# Tag Info

11

You need to use the correct notion of isomorphism here: namely, you need to choose an isomorphism between $G$ and $T$ which sends $H_i$ to $S_i$. Such an isomorphism need not exist given only that $H_i$ and $S_i$ are abstractly isomorphic. In other words, the problem here is that $H_1 H_2$ is not a "property of a group"; it depends on three things, namely ...

8

Note that it suffices to show that $a_i - b_i$ is even for some $i$. Using $$\begin{split} (a_1-b_1) + (a_2 - b_2) +\cdots (a_n - b_n) &= (a_1+ \cdots +a_n)-(b_1+ \cdots +b_n) \\ &=(1+\cdots +n) -(1+\cdots + n)\\ &= 0. \end{split}$$ As $n$ is odd, one of them $a_i - b_i$ must be even (as adding $n$ odd terms will give you an odd number, not ...

5

It does. If $C$ is any chain complex in your additive category and $F = 0$ the zero functor, then $H_n(FC) \cong F\bigl(H_n(C)\bigr)$ for each $n \in \mathbf Z$, as both sides are zero.

5

$\newcommand{\ZZ}{\mathbb{Z}}$ The groups may be isomorphic, but when you write something like $H_1 H_2$, you're omitting some information. To talk about this object, you need to know about their parent group $G$, and how $H_1$ and $H_2$ embed into it. And if the way they do so is different from how $S_1$ and $S_2$ embed into $T$, then you'll get a different ...

4

One way to proceed is to notice that $\mathbb Z_8$ is a local ring, so that its finitely generated projective modules are in fact free. Of course, this implies that finitely generated modules have at least $8$ elements.

4

One of the properties of an inner product is positive-definiteness, which requires the field of scalars to contain an ordered sub-field; in particular, finite fields, and fields of finite characteristic will not work, as it is not possible to define an order for them compatible with the field operations. If we wish the inner product to define a norm via ...

3

I believe you have the wrong copy of $\mathbb{C}$ inside $\mathbb{C}[G]$. Instead, you want to take $\mathbb{C}\cong\{c\cdot\delta_e\mid c\in\mathbb{C}\}$, where $\delta_e$ is the function $$\delta_e(g)=\begin{cases}1&\mbox{if }g=e\\0&\mbox{otherwise.}\end{cases}$$ In fact, there is an isomorphism of algebras ...

2

Let's examine more closely how $p$ acts on the different $\pi_i$s: $$p(\pi_1)=p \left( \{1,2\} \cup \{3,4\} \right)=\{p(1),p(2)\} \cup \{p(3),p(4)\}=\{2,3\} \cup \{4,1\}=\{1,4\} \cup \{2,3\}=\pi_3$$ Similarly, one can show that $$p(\pi_2)=\pi_2$$ and $$p(\pi_3)=\pi_1.$$ Thus, when viewed as a permutation of $\{\pi_1,\pi_2,\pi_3\}$, $p$ has the cycle ...

2

$\newcommand{\Z}{\mathbb{Z}}\Z_4$ is not projective over $\Z_8$. Indeed, consider the following exact sequence: $$0 \to \Z_2 \to \Z_8 \to \Z_4 \to 0.$$ The map $i : \Z_2 \to \Z_8$ maps $1$ to $4$, and the map $p : \Z_8 \to \Z_4$ is the quotient map. Then this exact sequence is not split, i.e. there's no $s : \Z_4 \to \Z_8$ such that $p \circ s = ... 2 Fix an element$h \in H$(we can do this, since$H$is non-empty). Consider the function:$L_h: H \to H$given by$L_h(x) = h\circ x$(we know the image of$L_h$is a subset of$H$by closure). Now$L_h$is injective, since the multiplication in$H$is "inherited" from$G$, and the group multiplication is injective. Since$H$is finite, it follows that ... 2 Let the pair be$\{e,a\}$and$\{e,b\}$. Then$\{e,a,b,ab\}$is the order$4$subgroup you want. It is a subgroup because$G$is abelian and$bab=abb=ae=a$and$aba=b$similarly. 2 Conjugation in$S_n$is about relabelling the objects that you are permuting. If$\sigma, \tau \in S_n$, then$\sigma^{-1}\tau\sigma$is the permutation that works by (1) relabelling everything using$\sigma$, (2) permuting the relabelled objects with$\tau$and then (3) undoing the relabelling using$\sigma^{-1}$. So in$S_n$(for$n \ge 3$), the subgroup ... 2 Proof: We prove$HK = KH$. You are trying to prove that$HK$is a subgroup of$G$. Proposition 14 says that for any subgroups$H$and$K$,$HK$is a subgroup if and only if$HK=KH$. So if you can prove that$HK=KH$, Proposition 14 implies that$HK$is a subgroup, which is what you are trying to prove. Let$h \in H$,$k \in K$. By assumption, ... 2 The simplest counterexample is for$n=2$with$G=C_2\times C_2$, with$H_1=H_2=C_2\times\{1\}$. But maybe you want$H_1\cap H_2=\{1\}$and$G=H_1H_2$, in which case another example is$G=S_3\times C_2$with$H_1=\{(\sigma,x)\}\cong S_3$where$x=1$iff$\sigma\in A_3$, and$H_2$a cyclic subgroup of order$2$of$S_3\times\{1\}$. Since$H_2$is not normal ... 2 Now$G$acts on the 3 Sylow 2-subgroups by conjugation. Note that we assumed here that all three Sylow 2-subgroups are not normal (or we are done). This implies in particular that the image$G \to S_3$has more than 2 elements since: If the image has only one elements, all Sylow 2-subgroups are normal, and If the image has two elements, the image must be ... 2 Look at the proof for integers. In this case, one looks at the set$I=\{a-bq\mid q\in\mathbb{Z}\}$, and uses the fact that$I^+=I\cap\mathbb{Z}_{\geq 0}$is nonempty and bounded below (hence, has a smallest element). For$K[x]$, consider the set$I=\{f-gq\mid q\in K[x]\}$. Note that if$0\in I$you are done. Otherwise consider$I^+=\{\deg(r)\mid r\in I\}$... 1 We know that a module is semisimple if and only if every submodule is a direct summand. Therefore, what you are asking is whether$\mathcal M _2 (\Bbb Z)$is a semisimple$\Bbb Z$-module. Note that$\mathcal M _2 (\Bbb Z)$is a free$\Bbb Z$-module (of rank$4), having for basis the subset $$\left\{ \left( \begin{array}{cc} 1 && 0 \\ 0 && ... 1 \forall s \in C(a) we have that$$ \begin{align} sa &= as \\ a^{-1}(sa)a^{-1} &= a^{-1}(as)a^{-1} \\ (a^{-1}s)(aa^{-1}) &= (a^{-1}a)(sa^{-1}) \\ (a^{-1}s)(e) &= (e)(sa^{-1}) \\ a^{-1}s &= sa^{-1} \\ \end{align} $$thus s \in C(a^{-1}). 1 Look in a slightly different way. Let H be a Sylow-2 subgroup. It has 3 cosets in G, and by picking an element x of order 3, the three cosets of H will be \{H,xH,x^2H\}. G acts on this set of cosets naturally. Now how x permutes these cosets? It will move them like 3-cycle:$$H\mapsto x.H \mapsto x.(xH)\mapsto x.(x^2H)=H.$$On the other ... 1 The class equation o(G)=o(Z(G))+sigmaq o(G)/o(C(g)), g is not in the center. Now o(Z(G)) is in fact the sum of order of sigletons if we go back to the definitions of the defining equivalence relation. 1 As (23)(23)=(1) the only elements in H are (1) and (23). 1 You have$$(1 \ 2) (2 \ 3) (1 \ 2 )^{-1} = (1 \ 2) (2 \ 3) (1 \ 2 )= (1 \ 3) \notin H$$So H is not normal. 1 It helps to think of isomorphisms not only as an equivalence of groups, but (in the same (mental) breath) also as a bijective function between the groups. So, when you say H_1 \cong S_1, you should also imagine f_1:H_1 \rightarrow S_1, a bijective group homomorphism. Similarly for f_2:H_2 \rightarrow S_2. With those maps, you can ask the question, ... 1 Let \{a, b\}\cup \{c, d\} be one of \pi_1, \pi_2, \pi_3, where \{a, b, c, d\} = \{1, 2, 3, 4\}. Then \phi is defined by$$\phi(p) (\{a, b\}\cup \{c, d\}) = \{p(a), p(b)\} \cup \{p(c), p(d)\}.$$For example, if p = (1234), then$$\begin{split} \phi(p) (\pi_1) &= \phi(p) (\{1, 2\} \cup \{3, 4\}) \\ &= \{p(1), p(2)\} \cup \{p(3), p(4)\} ... 1G$operates on$X$by permutation. Hence$p\cdot 10=11$,$p^2\cdot 10= p\cdot 11=5$, &c. Thus$\;G\cdot 10=\{4,5,6,8,10,11\}$.$G$is a group of order$6$, since the disjoint cycles making up$p$have order$3,6,1,1$. The orbit of$10$has cardinal$6$, and we have a bijection from$G/G_{10}$onto$G\cdot 10$, hence$ G_{10}=1$. 1 You can solve this problem by considering only$\langle x\rangle$, the subgroup generated by$x$. Using Lagrange's theorem tells us that the order of this subgroup divides$O(x)$and is hence coprime to$n$as well. This reduces the problem to the case of an abelian group generated by a single element, which is reasonably easy to solve. Notice that this ... 1 Bachet-Bézout theorem: if$\gcd(n,m)=1$, then$1=an+bm$for some integers$a$and$b$. Thus $$x=1x=(an+bm)x=n(ax)+b(mx)=n(ax)$$ and you can take$y=ax$. 1 We have $$(a_1-b_1)^2\cdots (a_n-b_n)^2=(1-b_{a^{-1}_1})^2\cdots (n-b_{a^{-1}_n})^2$$ so let us denote$c = b\circ a^{-1}$. In order to$(1-c_1)^2\cdots(n-c_n)^2$to be odd,$c_i$must be of opposite parity of$i$, but,$n$is odd so$\{1,2\ldots,n\}$has exactly$\frac{n-1}2$even and$\frac{n+1}2$odd numbers, so by pigeonhole principle there is at least ... 1 For the product to be even we require one pair$(a_i, b_i)$to have the same sign. Since$n$is odd, there are$\frac{n-1}{2}$even numbers and$\frac{n+1}{2}$odd numbers in$\{1, 2, \dots, n\}$, so it is impossible to pair each even number with an odd number, so one pair must have the same sign and we are done. 1 You should find it in Fuchs' “Infinite Abelian Groups” (volume 1). However, the proof is not really so difficult. Here's a sketch. Let$G$be a divisible abelian group. Then the torsion part$t(G)$is divisible as well and so it splits:$G\cong t(G)\oplus (G/t(G))$. Since$G/t(G)\$ is torsionfree, it has unique division by integers and so it becomes a ...

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