1
vote
1answer
69 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
40 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
70 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...
2
votes
2answers
54 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
37 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
27 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
35 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
68 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
votes
0answers
47 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 ...
0
votes
1answer
61 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
35 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
49 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
46 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
44 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
35 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 ...
3
votes
0answers
85 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
34 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
88 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
59 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 ...
0
votes
1answer
63 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
1answer
56 views

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

Let $\phi: G \to G'$ be a group homomorphism with $K = \ker\phi$. Then $\Phi: G/K \to \phi[G]$ given by $\Phi(gK) = \phi(g)$ is an isomorphism. If $\nu : G \to G/K$ is the homomorphism given by ...
6
votes
2answers
145 views

Intuition - Normal Subgroup Test - 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
37 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
43 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
28 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 ...
4
votes
0answers
58 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
72 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
47 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
votes
1answer
75 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
votes
1answer
39 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
33 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
92 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}[ \; \{ ...
4
votes
1answer
29 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
115 views

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

(32.) Let H be a subgroup of a 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
32 views

Magical Counterexamples to Nonidentities - Power of Cosets and Right Coset - Fraleigh p. 103 10.30, 33

Let H be a subgroup of a group G and let $a, b \in G.$ Prove the statement or give a counterexample. If $aH= bH,$ (30.) then $Ha= Hb.$ (33.) then $a^2 H = b^2 H.$ I understand p. 3: Let $G = S_3$ ...
5
votes
1answer
120 views

For $n \ge 3$, the only central element of $S_n$ is the identity - Fraleigh p. 86 8.47

Strengthening Exercise 46, show that the only element of $S_{\large{n \ge 3}}$ satisfying $\sigma\gamma = \gamma\sigma$ for all $\gamma \in S_n$ is the identity permutation. I call this $i$. The ...
6
votes
1answer
45 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
108 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, ...
5
votes
2answers
109 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
votes
2answers
60 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
votes
2answers
100 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 ...
5
votes
1answer
43 views

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 ...
4
votes
2answers
82 views

Suppose a unique a generates a cyclic subgroup of order. Show ax = xa. - Fraleigh p. 67 6.50

(1.) I don't understand above. How do you magically envisage and envision to let $b = xax^{-1}$? What I did was to start from the answer and see if I can get a chain of equivalences. $ax = xa ...
6
votes
3answers
136 views

Intuition and Tricks - Crafty Short Proof - Generators, Order of a Cyclic Group - Fraleigh p. 64 Theorem 6.14

This stronger result and easier proof is based on p. 58. Hence it isn't a duplicate of this. Theorem 206 and 207. Let $G$ be a group, $k \in \mathbb{N}$ and $a \in G$ such that $|a| = n$. Then: 206. ...
5
votes
3answers
167 views

Intuition and Tricks - Hard Overcomplex Proof - Order of Subgroup of Cyclic Subgroup - Fraleigh p. 64 Theorem 6.14

Update Dec. 28 2013. See a stronger result and easier proof here. I didn't find it until after I posted this. This isn't a duplicate. Proof is based on ProofWiki. But I leave out the redundant $a$. ...
2
votes
1answer
81 views

Solvable and nilpotent groups, normal series and intuition

I'm reading Hungerford's algebra and I'm on Nilpotent and solvable groups chapter. Hungerford starts with: Consider the following conditions on a finite group G: i) G is the direct product ...
4
votes
2answers
218 views

Intuition - An inverse of a generator is a generator and powers of generators - Fraleigh p. 58 5.46

(1). How do you envisage or envision a Cyclic Group with only one generator can have at most 2 elements? Solution is based on this. The integers have two generators, 1 and -1. Which should give you ...
5
votes
3answers
114 views

Intuition, proof, one-sided group definition - Any set with Associativity, Left Identity, Left Inverse is a Group - Fraleigh p.49 4.38

There's a similar post on this question but the third paragraph there is almost impossible to figure out. I explain in my profile why I have to use the big word "prognosticate." Proof that left ...
13
votes
1answer
189 views

Intuition about the class equation (and flowers).

I apologize in advance for the size of the images I've devoted a lot of time and effort to draw them on the computer and i didn’t manage to re-size. I'd be really thankful if anyone would edit this ...
6
votes
1answer
124 views

Symmetric groups and the “field with one element”

I have heard several times that one may regard the symmetric group on $n$ letters as the general linear group in dimension $n$ over the "field with one element". In particular this heuristic would ...