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0

In your case, both $s$ and $t$ are of order $6$, hence $\langle s\rangle$ and $\langle t\rangle$ are both isomorphic to $\mathbb Z/6\mathbb Z$. The intersection is trivial, since every permutation in $\langle s\rangle$ fixes $6$, and only trivial one in $\langle t\rangle$ fixes $6$.


1

For $n=4$, the eigenvalues are $4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2$. For $n=5$, the eigenvalues are the roots of $(x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdot x - 4)^{10} \cdot (x^2 + 2\cdot x - ...


1

A simple proof for seeing there is no primitive root $\bmod 2^n$ is that the powers of $x$ are congruent to $1$ or $x\bmod 8$, so you are missing at least half of the elements. Now lets prove all elements $\bmod 2^n$ are generated by $5$ and $-1$. There are $2^{n-1}$ congruence classes relatively prime to $2^n \bmod 2^n$. We will prove that the order of $5$ ...


0

No such subgroup exists. Pro-$p$ groups with the second definition (i.e. inverse limit of discrete, finite $p$-groups) can be easily shown to be equivalent to the first definition: $$G/N\cong P$$ where $N$ is open, normal and $P$ is a finite $p$-group. How so? Since $N$ is normal it is the union of some open subgroups from a basis of open normal subgroups ...


0

For each $n\in\mathbf{N}^{*}$ we note $\pi_n : \mathbf{Z} \to \mathbf{Z} / n \mathbf{Z}$ the canonical surjective morphism. If $n,m\in\mathbf{N}^{*}$ are integers such that $m|n$ we have $n \mathbf{Z} \subseteq m \mathbf{Z}$ so that we can define a canonical morphism $\varphi_{n,m} : \mathbf{Z} / n \mathbf{Z} \to \mathbf{Z} / m \mathbf{Z}$ sending the class ...


2

General representations of a finite group are of the form $$\chi=\bigoplus_{i=1}^k\chi_i^{n_i}$$ with $\chi_1,\ldots, \chi_k$ the finitely many irreducible characters which uniquely determine the representation by standard theory. But then as $$\deg \chi=\sum_{i=1}^k n_i\deg\chi_i$$ and $\deg \chi_i\ge 1$, we have $$\deg \chi\ge \sum_{i=1}^k n_i$$ ...


1

It's $H_1(G, \mathbb{Z})$ that's the abelianization of $G$. $H_1(G, \mathbb{R})$ is the tensor product of the abelianization of $G$ with $\mathbb{R}$, which is in particular a real vector space. It is much smaller than what you said.


0

As the book says, the orbit of $1$ under $\sigma$ is the set $$\{\sigma^n(1) \mid n\in \Bbb Z\}$$ so you just need to look at $$\mathcal O_{1,\sigma} = \{\sigma^0(1) = 1, \sigma(1), \sigma^2(1), \ldots\}$$ and since, as you observed, $\sigma$ maps $$ 1 \mapsto 3 \mapsto 4 \mapsto 5 \mapsto 6 \mapsto 2 \mapsto 1$$ the orbit of $1$ is the set containing all ...


0

I will employ here again the same web resource by Eric Moorhouse I linked yesterday in this answer to a related question. This finite group and this other have both order $27$: since there's no $n$ such that $n!=27$, they are not symmetric; since there's no $m$ such that $2m=27$, they are not dihedral. Just a glance at the bottom right corner of the two ...


2

A central product of countably infinitely many copies of $D_8$ (or more generally of any extraspecial group) is a counterexample. This group has the presentation $$\langle x_i,y_i,z\ (i \in {\mathbb N}) \mid z^2=1,\ x_i^2=y_i^2=1,\ [x_i,y_i]=z\ (i \in {\mathbb N}),\ {\rm all\ other\ pairs\ of\ generators\ commute}\ \rangle.$$ Its centre is the finite ...


1

The problem is that the definition is for multiplicative notation, while you consider an additive group (what makes the situation worse is that multiplication also makes sense for residue classes). For a group where you use additive notation the defintion transforms to: For $(G,+)$ a group and $x\in G$ define the order of $x$ to be the smalles ...


1

Let $(G,\star)$ be a group with $x\in G$. Suppose you want to write $x \star x \star \cdots \star x$ ($n$ times). How would you write it with the multiplicative notation? $x^n$. How would you write it with the additive notation? You can't say $nx$ or $xn$. You have to use an awkward construction like $\sum_{i = 1}^n x$. Now lets assume we are working ...


8

You surely know that many groups are not commutative. Because of this, an additive notation would bring confusion in non-abelian context. Think about group of invertible matrices $\mathrm{GL}(\mathbf{K},n)$ over a field $\mathbf{K}$: this is a group under matrix multiplication: wouldn't be rather tricky to think $A+B$ as the result of $A\cdot B$, knowing ...


4

Generally, multiplicative notation is used for the operation in an arbitrary group, and additive notation is reserved for the operation in Abelian (commutative groups). For some intuition into this, the general linear group $GL_n(\mathbb R)$ is the set of all invertible $n\times n$ matrices, with the operation of matrix multiplication. If $A,B\in ...


0

Although much of facts have been written down. Just as a summary, proof of all can be found in paper "A System Arising from a Weakened Set of Group Postulates" mentioned by user83548 Let (G, *) be a semigroup with a non empty set E consisting of left identities of G. for all g in G and for all e in E there exist h in h such that h*g=e, then it is group ...


1

The order of an element $g$ in $S_n$ is the LCM of the length of cycles in the factorizetion of $g$ into disjoint cycles. From here it is a question in number theory. I thing for $n=15$ the answer $105=7\cdot 5\cdot 3$.


0

First, the program IsNormalIntermediate is the following: IsNormalIntermediate:=function(G,H,K) local D1,D2,s,i,j,E1,E2,c; D1:=DoubleCosets(G,H,K); s:=Size(DoubleCosets(G,H,K)); c:=0; if s=Size(DoubleCosets(G,K,K)) then return true; else D2:=DoubleCosets(G,K,H); for i in [1..s] do for j in ...


1

I presume that $\epsilon$ is the sign of a permutation, and that you know already that this is well-defined(?) Suppose, then, that $\sigma$ and $\tau$ have sign equal to $1$. This means one can write each of them as a product of an even number of transpositions; let $k$ (resp. $\ell$) be the number of transpositions in some representation of $\sigma$ (resp. ...


-1

Yes and it's the kernel of the homomorphism of groups $$\epsilon: (S_n,\circ)\to (\{\pm1\},\times),\;\sigma\mapsto\epsilon(\sigma)$$ and it's called the alternating group.


4

For $k\in\mathbf{N}^{*}$ I note $k = \prod_{p\in\mathscr{P}} p^{v_p(k)}$ the unique decomposition of $k$ in product of prime numbers. Then you have by the chinese reminder theorem, and as a product is a special kind of projective limit and as projective limits commute (i.e., can be interverted) : $$\varprojlim_{n\geq 1} \mathbf{Z} / n \mathbf{Z} \simeq ...


1

Your map is surjective. It follows from one of the isomorphism theorems that its codomain is isomorphic to the quotient of its domain by its kernel.


1

Two elements in $S_n$ are conjugated if and only if they have the same factoriziation to disjoint cycels. Hence $K$ is normal in $S_4$ and therefore in $A_4$. From the same reason $H$ is not normal in $S_4$. $H$ is normal in $K$ since it is a subgroup of index $2$. I left for you to cheack if $H$ is normal in $A_4$.


0

First, the $p$-group $P$ is normal and hence your group $G$ is always of the form $$G=P\rtimes C_2.$$ By the fact that all matrices of order $2$ in $GL_2(P)$ are conjugate to -I, then groups $G=P\rtimes C_2$ which corresponding to different non-trivial $C_2$ action are all isomorphic. (show it). Hance, for any $P$ you have two non-isomorphic groups and ...


2

The group is (by explicit computation in GAP) the (unique) perfect group of order 1080. Its elements of order 3 are the elements in the centre, C3, as well as two conjugacy classes of order 120 each (they consist of elements that in the A_6 image are 3-cycles, respectively double 3-cycles). Representatives of these two classes are given by $PZ$ and $PQP$. A ...


-1

You can find a general formula in the following reference: Jesús Leanos, Rutilo Moreno, and Luis Manuel Rivera-Martınez."On the number of mth roots of permutations." AUSTRALASIAN JOURNAL OF COMBINATORICS 52 (2012): 41-54.


0

I think one of the most surprising results of this type is the following (and I could just be very naive): Kervaire and Milnor showed that diffeomorphism classes of oriented exotic spheres form the non-trivial elements of a finite abelian group under the connected sum for dimension not equal to $4$.


0

A set of permutations closed under composition. These were historically the first thing called groups, every group algebra is instantiated by some set of permutations closed under composition and most of the theorems you see in group theory books have been developed to understand permutations closed under composition better and their relationship with ...


2

For the mere purpose of counting the number of positions, I can't see why a group theoretical method would be more efficient as the conventional approach. However, if you are interested to know what those positions "look like", then perhaps a group theoretical approach can give you the details you are looking for. One thing that comes to my mind is some ...


0

Since associativity is not immediate to be verified for finite structures whose Cayley table is given, it is probably not obvious that magmas like this one or this other are groups. It is easier instead to catch at a first glance that both have identity or that none of these two is commutative.


1

If I assume that x is just a scalar when I'm trying to find Ker f like I did with Tobias Kildetoft using the nth roots of unity group. Isn't easier to assume directly that x can be written as $e^{i \varphi}$ ? We show that $Ker f = \mathbb{Z}_{n}$ $$Ker f = \lbrace (S,x) \in (SU(n) \times U(1)) \: \vert \: f(S,x) = Sx = e_{U(n)} = I \rbrace$$ Let's find ...


2

I'm writing the answer to my question but I'm not completely sure though Intro What do we want to show ? $$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$$ Groups definition $U(n)$ = the group of $n\times n$ unitary matrices $\Rightarrow$ $U \in U(n): UU^{\dagger} = U^{\dagger}U = I \Rightarrow \mid det (U) \mid ^{2} = 1$ $U(1) =$ the group of ...


1

If $d\theta$ is surjective, then $\dim G = \dim T_e G \geq \dim T_{\theta(e)}X \geq \dim X$. So at least if $X$ is connected, then $\theta$ has dense image in $X$. If $X$ is not connected, then clearly this is not the case. Now say that we are working over a perfect field of characteristic $p > 0$. Consider a map like $\theta: x \mapsto x^p$ from $G = ...


2

Actually, in a group the only element $x$ with $x*x=x$ is the neutral element. So by the very fact that $A\cap A=A$ for all elements and there are several $A$ possible, we see that this is not a group


1

The fact that $A\cap A=A$ does not in fact mean that $A$ is the identity. You need to find a $\textit{single}$ element $x$ such that $A\cap x=A$ for every $A\in P(N)$. Once you find the actual identity element, you can use it to find the inverse of an element (if such an inverse exists - if it doesn't exist, then you can't have a group). To make this ...


5

In $S_n$, the subgroup $S_a\times S_b$, $a+b=n$, $a\not=b$ is maximal. So taking the three maximal subgroups as stabilizers in $S_n$ of disjoint sets of different cardinalities looks like a good place to start for such configurations. Indeed, searching through alternating groups finds the following example (though it is not of this kind that motivated ...


2

Think of $S_6$ acting on itself by conjugation. Now, what is the size of the conjugation class of $(12)(34)$? So what does it tell you about the size of the normiaizor of this element? I will follow my advice. A simple combinatorial arrgument show that there are $$\frac{\binom{6}{2}\binom{4}{2}}{2}=45,$$ elements of the form $(ab)(cd)$ is $S_6$ and they ...


2

Obviously $P_1\cap P_2$ is a $p$-group so take an arbitrary Sylow p-subgroup $X$ of $N_G(P_1)\cap N_G(P_2)$ which contains $P_1\cap P_2$. Then $X\leq N_G(P_1)$ so $X$ is contained in a Sylow $p$-subgroup of $N_G(P_1)$, but this must be $P_1$ since $P_1\unlhd N_G(P_1)$ is the unique Sylow $p$-subgroup. So $X\leq P_1$ and similarly $X\leq P_2$. This implies ...


3

Sandpile groups have the curious feature that the identity element is very complicated. See http://www.ams.org/notices/201008/rtx100800976p.pdf.


2

I split this into many parts for you. The order of the parts is such that later parts may depend on earlier ones. Show that $a^k=a^\ell$, if and only if $k\equiv\ell\pmod9$. Review what you know about cyclic groups, if this is not crystal clear. Show by induction that for all natural numbers $k$ the equation $$ ba^k=a^{-k}b $$ holds in the group $G$. Show ...


3

Hint: Every non-identity element of $G$ is of the form $a^k$ or $a^kb, 0 \leq k \leq 8.$ Try to show that none of these elements can be in the center. For example, if $a^k \in Z(G)$ then $a^kb=ba^k.$ On the other hand by the given condition, $a^kb=ba^{-k}.$ So we must have $a^k=a^{-k} \Rightarrow a^{2k}=1 \Rightarrow 2k|9.$


11

Note first that $f(e) = e$. This is because $$f(f(e)) = f(e f(e)) = f(e) e = f(e),$$ and then also $$f(f(e)) = f(f(f(e))) = f(e f(f(e))) = f(e) f(e) = f(e)^{2}.$$ Then show that $f \circ f = 1$, the identity map. This is because $f(f(y)) = f( e f(y)) = f(e) y = e y = y$. Thus $f$ is bijective, and a homomorphism, as $$ f(x y) = f(x f(f(y))) = f(x) f(y). ...


10

We can start by proving that $f$ is a bijection. Let $y\in G$ be arbitrary. Then, $$f(e f(f(e)^{-1}y))=f(e)f(e)^{-1}y = y$$ holds, so $f$ is surjective. (Where $e$ is the identity element.) To prove that $f$ is also injective, suppose $f(x) = f(y)$. Then we have $$f(ef(x))=f(ef(y))$$ which means that $$f(e)x=f(e)y.$$ It follows that $x=y$, so $f$ is ...


2

There is proof of Schur-Zassenhaus, due to Wielandt, which does not make (explicit) use of group cohomology, but uses group actions instead. As has been noted in comments, it is easy to reduce to the case that $N$ is an elementary $p$-group for some prime $p.$ Let $[G:N] = h.$ We define an equivalence relation $\sim$ of transversals to $S$ by ...


2

First, there is no elements of order $6$ since your group is not cyclic. Then, any non-trivial element has order $2$ or $3$. It is easy to show that it is not possible that all the elements have order two or all the elements have order $3$ (show it). Then you have elements $a,b$ such that $$a^2=b^3=e.$$ Now since your group is non-commutative $$aba=b^2.$$ ...


1

It is a difficult question in general. You can look at some papers of Francis Oger, who worked on the question. In particular, there are links with model theory; for instance: Theorem: Let $G$ and $H$ be finitely generated finite-by-nilpotent groups. Then $G \times \mathbb{Z} \simeq H \times \mathbb{Z}$ iff $G$ and $H$ are elementary equivalent. See ...


4

If $Y_1,Y_2,Z_1,Z_2$ are finite groups, and if $Y_1\times Z_1\cong Y_2\times Z_2$ and $Y_1\cong Y_2$, then $Z_1\cong Z_2$. The following argument (which I believe is due to László Lovász) uses no group theory, so it applies to much more general kinds of finite algebraic structures. For finite groups $X,Y$ let $h(X,Y)$ denote the number of homomorphisms and ...


1

The Krull-Schmidt theorem implies that if $G_1 \times H_1$ and $G_2 \times H_2$ have both chain conditions and can be written as finite direct products of indecomposable groups (these conditions hold in true in particular if they are finite), then $G_1 \times H_1$ and $G_2 \times H_2$ are isomorphic if and only if the union of the multisets of isomorphism ...


3

Since the question itself has already been answered, this answer will add some references for further study. As Joseph Zambrano points out in the comments to his answer, "local rigidity" fails for two dimensional hyperbolic structures. This is a hint of the deep, beautiful theory of moduli spaces of geometric structures. Bill Goldman's survey paper ...


4

Arthur and Squirtle's answers are 100% correct, but just in case you want something more conceptual: This is an example of transport of structure. In detail, if $(G,\cdot)$ is a group, and $f$ is a bijection from $G$ onto a set $X$, then we can define an operation $\bullet$ on $X$ by setting $$ f(g) \bullet f(h) = f(g \cdot h) $$ for all $g,h \in G$. Note ...


2

Hint: if the degrees of the (complex) irreducible representations of $G$ are $$d_1, d_2, \ldots, d_l$$ then $$\mathbb{C} G \simeq \prod_{j=1}^l M_{d_j}(\mathbb{C})$$



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