphisms but this isn't a hard check ...


1

The claim is rather that for all $n\ge1$ and $a\in k$ with $a\ne 0$ the polynomial $X^n-a$ has a root. As $k$ is assumed algebraically closed, such root certainly exists.


1

As Taylor said, just find a generator (and it's the obvious one. If still confused, I can give a hint). In general, you can show that $\mathbb{Z}_m\times\mathbb{Z}_n\cong\mathbb{Z}_{mn}\iff \gcd(m,n)=1 $


2

Let $G$ be a group such that $\operatorname{Aut}(G)\cong\Bbb{Z}/8\Bbb{Z}$. Then the map $$\psi:\ G\ \longrightarrow\ \operatorname{Aut}(G):\ g\ \longmapsto\ (x\ \longmapsto gxg^{-1}),$$ is a group homomorphism with $\ker\psi=Z(G)$, the center of $G$. Every subgroup of $\operatorname{Aut}(G)\cong\Bbb{Z}/8\Bbb{Z}$ is cyclic, so this shows that $G/Z(G)$ is ...


1

No to both questions. For example for the first, in the generalized quaternion groups, there is a unique element of order $2$, which is thus contained in any non-trivial subgroup. For the second, consider either of the groups the group $S_3$ or $S_4$ which are solvable but have trivial centers.


2

You can relate the commutator for a Lie group $G$ to the (ring-type) commutator of its Lie algebra $\mathfrak{g}$. Say $g$ lies infinitesimally close to the group identity, $g = \exp(\epsilon\,X)$ for some $X \in \mathfrak{g}$. Likewise say $h = \exp(\epsilon\,Y)$ for some $Y \in \mathfrak{g}$. Then, $ghg^{-1}h^{-1}$ also lies infinitesimally near the ...


1

This is what a $G$-invariant bilinear form is: Let $f$ be a $G$-invariant bilinear form. Then $f(vg,wg)=f(v,w)$ for all $g\in G$. I'm going to work on a solution and post back.


0

$ \pi _i Q_{ij}= \pi_j P_{ji} $ From definition of $Q_{ij}$ Now if the Markov Chain is reversible then $Q_{ij}=P_{ij}$, and hence the result.


-1

Suppose $m,n\in\mathbb{Z}$ are such that $\mathbb{Z}\times\mathbb{Z}=⟨(m,n)⟩$, then there exists $x\in\mathbb{Z}$ s.t. $x(m,n)=(1,1)\implies xm=1=xn\implies x=\frac{1}{m}=\frac{1}{n} \implies m=n=1$ Well, now it must also be true that there exists $y \in \mathbb{Z}$ s.t.∶ $y(m,n)=(0,1)\implies ym=0,yn=1 \implies m=\frac{0}{y}=0 ,n=\frac{1}{y}\implies y=1, ...


2

In both cases, commutators measure how far the object is from being commutative. A group is commutative when its operation is commutative. A ring is commutative when its multiplication is commutative. (Addition in a ring is always commutative.) Hence, the definition of commutator in each case reflects what it means to commute, as it should.


6

Another perspective on this comes from considering the relationship between a Lie group and its corresponding Lie algebra. Details are at On the relationship between the commutators of a Lie group and its Lie algebra, but the upshot is that the ring commutator $[A,B]=AB-BA$ can be regarded as an infinitesimal version of the group commutator ...


29

Here is a more abstract and unified approach to commutators. I could phrase this whole answer in terms of category theory (and this is probably the best way to think of it), but I won't in order to make the answer more accessible. Let $G$ be a group. We want to make it commutative. What does that mean? We want to find a commutative group $A$ and a ...


38

The commutator of a group and a commutator of a ring, though similar, are fundamentally different, as you say. In each case, however, the commutator measures the "extent" to which two elements fail to commute. That is, given elements $a,b$, we wish to "compare" $ab$ and $ba$. In the context of a group, we only have one operation: "multiplication". One ...


1

Clearly the factor group you give is abelian and thus the direct product of cyclic groups and Derek Holt's comment gives you that you do not need more than $d$. If $d$ is also the minimal number of generators, Burnsides basis theorem shows that for $M=[G,G]G^p$ the group $G/M$ is the direct product of $d$ cyclic groups of order $p$. Since your $N\le M$, ...


3

I assume that your notation $\left|x\right|$ means the order of $x \in S_n$, and that by "cyclic decomposition" you mean the cycle type. Then, $\left|x\right|$ is the lcm of the lengths of all cycles of $x$. Thus, this lcm is $p$, so that each cycle must have either length $p$ or length $1$ (since $p$ is prime). Moreover, $x$ must have at least one cycle of ...


1

As Mariano noted, you can tell immediately that they are not isomorphic by considering a number of their algebra invariants. To know that they are isomorphic is usually more complicated. Though note that, as Ofir noted, all abelian groups of the same order over $\mathbb{C}$ (or any field with a primitive root with order the order of the groups in question) ...


4

The game is taking you through all pairs of $(a,b)$ such that $a ,b > 0 $ and $\gcd(a,b)=1$. Proof: Take $A$ be the set of all pairs that we get from the game. We first show that if $(a,b) \in A$ then $\gcd(a,b)=1$: it follows form Euclidean algorithm. Now by induction we show that if $a,b > 0$ and $\gcd(a,b)=1$ then $(a,b) \in A$: we know this ...


1

Perhaps what was meant is that two matrices in this subgroup are conjugate if and only if they have the same number of linearly independent eigenvectors, which can be seen by the fact that this determines the "nilpotent part" of Jordan canonical form.


1

Symmetric Groups Let $G=S_n$. Then as $S_n/A_n \cong \Bbb{Z_2} \implies G' \subseteq A_n$. Claim- $G'=A_n$. Proof- Let $\sigma \in A_n$. Then by a standard resule, we know $A_n$ is generated by $3$-cycles, hence cycle $\sigma= t_1t_2\dots t_n$ where $t_i's$ are $3$- cycles. Now let any arbitrary three cycle $(abc)$. It can be rewritten as ...


1

This problem is old but quite interesting. I have an answer to (I) which depends on some calculations in $\textsf{GAP}$ and Mathematica. I haven't thought about (II). Suppose an irreducible septic has roots $x_1,\ldots,x_7$ that satisfy $$ x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 – (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0. $$ I claim that the Galois group is ...


2

You are correct that, if $C$ is a subgroup, then it is a normal subgroup. However, $C$ is only a subgroup when $x$ is the identity element; to see this quickly, one may note that the relation defined by: $$x\sim y \Longleftrightarrow \exists g\in G [x=gyg^{-1}]$$ is an equivalence relation, dividing the group into conjugacy classes. So, $C$ happens to be the ...


0

Notice that if $C$ is a subgroup the identity must belong to $C$! The only conjugacy class containing the identity is the identity itself!.


1

$gaH=aH$ if and only if $ga$ is in $aH$ that is if $ga=ah$ for some $h\in H$. That is if $g=aha^{-1}\iff g\in aHa^{-1}$


0

You have missed something. You might check through an example. Consider $G = \mathbb{Z}$ the integers, and $H = 5\mathbb{Z}$ the multiples of $5$. These are both groups under addition, so we'll write cosets additively. Then $G/H = \mathbb{Z}/5\mathbb{Z}$ is the additive group of $5$ elements. Let's consider the coset $a+ H = 1 + 5\mathbb{Z}$. You've ...


2

Not quite. Especially, your solution $aH$ fails to be a group in general. $$gaH=aH\iff a^{-1}gaH= H\iff a^{-1}ga\in H\iff g\in aHa^{-1}$$ Alternatively: "Clearly" the stabilizer of $H$ is $H$. And for any action with $a\cdot x_1=x_2$ we have $G_{x_2}=aG_{x_1}a^{-1}$


0

You might find this video lecture series useful: Group Theory by LadislauFernandes. He has quite an extensive set of videso on introductory group theory. I haven't watched everything but from what I have, the quality is ok but a teensy bit slow.


4

Transitivity is clear because you can move the top left square into any other place by first moving it to the desired column using the top row alone, and then working with that column alone. Let us look at the commutator of the top row and leftmost column operations. We easily see that the move sequence $\leftarrow\uparrow\rightarrow\downarrow$ is a ...


8

This answer is almost complete and I think (read hope) that it's correct up to the point where I get stuck. (I've posted this as an answer because it's too long for a comment and I should go to bed.) It suffices by the first isomorphism theorem to find a group $H$ and a surjective homomorphism $\theta : G \to H \times H$ for which $\ker \theta = Z(G)$. ...



Top 50 recent answers are included