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1answer
21 views

Elements in the same coset and the Cayley Diagram

A question from Visual group theory, by Nathan Carter. In a Cayley diagram, if $aH$ is a coset of a subgroup $H$ of a group $G$ and $b$ belongs to $aH$, why is it that every node that can be reached ...
3
votes
2answers
77 views

The semidirect product as a deformation of the direct product

The way I think of the semidirect product is as a "deformation" of the direct product. Is there a way of making this intuition precise? Perhaps using some certain (co-) homology theory of groups?
10
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1answer
94 views

Geometric Intuition for Dihedral Group Automorphisms

I noticed the other day that the automorphism group of the dihedral group $D_{2n}$ (of order $2n$) is $\operatorname{Aff}(\mathbb Z/n\mathbb Z)$, the group of affine transformations of the $\mathbb ...
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votes
1answer
26 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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2answers
34 views

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 ...
4
votes
2answers
95 views

How can one visualize a homomorphic mapping.

It has been a year or so studying Group theory and Ring theory. Funnily enough, this is the part where i am able to solve most of the questions of the book quite easily, yet not fully understanding ...
4
votes
3answers
165 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 $K$ ...
1
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1answer
50 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 ...
3
votes
4answers
45 views

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, ...
2
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3answers
85 views

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 ...
2
votes
2answers
43 views

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 ...
1
vote
1answer
98 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) ...
2
votes
1answer
57 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 ...
3
votes
1answer
137 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...
3
votes
2answers
68 views

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 ...
2
votes
1answer
55 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
votes
0answers
60 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 ...
3
votes
3answers
44 views

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 ...
2
votes
2answers
92 views

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 ...
2
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0answers
64 views

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 ...
1
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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. ...
2
votes
2answers
41 views

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 ...
3
votes
1answer
60 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) ...
2
votes
0answers
73 views

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
votes
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
votes
1answer
58 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
votes
1answer
215 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 ...
2
votes
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
200 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 ...
4
votes
1answer
86 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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votes
1answer
76 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 ...
6
votes
2answers
85 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
votes
2answers
186 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
votes
1answer
50 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? ...
3
votes
0answers
64 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
votes
0answers
34 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 ...
5
votes
0answers
84 views

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 $N_G(H)$, ...
7
votes
2answers
109 views

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: ...
4
votes
1answer
63 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 ...
5
votes
1answer
106 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. ...
6
votes
1answer
48 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 = ...
5
votes
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.} ...
5
votes
1answer
105 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}[ \; \{ ...
5
votes
3answers
158 views

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 ...
4
votes
1answer
33 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 ...
6
votes
1answer
146 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\} = ...
4
votes
0answers
46 views

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 $H = \{(1), (1,3)\}$. ...
8
votes
1answer
181 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
votes
1answer
50 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 ...
5
votes
1answer
111 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, ...