Generated subgroups
Definition: Let G be a group, X is a subset of G, set \ (\ {H_i \} _ {i \ in I} \) of all subgroups containing X is G, the \ (\ bigcap_ {i \ in I} H_i \) a subgroup of the G, called G generated by X subgroup, referred to as
Proof.
\ (\ FORALL A, B \ in \ bigcap_ {I \ in the I} H_i, A, B \ in H_i, for \ Space All \ Space I \ in the I \)
Since \ (H_i \) are each a group, the \ (ab & ^ {-. 1} \ in H_i, for \ Space All \ Space I \ in the I \)
\ (\ THEREFORE ab & ^ {-. 1} \ in \ bigcap_ {I \ in the I} H_i \)
\ (\ therefore \ bigcap_ {i \ in I} H_i \) is a subgroup of G
Notation: the element X is called a subgroup
Generating a given subgroup elements shown below represented:
Let G be a group, X-= < \ (A_1, \ cdots, a_t \) > is a subset of G then,
Order elements: Let G be a group, a \ (\ in \) G, the subgroup < \ (a \) > is called the order of element a step, referred to as the ord < \ (a \) >.
Indeed, the order of element a, the ord < \ (a \) >, is to satisfy \ (a ^ n = e \ ) is the smallest positive integer n.
Proof: meet ord (a)> a number of elements 2 must be an even number.
When \ (a = a ^ {- 1} \) when, \ (A ^ 2 = E \) , then ord (a) = 2
is satisfied ord (a)> 2 of the elements a, satisfies \ (A \ ne a ^ {- 1} \
) provided the ord (A) = n-
\ ((A ^ {-. 1}) ^ n-= (A ^ n-) ^ {-. 1} = E ^ {-. 1} = E \)
Suppose \ (\ exists n ', 1 \ le n' <\ n) such that, \ ((A ^ {-. 1}) ^ {n-'} = E \)
then: \ (A ^ {n-'} = ( (a ^ {- 1}) ^ {- 1}) ^ {n '} = ((a ^ {- 1}) ^ {n'}) ^ {- 1} = e ^ {- 1} = e \ )
which the ord (provided a) = n contrary
it \ (ord (a ^ {-
1}) = ord (a) = n \) order summary, a group of elements if more than 2 then this element and its inverse is not necessarily the same, and the inverse order are equal.
Is satisfied ord (a)> 2 elements are present in pairs.
Cyclic group
In discussing the structure of the group, the cyclic group is the most simple structure of a population.
The description of the cyclic group, the cyclic group may be represented as a set of the form: < \ (A \) > = { \ (n-A ^ | n-\ in the Z \) }
for each of a cyclic group that is elements can be written in the form of a n-th power, wherein a \ (\ in \) G, n is an integer.
Theorem: a group formed of the adder Z integer that each sub-group H are the cyclic group, and there H = <0> or = H < \ (m \) > = mZ of {km = | K \ ( \ in \) the Z}, where m is the smallest positive integer in H. And if H \ (\ NE \) <0>, then H is unlimited.
Proof.
When H = <0> = 0 while {}, is an integer of 0 because the identity element of the additive group, so H is a subgroup of Z.
When H \ (\ NE \) when <0>, provided \ (A \ in H, A ^ {-. 1} = - A \ in H \) , so that H is a positive integer, the set H of the smallest positive integer m, may assume a> 0, the \ (\ EXISTS \) R & lt \ (\ in \) the Z, such that a = qm + r, where Q \ (\ in \) the Z, 0 \ (\ Le \ ) R & lt <m.
$ If R & lt \ NE \ (0, then A = R & lt - QM = Q + A (-m) \) \ in \ (H (group closure operation), which is a prerequisite for a positive integer m smallest H contrary, the r = 0. and \) \ FORALL A \ in H, A = km, K \ in the Z $, so H is a cyclic group.
Further properties of the group: Let G be a group, A \ (\ in \) G, then
1. When < \ (A \) > group is infinite when there are:
I) \ (A ^ K = E \) , k = 0 iff
ii) an element \ (a ^ k (k \ in Z) \) twenty-two unequal
Proof.
Consider the mapping additive group Z group G \ (f: n-\ mapsto ^ n-A \) , and f is not difficult to prove the same state.
By decomposition theorem obtained with: \ (Z / Ker (F) \ Cong F (Z) = <\) A \ (> \)
be apparent from the foregoing theorem, ker (f) either as a subgroup of Z is <0> either= mZ.
\ (\ Because \) < \ (A \) > is infinite
\ (\ THEREFORE \) Ker (F) = <0>, and < \ (A \) > and Z / ker (f) is one Relationship
2. When < \ (A \) > is a finite group, the set ord (a) = m, this case has:
I) is such that m \ (a ^ m = e \ ) is the smallest positive integer
II) \ (A ^ K = E \) , if and only if m | K
III) \ (A ^ R & lt = A ^ K \) , if and only if R & lt \ (\ equiv \) K (MOD m)
IV) element \ (A ^ k (k \ in Z / mZ) \) twenty-two ranging
V) < \ (A \) > = { m}. 1-, A ^ m \ (A, A ^ 2, \ cdots, A ^ = { E \) }
VI) for any integer. 1 \ (\ Le \) D \ (\ Le \) m, there are \ (ord (a ^ d) = \ frac {m} {(d, m)} \)
Proof.
Similarly, the configuration of the mapping \ (F: n-\ mapsto A ^ n-\) , then \ (the Z / Ker (F) \ Cong F (the Z) = <\) A \ (> \)
and where < \ ( a \) > is finite, and ord (a) = m
so that m is \ (a ^ m = e \ ) is the smallest positive integer
and \ (a ^ k = e \ ) is equivalent to \ (k \ in Ker (F) \) , equivalent to the k | m, similar, \ (a ^ a ^ K = R & lt \) equivalent to> \ (RK \ in Ker (F) \) , equivalent to R & lt \ ( \ equiv \) K (MOD m)
as Z / ker (f) and < \ (a \) > is one to one, so \ (a ^ k (k \ in Z / mZ) \) twenty-two unequal
final properties:
for group \ (<D a ^> \) , \ ((a ^ D) = E ^ K \) is equivalent to \ (DK \ in Ker (F) \) , equivalent to m | dk equivalent to \ (\ frac {m} { (d, m)} | \ frac {d} {(d, m)} k \)Obviously \ (\ frac {m} { (d, m)} \) and \ (\ frac {d} { (d, m)} coprime \) , whereby it is possible to obtain \ (\ frac {m} {(d, m)} |
k \) thus \ (ord (a ^ d) = \ frac {m} {(d, m)} \)
Properties of the cyclic group: Let G be a cyclic group.
I) If G is infinite, the generator of G is \ (A \) or \ (A ^ {-. 1} \) .
II) If G is a finite order m, the \ (a ^ k \) is a generator of G if and only if (k, m) = 1.
Theorem: Each group is isomorphic to the infinite loop additive group Z, each of the finite group of order m is isomorphic to the additive group Z / mZ.
This theorem can also be added from the group to an integer isomorphism cyclic group G by the previous structure The proven.
Permutation group
Definition: set S = {1,2, ..., n }, said S to its own mapping σ is a substitution, if σ is bijective, i.e.
\ (\ Sigma: S \ rightarrow S \) ( \ ( K \ mapsto i_k \) )
typically n permutations σ written \ ((\ begin {matrix} 1 & 2 & \ cdots & n \\\ sigma (1) & \ sigma (2) & \ cdots & \ sigma (n) \ end {matrix }) \)
Multiplication of permutations: set \ (\ Sigma \) and \ (\ sigma '\) is replaced with the two S, then their product \ (\ sigma \ sigma' \ ) a substitution is in the S, and \ ( (\ sigma \ sigma ') (
i) = \ sigma (\ sigma' (i)) \) If S substituted considered to function of itself, the permutation function is the complex multiplication operation.
Example: Let \ (\ sigma = (\ begin {matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 4 & 2 & 1 & 3 \ end {matrix}) \) and \ (\ sigma '= (\ begin {matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 6 & 4 & 2 & 3 & 1 \ end {matrix}) \ ) is S = {1, 2, 3 , 4, 5, 6} substitution on, the \ (\ sigma \ sigma '=
(\ begin {matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 3 & 2 & 5 & 4 & 6 \ end {matrix}) \) i.e. first do \ (\ sigma '\) replacement, do \ (\ sigma \) replacement
Substituted inverse transform: set \ (\ Sigma = (\ the begin {Matrix}. 1 & 2 & \ cdots & n-\\ \ Sigma (. 1) & \ Sigma (2) & \ cdots & \ Sigma (n-) \ End {Matrix}) \) , the inverse conversion is \ (\ sigma ^ {- 1 } = (\ begin {matrix} \ sigma (1) & \ sigma (2) & \ cdots & \ sigma (n) \\ 1 & 2 & \ cdots & n \ end {matrix} ) \)
Permutation group: the n-permutations collection was composed of all \ (S_n \) on the replacement multiplication constitute a group, whose order is \ (the n-\!) .
Rotation: set permutations σ n is a, if present = {the I \ (i_1, i_2 are used, \ ldots, I_n \) } \ (\ Subset \) {1,2, ..., n}, such that \ (\ Sigma (i_j) =. 1 + I_ {J}, \ Sigma (i_k) i_1 = \) , where j = 1,2, .., k- 1, and for any \ (J \ in \) {1,2 , ..., n} \ I, have \ (\ Sigma (J) J = \) , then σ is called a k- rotation, referred to as ( \ (i_1, i_2 are used, ..., i_k \) ) .
Theorem: arbitrary replacement product can be expressed rotation do not intersect, and without considering the order of multiplication, this representation is unique.
Example: Let \ (\ sigma = (\ begin {matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 1 & 2 & 4 & 3 \ end {matrix}) \) is S = {1, 2, 3 , 4, 5, 6} a permutation on, the σ It can be expressed as two rotation product, i.e.
\ ((\ begin {matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 1 & 2 & 4 & 3 \ end {matrix}) = (1,6,3) (2,5,4) \)
Rotation of the product of example
seeking: (1, 3) (1, 2)
Solution: Let \ (\ sigma_1 = (1, 3), \ sigma_2 = (1,2) \)
(1, 3) (1, 2) = \ (\ sigma_1 \ sigma_2 \)
\ (\ sigma_1 \ sigma_2 (. 1) = \ sigma_1 (\ sigma_2 (. 1)) = \ sigma_1 (2) = 2 \)
\ (\ sigma_1 \ sigma_2 (2) = \ sigma_1 (\ sigma_2 (2)) = \ sigma_1 (1) =. 3 \)
\ (\ sigma_1 \ sigma_2 (. 3) = \ sigma_1 (\ sigma_2 (. 3)) = \ sigma_1 (. 3) = 1 \)
so that (1, . 3) (. 1, 2) = \ ((\ the begin {Matrix}. 1 &. 3 \\. 3 &. 1 \ End {Matrix}) (\ the begin {Matrix}. 1 & 2 \\ 2 &. 1 \ End {Matrix}) = (\ the begin {Matrix}. 1 & 2 &. 3 . 3. 1 & 2 & \\ \ Matrix End {}) \) = (. 1, 2,. 3)