Gu Pei abstract algebra 1.4 "group homomorphic and homogeneous" Exercise answers

Exercise:

7. rewrite theorem 1.4.10 to a more general language. The first sentence is: "set $ F $ to the full homomorphic of the group $ g _ {1} $ to $ g _ {2} $, and $ H <G _ {1} $, and remember $ n ={\ RM Ker} f $, then ...... "

Similar to this theorem, we provide the following solutions:

(1) $ HN $ is a Subgroup containing $ N $ in $ g _ {1} $ and

$ HN = f ^ {-1} (f (H) $

That is, $ HN $ is the full original image of $ F (h) $;

(2) $ (H \ cap n) \ LHD h $ and $ {\ RM Ker} f | _ {H} = H \ cap n $;

(3) Consider homomorphic full shot $ f |_{ h}: H \ To F (H) = HN $, which is known according to the basic theorem of homomorphic.

$ HN \ simeq H/(h \ cap n) $

 

9. The following proposition is incorrect:

Incorrect $ g _ {1}, G _ {2} $, $ N _ {1} \ LHD g _ {1 }, N _ {2} \ LHD g _ {2} $,

$ G _ {1} \ simeq g _ {2}, N _ {1} \ simeq N _ {2} $

$ G _ {1}/N _ {1} \ simeq g _ {2}/N _ {2} $.

For example, $ g _ {1} = g _ {2} = \ mathbb Z :\{\ mathbb Z, ++ \}$, take $ N _ {1} = 2 \ mathbb Z, N _ {2} = 3 \ mathbb Z $, then $ N _ {1} \ simeq N _ {2} $. as a matter of fact, we can see from the homogeneous structure below:

\ Begin {Align *} \ Phi: N _ {1} & \ to N _ {2} \ 2a & \ mapsto 3A \ end {Align *}

However, $ g _ {1}/N _ {1} = \ mathbb Z _ {2 }, g _ {2}/N _ {2} = \ mathbb Z _ {3} $, while

$ | \ Mathbb Z _ {2} | = 2, | \ mathbb Z _ {3} | = 3 $

Obviously, the two are not homogeneous.

 

Additional questions:

1. proof theorem 1.4.8. The content is as follows:

If $ F $ is the full homomorphic of the group $ g _ {1} $ to $ g _ {2} $, remember $ n ={\ RM Ker} f $, then

(1) $ F $ creates a bishot between a subgroup of $ N $ and a subgroup of $ g _ {2} $ in $ g _ {1} $;

(2) The above ing maps normal subgroups into normal subgroups;

(3) If $ n \ subset H \ LHD g _ {1} $

$ G _ {1}/h \ simeq g _ {2}/F (h). $

Proof (1) make $ \ gamma $ represent the whole of the $ g _ {1} $ Subgroup containing $ N $, $ \ Sigma $ indicates the whole group of $ g _ {2} $.

\ Begin {Align *} \ Phi: \ gamma & \ To \ Sigma \ H & \ mapsto F (h) \ end {Align *}

Because $ F $ is full homomorphic, it is clear that $ \ Phi $ is indeed a ing, and $ \ Phi $ is obviously full. $ \ Phi $ ticket, otherwise $ H _ {1} exists }, H _ {2} \ In \ gamma $ and $ H _ {1} \ neq h _ {2} $ make

$ F (H _ {1}) = f (H _ {2}) $

Therefore, $ B \ In H _ {2} $ and $ B \ notin H _ {1} $ exist, and $ A \ In H _ {1} $ make

$ F (a) = F (B) $

Yi Zhi $ AB ^ {-1} \ In N \ subset H _ {1} $, and thus $ B \ In H _ {1} $. Conflict! This indicates that $ \ Phi $ is indeed single.

$ \ Phi $ indicates dual-shot.

(2) It is obvious that it is full of homomorphic by $ F $.

(3) ing between (2) $ F (h) \ LHD g _ {2} $

\ Begin {Align *} \ varphi: G _ {1}/H & \ To g _ {2}/F (h) \ Ah & \ mapsto F () F (h) \ end {Align *}

First, we will explain that $ \ varphi $ is indeed a ing, set $ Ah = BH $, that is, $ B ^ {-1} A \ in h $, then $ \ left (F (B) \ right) ^ {-1} f (a) \ In F (h) $

\ Begin {Align *} f (a) f (h) & = F (B) f (h) \ rightarrow \ varphi (AH) & = \ varphi (BH) \ end {Align *}

Therefore, $ \ varphi $ is indeed a ing.

Then it is not difficult to verify that $ \ varphi $ is homogeneous, so

$ G _ {1}/h \ simeq g _ {2}/F (h). $

 

2. Set $ \ Sigma $ to the self-homogeneous structure of the group $ G $.

$ \ Sigma (G) = g \ rightarrow G = e $

Proof:

(1) $ F: G \ To \ sigma (g) G ^ {-1} $ is a single shot;

(2) If $ G $ is a finite group, each element of $ G $ can be written as $ \ sigma (g) G ^ {-1} $;

(3) If $ G $ is a finite group and $ \ Sigma ^ 2 = {\ rm id }_{ g} $, then $ G $ is an odd-order Abel group.

Proof (1) set $ F (G) = f (h) $, then

\ Begin {Align *} \ sigma (g) G ^ {-1} & = \ sigma (h) H ^ {-1 }\\\ rightarrow \ sigma (H ^ {-1} g) & = H ^ {-1} G = e \ end {Align *}

Thus $ G = h $, that is, $ F $ single.

(2) because $ | G | <\ infty $, (1) knows

$ | F (G) | \ geq | G | $

Note $ F (g) \ subset G $ to $ F (G) = G $. in $ F (g) $, any element has the form of $ \ sigma (g) G ^ {-1} $, that is, $ G $ also has this form.

(3) By (2) $ \ forall g \ in G $, with the Form $ \ sigma (a) a ^ {-1} $, thus

\ Begin {Align *} \ sigma (g) & = \ Sigma ^ 2 (a) \ sigma (a ^ {-1}) = \ left (\ sigma () A ^ {-1} \ right) ^ {-1} = G ^ {-1} \ end {Align *}

If $ g \ NEQ e $ is available, then $ \ sigma (G) = G ^ {-1} \ neq g $ is available, therefore, the non-zero element in $ G $ must appear in pairs, so $ | G | $ is an odd number.

In addition, $ \ forall g, h \ in G $ has

$ GH = \ Sigma \ left (GH) ^ {-1} \ right) = \ sigma (H ^ {-1 }) \ sigma (G ^ {-1}) = Hg $

Therefore, $ G $ is an Abel group.

 

4. Verify that $ \ {\ mathbb Q ^ *, \ cdot \} $ and $ \ {\ mathbb Q, ++ \} $ are not homogeneous.

If the homogeneous $ F: \ mathbb Q ^ * \ To \ mathbb Q $ exists

$ F (-1) ^ 2 = F (1) = 0 $

Therefore, $ F (-1) = 0 $ is in conflict with $ F $!

 

5. Evaluate the $ \ {\ mathbb C ^ *, \ cdot \} $ sub-group $ N $

$ \ {\ Mathbb C ^ *, \ cdot \}/n \ simeq \ {\ mathbb R ^ +, \ cdot \}. $

Decoding ing \ begin {Align *} \ Phi: \ mathbb C ^ * & \ To \ mathbb R ^ + \ Z & \ mapsto | z | \ end {Align *}

It is not difficult to verify that $ \ Phi $ is a full homomorphic, and

$ {\ RM Ker} \ Phi =\{ e ^ {I \ Theta}:-\ pI <\ Theta \ Leq \ PI \}$ according to the basic theorem of homomorphic Recognition

$ \ Mathbb C ^ */{\ RM Ker} \ Phi \ simeq \ mathbb R ^ + $

Therefore, use $ n ={\ RM Ker} \ Phi $.

 

6. set $ H \ lhd g $ and $ | H | = n $, $ | g/h | = M $, and $ (m, n) = 1 $. proof: $ h $ is the only $ N $ subgroup of $ G $.

It is proved that if another $ N $ sub-group $ H _ {1} <G $ of $ G $ exists, it can be seen from theorem 1.4.10.

$ H _ {1}/(H _ {1} \ cap h) \ simeq (H _ {1} h)/h $

If $ | H _ {1} \ cap H | = L $ is set, it is clear that $ L \ mid N $. and $ (H _ {1} h)/H <g/h $, according to the Laplace Theorem

$ \ Vert H _ {1}/(H _ {1} \ cap h) \ vert \ big | \ vert g/h \ vert $

Thus \ begin {Align *} \ frac {n} {L} \ mid m \ rightarrow N & \ mid mL \ rightarrow n \ mid L \ end {Align *}

So $ n = L $, while $ H _ {1} \ cap H \ subset h $ and $ H _ {1} \ cap H \ subset H _ {1} $, yi Zhi

$ H = H _ {1} $

Therefore, $ h $ is the unique $ N $ subgroup.