Gu Pei abstract algebra 1.3 "subgroups and business groups" Exercise answers

Exercise:

4. It is proved that the subgroups with an exponent of $2 $ must be regular.

It is proved that if $ G $ is set as a group and $ H is <G $ and $ [G: H] = 2 $, there is a left companion set decomposition.

$ G = H \ cup ah, A \ notin h $

There must be right companion set decomposition.

$ G = H \ cup ha $

Obviously, $ Ah = ha $. We know $ H \ lhd g $ from the arbitrary representation of the equivalence class.

 

5. Set $ G $ as a group, $ H \ LHD g, k \ lhd g $, and $ H \ cap K =\{ e \} $. Proof

$ HK = KH, \ forall H \ In H, k \ In K $

It proves that normal subgroups are known.

\ Begin {Align *} HK & = K _ {1} H = KH _ {1} \ rightarrow K ^ {-1} k _ {1} & = H _ {1} H ^ {-1} \ In H \ cap k \ end {Align *}

$ K _ {1} = K, H _ {1} = h $, that is, $ HK = KH $.

 

6. Prove that no group can represent the union of its two true subgroups.

It is proved that the group $ G = A \ cup B $, where $ A and B $ are true subgroups of $ G $. therefore, $ g \ In A, G \ notin B $ and $ H \ notin a, H \ in B $ exist. Therefore, consider $ GH $. obviously, $ GH $ is neither in $ A $ nor in $ B $. Therefore, $ GH \ notin G $ is a group conflict with $ G $!

 

8. If the group $ G $ has only one sub-group with the order of $ N $ h $, then $ H \ lhd g $.

Proof $ \ forall g \ in G $, evaluate $ G ^ {-1} Hg $, it is not difficult to verify that it is still a group, and

$ \ Left | G ^ {-1} HG \ right | = | H | $

$ G ^ {-1} Hg = h $ is known by the uniqueness of $ h $, so that $ h $ is regular.

Additional questions:

1. Set $ h $ to a subgroup of the integer plus group $ \ mathbb Z $ to prove that $ m \ In \ mathbb Z $ makes $ H = m \ mathbb Z $.

It is proved that because $ \ mathbb z = <1> $ is a cyclic group, any of its subgroups $ h $ must also be a cyclic group. Therefore, $ m \ In \ mathbb Z $ makes

$ H = <m> = m \ mathbb Z $

Since the loop group is followed by the content, another method can be used here: if $ H =\{ 0 \}$, the conclusion is obvious; if $ H \ NEQ \ {0 \} $, consider the set

$ S =\{ | T |: T \ in h, t \ neq0 \} $

The set $ S $ has a minimum value of $ M $ based on the principle of the lowest natural number. Let's say $ H = m \ mathbb Z $, otherwise, there must be a $ n \ in h $ and $ n> m> 0 $, but $ m \ nmid N $, then perform a division with remainder.

$ N = MQ + R, 0 <r <M $

Because $ h $ is a group, $ r \ in h $ is an extremely small conflict with $ M $!

 

2. Set $ H and K $ as two finite subgroups of the group $ G $.

$ | HK | =\ frac {| H | \ cdot | K |}{| H \ cap K |}. $

Proof that $ (H \ cap K)

\ Begin {Align *} H & =\ bigcup _ {I = 1} ^ {n} H _ {I} (H \ cap K) \\& =\ bigcup _ {I = 1} ^ {n} \ left (H \ cap H _ {I} k \ right) \ tag {1 }\\& = H \ bigcap \ left (\ bigcup _ {I = 1} ^ {n} H _ {I} k \ right) \ end {Align *}

The following facts are used in formula (1:

$ G (H \ cap K) = GH \ cap GK, G \ in G. $

We can see that $ H \ subset \ bigcup _ {I = 1} ^ {n} H _ {I} k $, further

$ HK \ subset \ bigcup _ {I = 1} ^ {n} H _ {I} k $

On the other hand, each $ H _ {I} k \ subset HK $

$ \ Bigcup _ {I = 1} ^ {n} H _ {I} k \ subset HK $

Therefore, for $ HK = \ bigcup _ {I = 1} ^ {n} H _ {I} k $, the Order on both sides is obtained.

$ | HK | = n | K | = \ frac {| H | \ cdot | K |}{| H \ cap K |}$

 

5. Set $ A and B $ to the two subgroups of the group $ G $. proof:

(1) $ AB <G $ is equivalent to $ AB = BA $;

(2) If one of $ A and B $ is formal, then $ AB <G $;

(3) If $ A and B $ are both formal, then $ AB \ LHD G $.

Proof (1) necessity: If $ AB <G $, then

$ AB = (AB) ^ {-1} = B ^ {-1} a ^ {-1} = BA $

Adequacy: Set $ AB = BA $. Therefore, for any $ A _ {1} B _ {1}, A _ {2} B _ {2} \ In AB $ has

\ Begin {Align *} A _ {1} B _ {1} B _ {2} ^ {-1} A _ {2} ^ {-1} & = _ {1} B _ {1} A _ {3} B _ {3} \ & = A _ {1} A _ {4} B _ {4} B _{ 3 }\\& = A _ {5} B _ {5} \ In AB \\\ rightarrow AB & <G. \ end {Align *}

(2) Not General. Set $ A \ lhd g $.

\ Begin {Align *} A _ {1} B _ {1} B _ {2} ^ {-1} A _ {2} ^ {-1} & = _ {1} B _ {1} A _ {3} B _ {2} ^ {-1} \ & = A _ {1} A _ {4} B _{ 1} B _ {2} ^ {-1} \ In AB \ rightarrow AB & <G. \ end {Align *}

(3) By (2) zhi $ AB <G $, $ \ forall g \ in G $, consider

\ Begin {Align *} gabg ^ {-1} & = gag ^ {-1} GBG ^ {-1} = AB \ end {Align *}

Thus $ AB \ LHD G. $