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25 views

Meaning of “Identify a set with another set” in group theory

There is a exercise problem that asks "Identify a set with another set ". I don't understand what I should do. Do I need to establish a bijection between them? Thanks EDIT-I: Actual question: G is a ...
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Finding/Recognising non-cyclic proper subgroups.

$Q$ is a multiplicative group of order $12$. You are given that two elements of $Q$ are $a$ and $r$ and that $r$ has order $6$ and $a^2=r^3$ You are also given that $a$ has order $4$, $a^2$ has order ...
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159 views

Intuition behind normal subgroups

I've studied quite a bit of group theory recently, but I'm still not able to grok why normal subgroups are so important, to the extent that theorems like $(G/H)/(K/H)\approx G/K$ don't hold unless $K$ ...
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1answer
45 views

Intuition - Normal Subgroup Test - Fraleigh p. 141 Theorem 14.13

(1.) Not querying proofs or formality. I do this in my other question. Normal Subgroup Test says H is normal in G $\iff gH{g}^{-1}\subseteq H$ for all $g \in G$. What's the intuition of this ...
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4answers
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Equivalence Relation definitions of Coset - looks like 1-step Subgroup Test? [Fraleigh p. 97 theorem 10.1]

p. 4 We are especially interested in the case where the set is a group, and the equivalence relation has something to do with a given subgroup. That is, we want to partition a group G into subsets, ...
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Any group of prime order is cyclic - Proof blueprint [Fraleigh p. 100 Cory 10.11] [closed]

Not querying the proof or formality. I include only part of the proof. The order of the group is a prime number. Call it p. Hence by means of the definition of prime number, $p > 1$. Since the ...
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2answers
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Intersection of Groups is a Group? Is a Union of Groups? - Fraleigh p. 66 Exercise 6.32h

This is a true or false question, hence are the answers supposed to follow quickly? Because the empty set has no identity element, hence $\emptyset$ is not a group. Hence I'm inquiring for ...
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92 views

What is the relationship between the second isomorphism theorem and the third one in group theory?

The second isomorphism theorem [wiki] in group theory is as follows: Let $G$ be a group. $H \triangleleft G, K \le G$. Then: $HK \le G$, $(H \cap K) \triangleleft K$, and $K/(H \cap K) ...
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56 views

Action of a group on itself by conjugation is faithful $\iff$ trivial center

p. 5: A group action of G on X is called faithful (or effective) if different elements of G act on X in different ways: when $g_1 \neq g_2$ in G, there is an $x \in X$ such that $g_1 \cdot x \neq ...
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122 views

A Nonabelian group of order of product of primes G has a trivial center - Fraleigh p. 153 15.18

Using Exercise 37, show: A nonabelian group G of order pq where p and q are primes has a trivial center. Reference: http://users.humboldt.edu/pgoetz/Homework%20Solutions/Math%20343/hw...
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2answers
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Proof blueprint - If $G/Z(G)$ cyclic then $G$ Abelian - Fraleigh p. 153 15.37

(1.) Why didn't Fraleigh state the result in the direct form like in my title? Why state it with the negations and then prove the contrapositive? Isn't this extra unnecessary work? (2.) How do you ...
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1answer
51 views

Commutator subgroup of a simple group - Fraleigh p. 152 15.19(h)

True or False: (19h.) The commutator subgroup of a simple group G must be G itself. Answer: http://www.auburn.edu/~huanghu/math5310/alg-hw-ans-i think 3.pdf ...
3
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0answers
54 views

Center/Commutator Subgroup of Direct Product = Direct Product of these Subgroups - - Fraleigh p. 64 Theorem 6.14

(1.) What's the intuition? Full proof for Center Subgroups (2.) What's the proof blueprint? I know proof's using $A = B \iff A \subseteq B \wedge B \subseteq A$. But where did $(ga,hb)$ in ...
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3answers
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Intuition - $fr = r^{-1}f$ for Dihedral Groups - Carter p. 75

Name $r$ = clockwise 90 deg. rotation and $f$ = flip across the square's vertical axis = the brown $\color{brown}{f}$ in my picture underneath. Zev Chonoles's $f$ is different. Carter fleshes out why ...
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2answers
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Intuition, Questions on Commutator Subgroup $\neq$ Set of All Commutators - Fraleigh p. 150 Theorem 15.20

This is too advanced for me. Not asking about proofs here. Theorem 15.20: The set of all commutators $= \{aba^{-1}b^{-1} : a,b \in G \} $ generates $ \color{red}{\text{ and hence $\neq$} }$ (but ...
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Intuition and Proof - $H$ is a maximal normal subgroup of $G \iff$ $G/H$ is simple. - Fraleigh p. 150 Theorem 15.18

I don't understand some steps in the proof by B.S.. Start with some definitions. http://en.wikipedia.org/wiki/Maximal_subgroup#Maximal_normal_subgroup: $H \unlhd G$ is a maximal normal subgroup ...
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1answer
80 views

Intuition - Theorem - A group homomorphism preserves normal subgroups - Fraleigh p. 149. Theorem 15.16

p. 128, 129. Theorem 13.12. Let $h$ be a homomorphism of groups $G \to G'$. III. If $S \le G$, then $h[S] \le \color{red}{G'}$. IV. If $S' \le G'$, then $h^{-1}[S'] \le G$. p. 149. ...
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2answers
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Intuition - Quotient Group of Direct Products - Fraleigh ch. 15

Tried http://www.proofwiki.org/wiki/Quotient_Group_of_Direct_Products Proof on p. 3 and 4 . For the case $n = 2$. Define $h: A_1 \times A_2 \rightarrow \dfrac{A_{1}} {B_{1}} \times \dfrac ...
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1answer
57 views

Collapsing a Factor to the identity element - Fraleigh p. 14 Theorem 15.8

p. 146: We should acquire an intuitive feeling for this theorem in terms of $\color{red}{collapsing}$ one of the factors to the identity element. p. 147 15.8 Theorem: $\hat{H} = \{(h, e) ...
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0answers
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Intuition of Picture - Collapse, Factor Group, Homomorphism, Normal Subgroup - Fraleigh p. 144 Figure 15.1

Let $N \unlhd G$. In the factor group $G/N$, the subgroup $N$ acts as identity element. Regard N as being collapsed to a single element, to the identity element. This collapsing of N together ...
2
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1answer
60 views

Nontrivial Homomorphism(s) from $\mathbb{Z_3}$ to $S_3$ - Fraleigh p. 134 13.37

Reference: http://users.humboldt.edu/pgoetz/Homework%20Solutions/Math%20343/hwsome number 1 to 17 that I forgotsolns.pdf There are exactly two nontrivial ...
2
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1answer
51 views

Characterize normal subgroups - Find all subgroups of $S_3$ conjugate to $\{id, (1,3) \}$ - Fraleigh p. 143 14.29

(27.) A subgroup H is conjugate to a subgroup K of a group G (viz. p. 141 $K \le G$ is a conjugate subgroup of $H$), if $i_g[H] = gHg^{-1} =K$ for some $g \in G$. Show that conjugacy is an ...
2
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1answer
205 views

In a finite cyclic group of order n, number of solutions to $x^m = e$ - Fraleigh p. 68 6.53,54

(53.) Show that in a finite cyclic group G of order n, written multiplicatively, the equation $x^m = e$ has exactly m solutions $x$ in G for each $m \in \mathbb{N}$ that divides n. (54.) With ...
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1answer
37 views

If $\phi[H] \subseteq H'$, homomorphism from G to G' induces homomorphism from G/H to G'/H' - Fraleigh p. 143 14.39

Let $H \trianglelefteq \text{ group } G$ and let $H' \trianglelefteq \text{ group } G'$. Let $\phi$ be a homomorphism of G into G'. Show that if $\phi[H] \subseteq H'$, then $\phi$ induces a natural ...
2
votes
1answer
150 views

Intersection of Normal Subgroups is Normal in Subgroup but Not Group - Fraleigh p. 143 14.35

Show that if H is a subgroup of a group G, and N is a normal subgroup in G, then $H \cap N$ is normal in H. Show by an example that $H \cap N$ need not be normal in G. I can condone the proof hence ...
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1answer
78 views

Prove we can speak of the smallest normal subgroup containing any subset - Fraleigh p. 143 14.31,32

http://www.auburn.edu/~huanghu/math5310/alg-hw-ans-13 (I think).pdf Apologies if I missed some backslashes which are induced by InftyReader version 2.9.7.2. Does ...
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1answer
71 views

understanding into algebraic terms difference between homology and cohomology

my previous question understand quotient group was related to understanding of quotient group,i dont need to know too much detailed in group theore,just some part of algebraic topology,especially ...
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2answers
79 views

Intuition - Fundamental Homomorphism Theorem - Fraleigh p. 139, 136

Let $\phi: G \to H$ be a group homomorphism with $K = \ker\phi$. Then $G/K \simeq \phi[G]. $ The hinge to the proof is to define $\Phi: G/K \to \phi[G]$ given by $\Phi(gK) = \phi(g)$. Then we must ...
6
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2answers
180 views

Questions on Proofs - Equivalent Conditions of Normal Subgroup - Fraleigh p. 141 Theorem 14.13

(1.) Why did Fraleigh shirk the proof for $(2) \implies (1)$? By dint of Arthur's comment, $(2) \iff \color{crimson}{gHg^{-1} \subseteq H} \quad \wedge \quad gHg^{-1} \supseteq H \implies ...
4
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1answer
41 views

Visualize meaning of quotient in quotient map, group - etc?

What are the reasons for the name "Quotient" in Quotient map, group - etc? Overhead picture shows each of the three cosets in $A_4$ is mapped to a single - gray - node. But this isn't division? ...
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55 views

Visualize cosets of kernel of homomorphism, normal subgroup

Question 1. 'Since we know that the codomain is a group, this cannot happen.' I don't understand. Can someone elaborate? I know all homomorphisms are functions but not vice versa. Functions are ...
3
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0answers
32 views

Visualize every quotient map follows a pattern, subgroup and its left cosets

page 167. Because of the Fundamental Homomorphism Theorem, Nathan Carter calls non-embedding homomorphisms quotient maps. This is one of the key facts about homomorphisms: they come in ...
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Visualize normal subgroup, normalizer, cosets.

A few important aspects of the relationship $H \lhd N_G(H) \le G$ are highlighted in Figure 7.31. First, the size of $N_G(H)$ is some multiple of |H|, and the size of G is some multiple of $N_G(H)$, ...
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Visualize Fundamental Homomorphism Theorem for $\phi: A_4 \rightarrow C_3$

Question 1. How do you see $\ker\phi = V_4 $ = Klein 4 group ? Book doesn't give formula for $\phi$? Question 2. What's $H$ in $i(aH) = \phi(a)$? I think $H = \ker\phi$ ? Question 3. Why is $i: ...
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1answer
55 views

Image of Group Homomorphism is Finite and Divides |Domain of Group| - Fraleigh p. 135 13.44

Let $\phi: G \rightarrow G'$ be a homomorphism. Show that if $|G|$ is finite, then $|\phi[G]|$ is finite and divides $|G|$. Because $φ[G] = \{φ(g) \, | \, g ∈ G\}$, we see $|φ[G]| ≤ \quad |G|$ which ...
4
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1answer
94 views

Visualize left, right cosets and conjugation

I drew everything that's in orange. Figure 6.8. Left illustration - Each left coset gH is where H arrows can reach from g, which looks like a copy of H based at g, as in the left illustration. ...
5
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1answer
41 views

Visualize cosets of $\left<(0,1)\right>$ partition $C_3 \times C_3$

Page 105 says - A careful look at Figure 6.9 reveals that the cosets of $\left< \, (0,1) \,\right>$ partition $C_3 \times C_3$. How is this true? The picture shows $gH = left picture = ...
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1answer
38 views

Is there a nontrivial homomorphism for each of the given groups? - Fraleigh p. 134 13.38, 13.41, 13.43

(38.) $\mathbb{Z} \rightarrow S_3$? Let $φ(n) = \begin{cases} \mathrm{id} \in S_3 &, \text{for all $n$ even,} \\ \mathrm{transposition} (1,2) &, \text{for all $n$ odd integers.} ...
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1answer
99 views

Intuition - Homomorphic Image of Group Element is Coset - Fraleigh p. 135 13.52, p.130 Theorem 13.15

Theorem 13.15: Let $\phi: G \rightarrow G'$ be a group homomorphism, $g \in G$. Then $g\ker\phi = (\ker\phi)g = \operatorname{Im}^{-1} \left[ \; \{ \; \phi(g) \; \} \; \right] = \phi^{-1}[ \; \{ ...
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3answers
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Tricks - Prove Homomorphism Maps Identity to Identity - Fraleigh p. 128 Theorem 13.12(1.)

Let $\phi$ be a homomorphism of a group G into a group G'. If $e =$ the identity element in G, then $\phi(e) =$ the identity element in G'. Is this what Sharkos is trying to answer: About ...
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1answer
32 views

Intuition and Strategy - Index of Subgroup of Subgroup Proof - Fraleigh p. 103 10.38

This isn't a duplicate. I tried kb's answer and Answerer 1 but I'm still confounded. I like $\frac {\left| G\right| } {\left| H\right| }$ better than $[G:H]$ hence I write it as a fraction. Suppose ...
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1answer
138 views

A subgroup has the same number of left and right cosets - Tricks - Fraleigh p. 103 10.32, 35

(32.) Let $H \le$ group G and let $a, b \in G.$ Prove or disprove. If ${aH= bH},$ then $Ha^{-1} = Hb^{-1}.$ $\color{blue}{Ha^{−1}} = \{\color{magenta}ha^{−1} | h ∈ H\} = ...
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0answers
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Counterexamples to Nonidentities - Power of Cosets and Right Coset - Fraleigh p. 103 10.30, 33

Let $H \le$ group G and $a, b \in G.$ Prove or give a counterexample. If $aH= bH,$ (30.) then $Ha= Hb.$ (33.) then $a^2 H = b^2 H.$ I understand p. 3: Let $G = S_3$ and $H = \{(1), (1,3)\}$. ...
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1answer
178 views

The only element of $S_{\large{n \ge 3}}$ satisfying $\sigma y = y\sigma$ for all $y \in S_n$ is the identity permutation - Fraleigh p. 86 8.47

I don't want to type Greek letters hence I replaced $\gamma$ by $y$. Microsoft didn't replace them all. Call the identity permutation $id$. Prove the contraposition: $\sigma \neq id \implies ...
6
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1answer
49 views

If two powers of permutations are equal and have no common symbols, they're the identity. - Mulholland p. 44 Proof to Theorem 4.2

Theorem 4.2 (Order of a Permutation): The order of a permutation written in disjoint cycle form is the least common multiple of the lengths of the cycles. Proof: One cycle: As we noted above, a ...
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1answer
110 views

Intuition - Identities with 2-Cycles and 3-Cycles - Mulholland p. 69, 86 - Fraleigh p. 90

Jamie Mulholland p. 69 Theorem 6.1 or Fraleigh p. 90 Corollary 9.12 Any permutation of a finite set of at least two elements is a product of 2-cycles. $1. (a_1, a_2, ···,a_n)= (a_1, a_n)(a_1, ...
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143 views

How to Visualize Diagonally Opposite Vertices

Consider a cube that exactly fills a certain cubical box. As in Examples 8.7 and 8.10, the ways in which the cube can be placed into the box corresponds to a certain group of permutations of the ...
4
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2answers
68 views

Isomorphism of Group with the Image of the Group - Fraleigh p. 82 Lemma 8.15

I found multifarious duplicates that I listed at http://math.stackexchange.com/a/631364/53934. I edged the purple part because my answer proves it more efficiently. I remember that any function ...
5
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2answers
116 views

Visual Group Theory's Intuitive Proof - Cayley's Theorem - Nathan Carter pp. 85, Theorem 5.1

Theorem 5.1. Cayley's Theorem: Every group is isomorphic to a collection of permutations. Figure 5.31. A multiplication table for the group $V_4$, with nodes numbered 1 through 4 to facilitate ...
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1answer
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Connections and Differences between these Cayley Diagrams for $A_4$ and $S_4$ - Carter pp. 80, 82

Reference: Nathan Carter pp. 80, 82, ch. 5, Visual Group Theory Figure 5.24. As you will read in the next section, it is no coincidence that [the Cayley digram for $S_4$] looks cube-like. A Cayley ...