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## New answers tagged group-theory

0

Let's look at $\langle a\rangle$ which is a cyclic group of order $6$. Hence, $\langle a\rangle = \{a^0 = e, a^1, a^2 \cdots a^5\}$, by the definition of $\langle a\rangle$, where $|\langle a \rangle| = 6$. Now, the elements of this group that are generators of $\langle a \rangle$ are those elements $a^m$, where $m$ is an integer, $0\lt m \lt 6$, such that ...

1

Hint: If $G=\langle z\rangle$ is a cyclic group of order $n,$ say, then you know that the elements of $G$ are (in multiplicative notation) $1,z,z^2,...,z^{n-1}.$ For which $k\in\{0,1,2,...,n-1\}$ does $z^k$ have order $n$?

2

$\Rightarrow:$ Suppose $\alpha$ is an automorphism. Then $\alpha(g h)= \alpha(g) \alpha(h)$. So, by the definition of $\alpha$, $(gh)^{-1}= g^{-1}h^{-1}$. Since $(gh)^{-1}=h^{-1}g^{-1}$, $G$ is abelian because $h^{-1}g^{-1}= g^{-1}h^{-1}$ $\forall g,h\in G$. Does this help with the proof overall? Let me know if you have trouble with the converse.

0

Try "An introduction to the theory of groups/Joseph Rotman"- Theorem 3.18 - page_55 Or Rorman's "Advanced Modern Algebra" book page_99

0

I think with the complexity of the rules that the best you will do is a tree. Start with the empty necklace, then branch to one of each color, and continue by checking what colors can go next given the rules. Depending on how restrictive the rules are, this may stay small or it may explode. If they are contradictory, you would hope to find that out ...

0

Consider $E=\{ e \in \mathbb Z : x^e =1 \mbox{ for all } x\in G \}$, the set of exponents of $G$. Then $E$ is a subgroup of $\mathbb Z$ and so $E=m\mathbb Z$. By Lagrange's Theorem, $n\in E$ and so $m$ divides $n$. $m$ is the lcm of the orders of all elements of $G$. If $a,b\in G$ with $r=ord(a)$ and $s=ord(b)$ and $\gcd(r,s)=1$, then ...

0

You've shown that $d$ divides $n$, but you haven't shown that $d \neq n$ (i.e., you haven't shown that $d$ is strictly smaller than $n$). This is where the "non-cyclic" condition on $G$ comes in. Suppose for the sake of contradiction that $H = G$, and that $H$ is the only non-trivial subgroup of $G$, and hence $|H| = d\mid n = |G|$ and $d = n$ for every ...

0

Suppose $N$ is a normal abelian $p$-group of $G$, and that $p\mid |G/N|$. Then $G$ splits over $N$ if and only if $P$ splits over $N$, where $P$ is a Sylow $p$-subgroup of $G$. This is a theorem of Gashutz. So let's imagine we're in the following situation: For some Sylow $p$-subgroup $P$ of $G$ we have (i) $P$ normal in $G$, (ii) $[P,P]=[G,G]\cap ... 1 Well, question 1 is a little funny. There are two, in my mind, completely different things going on. The first is the difference between a left action and a right action, and how to convert between a left action of$G$and a right action of$G^{op}$. You understand that perfectly well already. The second is that a left action of$G$on a set$X$is the same ... 2 Consider the action of$G$on the cosets of$H$given by$g\cdot (xH)=gxH$. This gives you a homomorphism$G\to S_n$whose kernel is a normal subgroup contained in$H$and having index equal to the size of the image, hence a divisor of$n!$. 0 The nice thing about the group$\mathbb{Z}$is that it is the free group generated by one element. That is, you can map a generator (1 or -1) of$\mathbb{Z}$to any element and that defines a (unique) group homomorphism. One way of seeing the existence of such a homomorphism is as follows: Let$G$be a group and$g \in G$. Then the subgroup$\langle g ...

3

a) is fine. But for b) Note that $n\in\ker \phi$ if $g^n=1$ for $g:=\phi(1)$. As such $g$ can only have orders $2$ or $3$ (or of course $1$) in $S_3$, only $2\mathbb Z$ and $3\mathbb Z$ (and of course $\mathbb Z$ that was excluded) are possible kernels. c) is fine - though in the light of b) you were probably supposed to take $n\mapsto (1\,2\,3)^n$ as one ...

0

The kernel must be normal in $G$ and contain $H$. $$\bigcup_{H\le N\lhd G} N$$ is the smallest such normal group. Divide by it.

3

Hint. Can you make $y$ out of $x$ and $xy$?

1

The following is, imo, a rather nice though perhaps not so used way to answer (c): first, if $\;G\;$ is cyclic there's nothing to prove as a finite cyclic group has one unique subgroup of each and every divisor of its order, so let us assume $\;G\;$ is not cyclic. (1) Prove $\;G\cong C_p\times C_p\;$ , with $\;C_p=$ the cyclic group of order $\;p\;$ . (2) ...

1

Consider the subgraph with the following as vertices and edges. $\langle sr,r^2 \rangle------D_{16} ------ \langle s,r^2 \rangle$ $\langle sr,r^2 \rangle------ \langle r^2 \rangle ------ \langle r^4 \rangle$ $\langle sr,r^2 \rangle------ \langle sr^3,r^4 \rangle------\langle sr^3 \rangle ------ \langle 1 \rangle$ $\langle sr^2,r^4 \rangle ------ ... 4 So if you have no elements of order$p^2$, then all your elements have order$p$. Hence every group you have is cyclic of order$p$. Since they meet at the identity, there is a total of$x(p-1)+1=p^2$elements.$x$is what? 0 Here we know the group itself has order$p^{2}$, so if it is cyclic then you have a unique order$p$subgroup. Otherwise pick up another element$b$not in the$p$-group$A$we already chosen, and we may argue that$|b|=p$. Since you already proved that the intersection of$\langle b \rangle\cap A=e$, you can show$G=A\times \langle b\rangle$. So the group ... 0 The main reference is "Primer on mapping class group" by Farb and Margalit. The point is that for once punctured torus the abelianization of the fundamental group defined an isomorphism of the mapping class group to$GL(2,Z)$. Thus, under your assumption the automorphism has to be inner, conjugation by some element$g$of the free group. Now, use the ... 0 If$x\in B$, then$g_2(x)\in B$, and hence$g_1(g_2(x))\in B$– Prahlad Vaidyanathan Oct 28 at 18:01 To beef up this "answer", I'll add that the statement may fail for infinite sets. For example, if$A=\mathbb Z$and$B$is the set of positive integers, then$g(x)=x+1$has the property$x\in B\implies g(x)\in B$but its inverse does not. 1 Hint: You can use the fundamental theorem on finite abelian group. Write$G=\mathbb{Z}_{n_1}\oplus\dots\oplus \mathbb{Z}_{n_r}$with$n_i$powers of primes. Then for each$n_i$we have that$n_i\mid a$or$n_i\mid b$(Why?), and this or is exclusive. Meaning that both cannot happen (use the hypothesis of the problem). This should tell you how to break up ... 1 Well,$\langle x\rangle$is certainly a subgroup of its centralizer$Z(x)$(which does have order 7) and a group of order$7$can have only one nontrivial subgroup. If$x=1$, then its class size would be$1$, because the centralizer of$1$is the entire group. 2 Well, probably not that useful (or insightful), but it's guaranteed if$[G,G]$and$G/[G,G]$have coprime order. 0 consider that any Automorphism preserves the order, means order of Z(G) = order of τ(Z(G)). 4 Not sure how to "convince" you, but I suspect you are worried because some of the 2-cycles do the job and others do not. The trick is in picking the right 4-cycle to go with your 2-cycle or vice versa. (1,2) works with (1,2,3,4) because 1 and 2 are adjacent in (1,2,3,4) whereas 1,3 is not. But (1,3) will work with (1,3,2,4). For more info: ... 2 I'm not sure what are you up to, but observe that a rather "nice" (who decides?) group fulfills your conditions: the symmetric group$\;S_n\;$, since $$S_n=A_n\rtimes C_2\;,\;\;\text{and}\;\;S_n'=A_n\;,\;\;S_n/S_n'=C_2$$ So you can always say that at least, by Cayley's Theorem, any finite group is a subgroup of a group with the form you want. 3 The Brauer group of$\mathbb Q$(and more generally, of any finite extension of$\mathbb Q$) is described by class field theory. The answer is that it is isomorphic to the direct sum of$\mathbb Z/2 \mathbb Z$and a countable number of copies of$\mathbb Q/\mathbb Z$. More canonically, consider the direct sum $$\dfrac{1}{2}\mathbb Z/\mathbb Z \oplus ... 3 Note that the minimum size of a cycle for the graph for D_{16} is four. This means, if it were to be planar, any "polygons" in the graph would have to have 4 sides or more, and that \frac{4}{2}f\le e. Since v+f-e=2, v-\frac12e\ge 2\Longrightarrow e\le2v-4 must be satisfied for the graph to be planar. There are 19 vertices and 36 edges. ... 5 Hint: Let |G|=n. Take e \neq g_1 \in G and consider \langle g_1 \rangle. Since g_1 \neq e, then |\langle g_1 \rangle| \geq 2. If G = \langle g_1 \rangle then G is generated by 1-element set, else let g_2 \in G - \langle g_1 \rangle. Consider \langle g_1,g_2 \rangle. Since \langle g_1 \rangle is a proper subgroup of \langle g_1,g_2 ... 0 Hint: All elements of H are points on a line l with equation y=2x. So, geomtrically, the cosets should correspond to parallel lines. In each case, a displacement vector acts as a coset repesentative. Can you figure out from this how to describe the operation on the quotient group in geometric terms? 1 Hint: each normal subgroup of a group G is the intersection of the kernels of some of the irreducible characters of G. 0 Here is an idea : You need to first show that the orthogonal matrices in \mathbb R^2 are of the form either \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{bmatrix} or \begin{bmatrix} \cos \theta & \sin\theta \\ \sin\theta & -\cos\theta \\ \end{bmatrix} Hence you can conclude that ... 1 Like Derek Holt mentions in the comments, it seems okay, but you should consider the case N \cap H = N. You are also assuming that G is finite. This is not necessary. For a proof that works in general, you can apply following statement, which is very easy to prove but useful. (Dedekind's modular law) If A, B and C are subgroups such that B ... 1 We know that any subgroup must have the identity element in it. We also know that every subgroup must contain inverses for all of its elements. Suppose that H is a subgroup of order n-1. Let x designate the one element of G not in H. Then x must be its own inverse, as if x^{-1} \not= x we have that x^{-1}\in H, yet x is not in H (which is ... 1 Consider an element x in G and not in H. Since |H| = n - 1, x^{-1} must be in H. However, the inverse is unique so it's easy to see that x^{-1} has no inverse in H because x is not in H. 2 Take a look at gH, where g \in G, g \notin H. Then I claim gH \cap H = \varnothing; for if h \in gH \cap H, then we have h \in H and also h \in gH, whence h = gk for k \in H. Then g = hk^{-1} \in H, a contradiction. So gH \cap H = \varnothing; but then since gH and H each have n - 1 elements, H \cup gH has 2(n - 1) = 2n - ... 3 Let g be an element of G that is not in H. Let h be any nonidentity element of H. Now show that gh is not in H. 7 \DeclareMathOperator{\br}{Br} This can be done using the Brauer-Hasse-Noether theorem. One statement of this theorem is that for k a global field, there is a canonical exact sequence$$ 0 \to \br(k) \to \bigoplus_v \br(k_v) \to \mathbb Q/\mathbb Z \to 0 . $$where v ranges over all places (finite and infinite) of k. It is known (see for example ... 4 The Brauer group of \mathbb{Q} can be identified with a subgroup of \bigoplus_vBr(\mathbb{Q}_v), the direct sum taken over all places of \mathbb{Q} (including the Archimedean one). The map from Br(\mathbb{Q}) into this direct sum is given by tensoring any central simple algebra representing a class of the Brauer group with all the completions, to get ... 0 Order of ab need not be equal to the lcm of |a| and |b| if the gcd is not relatively prime. For example consider G=\langle x\rangle with a=x and b=x^3. Here |a|=4, |b|=4 but |ab|=1. Unless \gcd(|a|, |b|)=1 is given we can't say that order of ab is same as lcm(|a|, |b|) 2 Recall that the center of a group Z(G) consists of all elements that commute with every element in G. That's pretty much all you need, here: If H<Z(G), and h \in H, then h \in Z(G), and so for every g \in G,\; g^{-1}hg =g^{-1}gh = eh = h (h \in Z(B), so h commutes with g.) Since this is true of all h \in H, we have that g^{-1}Hg = ... 4 Since if H<Z(G), and h \in H, for every g \in G g^{-1}hg = h, (since h is central), it follows that g^{-1}H g = H (it is true elementwise). 1 Your initial analysis about a symmetric matrix being decomposable into the trace and a trace-less symmetric matrix is correct. But, your "naive" description of the trace is not correct. First of all, the trace is independent of the basis, and is a sum of the eigenvalues. This can be seen by considering a diagonal matrix, D, and its similar matrix, A, by ... 1 For another counterexample, consider the free group on two generators, G = \langle x,y\rangle. The abelianization of G (i.e., G/[G:G]) is \mathbb Z \times \mathbb Z, and in order to express G as a semidirect product, you need to find a complementary subgroup to [G,G] which is isomorphic to \mathbb Z \times \mathbb Z. However, the centralizer of ... 1 I don't know if you are familiar with Group actions, but this claim is a special case of the following general case: Theorem: If H\le G is a p- subgroup of a finite group such that p\mid [G:H] then p\mid [N_G(H):H]. Indeed, if |G|=p^m for a some m, then for H we have an strict inclusion there.i.e; H<N_G(H). It is good to know that ... 1 Hint: In fact ,a p- group is a nilpotent group. You can prove these step by step: A group G is nilpotent. Every subgroup of G is subsnormal. If H<G (strictly) then H<N_G(H) (strictly). We have that 1\longrightarrow 2, 2\longrightarrow 3 and if G is finite 3\longrightarrow 1. 1 Let G be a group of order 144 and assume that G is simple. We will argue through an analysis of the Sylow 3-subgroups of G to arrive at a contradiction. On the way we will use a result maybe less known. Lemma Let G be a group and p a prime dividing the order of G. Assume that for every pair P, Q \in Syl_p(G), P=Q or P \cap Q = 1. Then ... 1 Here is the start to one approach to this problem. Let h be a generator \mathbb Z\cong H \subset G, and let g\in G with g\not\in H. Then g and h generate G. Moreover, g^2=h^a for some a\in \mathbb N because G/H \cong \mathbb Z /2 \mathbb Z, and ghg^{-1}=h^b for some b\in \mathbb N because H is normal. Therefore, the groups you ... 2 It depends on the polynomial f. If x + (f(x)) is a generator, f is called primitive. For each finite base field and each degree, a primitive polynomial always exists. Example There are 3 irreducible polynomials of degree 4 over \Bbb F_2:$$x^4 + x + 1,\quad x^4 + x^3 + 1,\quad x^4 + x^3 + x^2 + x + 1$$The first two polynomials are primitive, but ... 2 Essentially, you wish to construct an isomorphism$\alpha : S_\mathbb{N} \to S_\mathbb{Z}$. But what are the elements of$S_\mathbb{N}$? They're just$f : \mathbb{N} \to \mathbb{N}$bijective. Likewise for$\mathbb{Z}$. Hint: there is some$g : \mathbb{N} \to \mathbb{Z}\$ that is bijective. Can you use this to construct an explicit isomorphism? What is a ...

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