# Tag Info

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Whenever $G$ is finite and its automorphismus is cyclic we can already conclude that $G$ is cyclic. Because as we already saw $G$ is abelian and finite, we can use the fundamental theorem of finitely generated abelian groups and say that wlog $G=\mathbb{Z}/p^k\mathbb{Z} \times \mathbb{Z}/p^j \mathbb{Z}$. But the automorphismgroup isn't abelian and ...

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Yes, the perfect group $\operatorname{SL}(2,5)$ of order 120 is one. By Cauchy's theorem you won't find an example with $G/N$ abelian, so this is more or less the typical example. You want a non-split extension, and you probably want $N$ to be small to avoid coincidentally containing $G/N$. A good way to do this are the non-simple quasi-simple groups. If ...

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Let $G$ act transitively on a set $A$, let $a \in A$, and let $B$ the set of points fixed by $G_{a}$. Let $b \in B$. Then there exists $g$ such that $a g = b$. Let $h \in G_{a}$. Then $a g h = b h = b = a g$, so that $g h g^{-1} \in G_{a}$. In other words, $g \in N_{G}(G_{a})$. Conversely, if $g \in N_{G}(G_{a})$, then $b = a g$ is fixed by $G_{a}$. Now ...

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Your list of expressions for group elements has the nice property that if the expression $u \cdot v$ is in the list, then so are $u$ and $v$. Such a list turns the Cayley graph of the group into a directed acyclic graph with a single source (the identity), or perhaps better a spanning tree with direction. The multiplication in the group is given by the ...

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If $q\mid p-1$ then $\rm{Aut}(C_p)$ has a unique subgroup of order $q$, and the map embedding $C_q$ in $\rm{Aut}(C_p)$ gives a semidirect product, which is not abelian (easy to check). On the other hand, if $G$ is some other non-abelian group of order $pq$ then it is an easy exercise that $G$ has a normal subgroup of order $p$ and since $G$ also has a ...

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The group of invertible elements of $\operatorname{Grp}(G,G)$ contains the automorphism group of $G$, and in particular the group of inner automorphisms, which is isomorphic to $G/Z(G)$. So take the monoid $\{ z \} \cup C_{p}$, where $p > 2$ is a prime, $C_{p}$ is a multiplicative cyclic group of order $p$, and $z$ is the zero. So $G/Z(G)$ is ...

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As you noted the group $G$ has an element of order $p$ and an element of order $q$ where $p,q$ are distinct primes, so we have $|G|=pq$. This is not necessarily happened. For example take $G=\mathbb Z_{18}\cong\mathbb Z_2\times\mathbb Z_{9}$.

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A coset of this subgroup is a set of all real numbers who's pairwise ratio is a power of $2$ (really a power of $\frac{1}{2}$, but that's the same thing), possibly with a sign change. For instance, the coset containing $\pi$ also contains $16\pi$ and $-\frac{\pi}{128}$, because $16$ and $\frac{1}{128}$ are both powers of $2$. Same thing, but this time the ...

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If I understand correctly, you mean the subgroups $$H_1=\langle\tfrac{1}{2}\rangle\subset\mathbb{R}^{\ast}\qquad\text{ and }\ H_2= \langle\tfrac{1}{2}\rangle\subset\mathbb{R}^+.$$ So as sets, these subgroups look like $$H_1=\left\{2^k:\ k\in\mathbb{Z}\right\}\qquad\text{ and }\qquad H_2=\left\{\tfrac{k}{2}:\ k\in\mathbb{Z}\right\}.$$ Can you now determine ...

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Consider the subgroups $A={\rm Sym}(2{\bf Z})$ and $B={\rm Sym}(1+2{\bf Z})$ sitting inside $G={\rm Sym}({\bf Z})$. So $G$ is the set of bijections from the set of integers to itself, $A$ is the set of permutations of the even integers or equivalently the permutations which fix all odd integers, and $B$ the set of permutations of the odd integers or ...

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First use $b^2=1$ to reduce $b^s$ to at most one $b$. Then write $b^{-1}ab=a^{-1}$ as $ba=a^{-1}b$ (again using $b^2=1$), and use that to successively move the single $b$ to the right.

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Suppose that $x,y\in D_{2n}$. Then for some $r,t \in \{0,1,...,n-1\}$ and for some $s,u \in \{0,1\}$, we know that $x=a^rb^s$ and $y=a^tb^u$. Recall that $D_{2n}$ is a group, and is thus closed under multiplication. Thus, $xy\in D_{2n}$ so that for some $i \in \{0,1,...,n-1\}$ and for some $j \in \{0,1\}$, we have $xy = a^ib^j$. Hence, as desired: $$... 2 \operatorname{Out}(G) is defined as the quotient \operatorname{Aut}(G)/\operatorname{Inn}(G), so$$\left|\operatorname{Out}(G)\right|=\left[\operatorname{Aut}(G):\operatorname{Inn}(G)\right]=\frac{\left|\operatorname{Aut}(G)\right|}{\left|\operatorname{Inn}(G)\right|}$$no matter what G is. Your confusion seems to be that you think that ... 2 Take A' \to A \to A/A', the first map is the inclusion and the second is the quotient projection. Take then B' \to B \to B/B', the maps as above. These are both exact sequences, which means \operatorname{im}(\text{inclusion}) = \ker(\text{projection}). You should notice that (\phi', \phi, \phi'') is a morphism of exact sequences, which means that ... 1 As I began to correct my former post: Hints$$\begin{align*}\bullet&\;\;\;G^2:=\langle x^2\;;\;x\in G\rangle\lhd G\\ \bullet&\;\;\;G^2\le H\\ \bullet&\;\;\;\text{The group}\;\;G/G^2\;\;\text{is abelian and thus}\;\;G'\le G^2\end{align*}$$10 H is a normal subgroup of G \iff \forall h\in H \forall g\in G:g^{-1}hg \in H g^{-1}hg=g^{-1}g^{-1}ghg=(g^{-1})^2h^{-1}hghg=(g^{-1})^2h^{-1}(hg)^2\in H(hg\in G \to (hg)^2\in H) then$$g^{-1}hg \in H$$3 Can you see that you have correctly answered your question? For abelian group G, we know \,\operatorname{Inn}(G) \cong e.\; So, we have that$$\operatorname{Out}(G)\cong \operatorname{Aut}(G)/\operatorname{Inn}(G)\cong \operatorname{Aut}(G)$$2 The trick is to first show that N is abelian and after that to show that all non-identity elements have order p for some fixed prime p. Note that any non-trivial group will have a simple quotient, so the condition that any simple quotient is abelian means that N' is a proper subgroup of N. But N' is a characteristic subgroup of N, so it is ... 2 Here is my comment as an answer with the details: We know that for all \varphi\in\rm{Aut}(G) and all h\in H we have h^{-1}\varphi(h)\in K. In particular, this holds for the automorphisms \varphi_g given by \varphi_g(x) = g^{-1}xg (for g\in G). Thus, we have that for all h\in H and g\in G, h^{-1}\varphi_g(h) = h^{-1}g^{-1}hg = [h,g]\in K ... 2 In general, if H is a normal subgroup of G, the elements of the quotient group G/H are by definition the cosets gH for g\in G. To find them, write them one at a time. First, we know we have$$H = \{(1,1),(2,2),(3,3),(0,0)\}$$Now take an element not in H, let's say (0,1). This gives us the coset$$(1,2) + H = \{(1,2),(2,3),(3,0),(0,1)\}$$... 2 If \Gamma<\Bbb Z^k has index n, then n(\Bbb Z^k/\Gamma)=(0). Therefore \Gamma contains the subgroup (n\Bbb Z)^k. This shows that there's a bijection between the wanted subgroups and the subgroups of$$\Bbb Z^k/(n\Bbb Z)^k\simeq(\Bbb Z/n\Bbb Z)^k$$of index n. Then one can simply apply the structure theorem for finite abelian groups. 4 Try SortParallel(D1, 3s); and then redo the steps D := []; for k in [1..35] do D1[k]:=Size(Orbit(G1,3s[k],OnSets)); od;; D1; Is this that you want? 1 A result commonly attributed to Clebsch and Gordan gives you the answer:$$ \varphi_2\otimes \varphi_3\cong \varphi_5\oplus\varphi_3\oplus\varphi_1. $$You can study the dimensions of weight spaces to see this. The weight five space is 1-dimensional (spanned by the tensor product of the highest weight weight spaces of both \varphi_2 and \varphi_3. The ... 2 In your hypothesis there is an element 1\neq a\in G whose order k is a divisor of 20 (proof: divide 20 with remainder by k and use minimality of k), i.e k\in\{2,4,5,10,20\}=:A. But the order of an element in a finite group must divide the order of the group (by Lagrange's theorem, for example) and none of the element of A divides 9, so (B) is ... 4 The key here is Lagrange's theorem which says that "The order of an element divides the order of a group (number of elements in G)." Now since a^{20} = 1, the order of a (denoted by o(a) ) must be a factor of 20. So then o(a) = 20, 4, 5, 2 (we are given o(a) \neq 1). Now by Lagrange's theorem, we know that o(a) must divide |G| \space ... 3 By request, posting a counterexample for question 2 as an answer. The group of rational numbers \mathbb{Q} under addition is clearly not simple. But the endomorphism ring \mathrm{End}(\mathbb{Q}) is isomorphic to \mathbb{Q} itself, and is a division ring, which means that all non-zero endomorphisms are automorphisms. 1 This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that ... 6 Nice question! The answer is no. The smallest potential counterexample works. Let G be the binary icosahedral group. This is a perfect group of order 120 (in fact the smallest nontrivial such group which is not simple). Its only nontrivial normal subgroup is its center \pm 1, hence its only nontrivial quotient is the icosahedral group A_5, which ... 2 Your answer has a couple of issues as discussed in the comments: You are using n to denote the pair (n,0) (recall that the elements of \mathbb Z\rtimes \mathbb Z_2 are defined as pairs). More importantly, you are basically arguing that \mathbb Z\rtimes \mathbb Z_2/\langle n\rangle and \mathbb Z_n\rtimes \mathbb Z_2 are equal as sets, which isn't ... -2 The way to show this, is to note that any of the dihedral groups have a cyclic condition. When this condition is removed, the elements are infinite. In essence, the cyclic groups can be written with a mod(Z) condition, while the infinite group is just z. Thus, every cyclic group becomes the infinite group with a cyclic remainder. 0 Here's another approach. You want to show for any p-subgroup K, K\subseteq aPa^{-1} where P is a Sylow p-subgroup and a\in G . Let X=\{aP \mid a\in G\} be the set of left cosets of P in G. Now let K act on X in the following manner k \cdot(aP)=(ka)P . Let |G|=p^n m where p \nmid m , we know that |P|=p^n since P is a Sylow ... 1 This is a very general fact, and does not need fancy extra structure like that of a Lie group (or even a Riemannian structure): If you have two smooth vector fields X,Y on a smooth manifold M (for simplicity, assume complete vector fields), and if \phi_t: M \to M denotes the flow of X, then$$\frac{d}{dt}\Big|_{t=0} (\phi_t)_* Y = [X,Y].$$... 1 HINTS You have G=\text{GL}(n,\mathbb{R}) with the operation of matrix multiplication and H=\mathbb{R}^* with the operation of multiplication. You gave \det : \text{GL}(n,\mathbb{R}) \to \mathbb{R}^*. You need to show that composing two elements of G and then moving it over to H gives the same as moving two elements of G over to H and ... 2 For the first part, you must show that if A\in\text{GL}_n(\Bbb R), then \det(A)\in\Bbb R and \det(A)\ne0, and that for A,B\in\text{GL}_n(\Bbb R), we have \det(AB)=\det(A)\det(B). You should already know of at least one way to show that a subset K of a group G is a subgroup of G. Use such a way to show that the kernel of \phi is a subgroup ... 0 First study the group U(2). This is the set of matrices A satisfying$$A \overline{A}^t=1,$$where the overline denotes complex conjugation of the entries and the t is the transpose. The map A \mapsto A \overline{A}^t is a differentiable map from the set of all 2 by 2 complex matrices to the set of all matrices C satisfying \overline{C}^t=C. ... 2 But what if x=b^2? Then b^x = b^{b^2} = b \neq b^{-1}. In other words, x and b do not commute, but b^2 and b do commute. Equality of x=b^2 implies x commutes with b, despite our assumption that x^{-1} b x = b^{-1}. In my notes, it says that any group G is isomorphic to Q_{2n} if the following conditions hold: This should say ... 1 No. Consider the case G=(\Bbb R,+),~B=F=S^1, and G acts on F via$$t\cdot z=\exp(it)\cdot z$$Then there is no map g:B\to G such that g\cdot 1=\mathrm{id}, where 1 is the constant map B\to F, z\mapsto 1. Indeed any map g: S^1\to\Bbb R is homotopic to a constant map because \Bbb R is contractible, however 1 and \mathrm{id} are not ... 1 Even if it can't be applied to your example, I would like to point out that in the abelian case (more generally in any abelian category) it's equivalent to have a split exact sequence: 0 \to \ker(\phi) \to G \to Im(\phi) \to0 2 Let N = ker(\phi) and K = im(\phi), then you're asking when, given an exact sequence 1 \to N \to G \to K \to 1 is trivial. First you need the extension to be split, that is, there must exist a morphism s : K \to G such that the composition \phi \circ s is the identity. In this case G \simeq N \rtimes K, the semidirect product of N and K : ... 1 Mathscinet gives a favourable review of the book you mention, namely Partially Ordered Groups by A. M. W. Glass. The review says that it `will surely be an instant classic', and it has apparently been cited 78 times. It concludes saying that the book will get the reader to the forefront of research in the field and would be suitable texts for students in ... 1 If it's finite groups you are talking about, then the following argument applies. Your hypothesis is that G_{\alpha} \le G_{\beta}. Since G acts transitively, there is g \in G such that \alpha g = \beta, and then g^{-1} G_{\alpha} g = G_{\beta}. If G is finite, this implies that G_{\alpha} and G_{\beta} have the same order, so they are ... 1 Considering bit strings of length n as n-tuples of elements of \mathbb Z/2\mathbb Z, the XOR operation is simply the addition in (\mathbb Z/2\mathbb Z)^n. So, yes, this "is" just the group (\mathbb Z/2\mathbb Z)^n. 2 From Wikipedia (http://en.wikipedia.org/wiki/Projective_linear_group#Finite_fields): The projective special linear groups PSL(n, F_q) for a finite field F_q are finite simple groups whenever n is at least 2, with two exceptions:PSL(2, F_2) and PSL(2, F_3). Since PSL is a quotient group of SL, then GL cannot be soluble. 1 To determine \mathrm{Inn}(D_4),first observe that the complete list of inner automorphisms is$$\phi_{R_{0}},\phi_{R_{90}},\phi_{R_{180}},\phi_{R_{270}},\phi_{R_{H}},\phi_{R_{V}},\phi_{R_{D}},\phi_{R_{D'}}$$Here R_0,R_{90},R_{180},R_{270} are rotations of square. H and V denote the reflection of square with horizontal and vertical reflection and D and ... 4 I don't think there is a very conceptual approach giving a meaning to the comparison of conjugacy class sizes, but I think there is a way to make the argument quite short. The first thing to do is to cancel the term a_1! for i=1 (for the points remaining fixed) in the denominator against a part of the numerator n!, which leaves the numerator equal to ... 5 We know that$$D_4=\langle x,y|x^4=y^2=1, (xy)^2=1\rangle = \{e, x, y, x^2, x^3, xy, x^2y, x^3y\}$$By some handy manipulation we can see that |x^2y|=2.$$(x^2y)^2=x^2yx^2y=x(xyx)(xy)=(xy)^2=1 In fact, $2$ is the smallest positive integer in which we can lead $x^2y$ to $1$. With the same way we can see that $|x^3y|=2$. On the other hand $|x^3|=4$. So ...

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Yes, your conjecture is correct. Here is proof: Suppose $g \in X$. Then, since $e \in X$, $e=g^{-1}g \in g^{-1}X$, so that by cosets of $X$ partitioning $G$, $g^{-1}X=X$, so that $e \in X \Rightarrow g^{-1}e \in g^{-1}X=X$. This proves that $X$ is closed under inversion map $G \rightarrow G$. For composition, if $g,h \in X$, then $h=he \in hX$, so similarly ...

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Mark A. Armstrong - Groups and Symmetry is definetely one of what you are looking for. http://www.amazon.com/Groups-Symmetry-Undergraduate-Texts-Mathematics/dp/0387966757

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This is only a partial argument that solves some cases. The idea is simple: instead of focussing on a particular conjugacy class, we consider that of an appropriate power which has simpler cycle structure. It follows from the fact that $\binom{n}{2}<\binom nl$ for all $2<l<n-2$ that for any permutation with total length $l$ (i.e. having $n-l$ fixed ...

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A group $G$ typically have many different generating sets, that is subsets $S\subseteq G$ with $\langle S\rangle = G$, and among these are sets of different cardinality (for example, trivially $S=G$ is a generating set). Therefore it makes little sense to speak of the number of generators of a group (or even of the set of generators). We do speak of the ...

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