# Subgroup

In group theory of mathematics , a subgroup of a group is a subset of , which is itself a group in terms of the linkage . There is the shorthand way of reading as " is a subgroup of ". ${\ displaystyle (U, \ circ)}$ ${\ displaystyle (G, \ circ)}$ ${\ displaystyle U}$${\ displaystyle G}$${\ displaystyle \ circ}$${\ displaystyle U \ leq G}$${\ displaystyle U}$${\ displaystyle G}$

The group is called the upper group of the subgroup , in characters . ${\ displaystyle (G, \ circ)}$${\ displaystyle (U, \ circ)}$${\ displaystyle G \ geq U}$

Subgroups are the substructures in group theory.

## Equivalent Definitions

A non-empty subset of if and is a subset of , ${\ displaystyle U}$${\ displaystyle G}$${\ displaystyle (U, \ circ)}$${\ displaystyle (G, \ circ)}$

1. if to any two elements in their connection is in, and with each element in also its inverse.${\ displaystyle U}$${\ displaystyle U}$${\ displaystyle U}$
2. if applies to all .${\ displaystyle a, b \ in U \ Rightarrow a \ circ b ^ {- 1} \ in U}$
3. when an equivalence relation is on .${\ displaystyle a \ sim b: \ Longleftrightarrow a \ circ b ^ {- 1} \ in U}$${\ displaystyle G}$
4. if applies to all .${\ displaystyle a \ in U, b \ in G \! \ setminus \! U \ Rightarrow a \ circ b \ notin U}$
 proofs If there is a subgroup, then all 4 criteria apply. ${\ displaystyle U}$ Let criterion 1 apply. Then contains the neutral element of , which also turns out to be a neutral element in . ${\ displaystyle U}$${\ displaystyle e = a \ circ a ^ {- 1}}$${\ displaystyle G}$${\ displaystyle U}$ Let criterion 2. Be . Then with me too . Because is also . Finally , it is due also . ■ ${\ displaystyle b \ in U}$${\ displaystyle a: = b}$${\ displaystyle b \ circ b ^ {- 1} = e \ in U}$${\ displaystyle e \ in U}$${\ displaystyle e \ circ b ^ {- 1} = b ^ {- 1} \ in U}$${\ displaystyle a, b \ in U}$${\ displaystyle b ^ {- 1} \ in U}$${\ displaystyle a \ circ (b ^ {- 1}) ^ {- 1} = a \ circ b \ in U}$ Let criterion 3 apply. The reflexivity means for according to criterion . Substituting , then follows because of the symmetry also . The transitivity means that out and at the end follows. ■ ${\ displaystyle a \ sim a}$${\ displaystyle a \ in G}$${\ displaystyle a \ circ a ^ {- 1} = e \ in U}$${\ displaystyle b: = e}$${\ displaystyle a = a \ circ e ^ {- 1} \ in U}$ ${\ displaystyle a \ sim e \ implies e \ sim a}$${\ displaystyle e \ circ a ^ {- 1} = a ^ {- 1} \ in U}$ ${\ displaystyle (a \ sim c) \ land (c \ sim b) \ implies (a \ sim b)}$${\ displaystyle (a \ in U \ Longleftrightarrow) \; a \ sim e}$${\ displaystyle (b \ in U \ Longleftrightarrow) \; e \ sim b}$${\ displaystyle a \ sim b \; (\ Longleftrightarrow a \ circ b ^ {- 1} \ in U)}$ Criterion 4 applies . There is a path . Be . Because of can not be in . Because of can not be in . Be too . Because of and can not be in . ■ ${\ displaystyle U \ neq \ emptyset}$${\ displaystyle a \ in U}$${\ displaystyle V: = G \! \ setminus \! U}$${\ displaystyle a = a \ circ e \ in U}$${\ displaystyle e}$${\ displaystyle V}$${\ displaystyle e = a \ circ a ^ {- 1} \ in U}$${\ displaystyle a ^ {- 1}}$${\ displaystyle V}$${\ displaystyle b \ in U}$${\ displaystyle a ^ {- 1} \ in U}$${\ displaystyle b = a ^ {- 1} \ circ (a \ circ b) \ in U}$${\ displaystyle a \ circ b}$${\ displaystyle V}$ The reference to items outside of criteria 3 and 4 is an apparent one. Criterion 3 is . And the evidence given for criterion 4 is based on the fact that the right factor cannot be if the product is. In this respect, all relevant links remain within . Criteria 3 and 4 are also completely independent of the size of . Viewed in this way, they are special formulations of the transitivity of the subgroup relation (see § #Properties ). ${\ displaystyle U}$${\ displaystyle U = \ {a \ in G \ mid a \ sim e \}}$${\ displaystyle \ notin U}$${\ displaystyle \ in U}$${\ displaystyle U}$${\ displaystyle G}$${\ displaystyle \ leq}$

Depending on the type of link, various criteria for verifying the subgroup property can be advantageous. The fourth criterion is formulated without the formation of an inverse and can therefore be used in cases in which the formation of an inverse causes difficulties.

## Examples

• The integers are a subgroup of the rational numbers in terms of addition .${\ displaystyle \ mathbb {Z}}$ ${\ displaystyle \ mathbb {Q}}$
• Each subgroup of has the form .${\ displaystyle (\ mathbb {Z}, +)}$${\ displaystyle n \ mathbb {Z}: = \ {nm \ mid m \ in \ mathbb {Z} \}}$
• The set of even permutations (cycle notation) is a subgroup of the symmetric group .${\ displaystyle \ {{\ mbox {id}}, (1 \ 3 \ 2), (1 \ 2 \ 3) \}}$ ${\ displaystyle S_ {3}}$
• The group of matrices with determinant 1 is a subgroup of the group of invertible matrices .${\ displaystyle n \ times n}$

## Special subgroups

• A group itself and the one-element group are always subgroups. These are called the trivial subgroups of . In the case , these two subgroups are the same and represent the only subgroup. All other groups have at least two subgroups, namely the two mutually different trivial ones.${\ displaystyle G}$${\ displaystyle G}$${\ displaystyle \ {e \}}$${\ displaystyle G}$${\ displaystyle G = \ {e \}}$${\ displaystyle G \ neq \ {e \}}$
• A different subgroup is called a real subgroup, in short .${\ displaystyle G}$${\ displaystyle U}$${\ displaystyle U
• A sub-group, the core of a group homomorphism the group is called normal subgroup of the group . With it one can factor group of are formed.${\ displaystyle G}$${\ displaystyle G}$${\ displaystyle G}$
• A subgroup that is mapped into itself among all automorphisms of the group is called a characteristic subgroup of the group. Apparently characteristic subgroups are normal subgroups.

## properties

The neutral element of a group is the neutral element of every subgroup and therefore it is particularly included in every subgroup.

The average of a family of subgroups of a group is a subgroup of . ${\ displaystyle G}$${\ displaystyle G}$

The subgroup relation is transitive . That is, if is a subgroup of a group which is itself a subgroup of , then is also a subgroup of . So briefly applies ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle C}$${\ displaystyle A}$${\ displaystyle C}$

${\ displaystyle A \ leq B, B \ leq C \ Rightarrow A \ leq C}$

It should be noted that the corresponding statement does not apply to normal parts.

The set of Lagrange provides a necessary criterion for the existence of a subgroup with a certain finite groups order . It follows from it that the order of a subgroup of a finite group shares the order of the group . For example, if a prime number, the order of a subgroup can only be 1 or . So in this case the trivial subgroups are the only subgroups of . Further statements about the existence of certain subgroups with a certain order can be obtained from the Sylow theorems . If the group order is a prime number and a divisor, there are subgroups of the order . The 12-element alternating group A 4 has no subgroup of order. 6 ${\ displaystyle U}$ ${\ displaystyle G}$${\ displaystyle G}$${\ displaystyle | G |}$${\ displaystyle U}$${\ displaystyle | G |}$${\ displaystyle G}$${\ displaystyle p}$${\ displaystyle p ^ {n}}$${\ displaystyle p ^ {k}, 0 \ leq k \ leq n}$

## Generated subgroups

Since the intersection of subgroups is a subgroup again, there are to each subset of a group a with respect to the inclusion minimum subset of which contains. This subgroup is denoted by and called the subgroup of generated by . So you define abstractly ${\ displaystyle E \ subseteq G}$${\ displaystyle (G, \ circ)}$${\ displaystyle G}$${\ displaystyle E}$${\ displaystyle \ langle E \ rangle}$${\ displaystyle E}$ ${\ displaystyle \ langle E \ rangle}$${\ displaystyle G}$

${\ displaystyle \ langle E \ rangle: = \ bigcap _ {E \ subseteq U \ leq G} U}$

One can show that the elements of are exactly the elements of , which are obtained by combining finitely many . Here denotes the set of the inverses of the elements of . The following applies: ${\ displaystyle \ langle E \ rangle}$${\ displaystyle G}$${\ displaystyle a_ {i} \ in E \ cup E ^ {- 1}}$${\ displaystyle E ^ {- 1}}$${\ displaystyle E}$

${\ displaystyle \ langle E \ rangle = \ {a_ {1} \ circ a_ {2} \ circ ... \ circ a_ {n} | a_ {1}, \ dotsc, a_ {n} \ in E \ cup E ^ {- 1}, n \ in \ mathbb {N} \}}$

For a subgroup , that is called a generating system of . The generating system of a subgroup is not unique. ${\ displaystyle U}$${\ displaystyle U = \ langle E \ rangle}$${\ displaystyle E}$${\ displaystyle U}$

A subgroup that has a finite generating system is called a finitely generated group. If a generating system has one element , it is called cyclic and one writes . If you want to describe explicitly by its elements, you get: ${\ displaystyle U}$${\ displaystyle U}$${\ displaystyle g}$${\ displaystyle U}$ ${\ displaystyle U = \ langle g \ rangle: = \ langle \ {g \} \ rangle}$${\ displaystyle \ langle g \ rangle}$

${\ displaystyle \ langle g \ rangle: = \ {g ^ {z} | z \ in \ mathbb {Z} \}}$,

The group order is called the order of the generating element . ${\ displaystyle | \ langle g \ rangle |}$${\ displaystyle g}$

The set of all subgroups of a group forms a complete association , the subgroup association. The two trivial subgroups and correspond to the zero and one element of the association. Here are the association operations ${\ displaystyle G}$${\ displaystyle \ {e \}}$${\ displaystyle G}$

${\ displaystyle U \ land V = U \ cap V}$ (Average),
${\ displaystyle U \ lor V = \ langle U \ cup V \ rangle}$ (subgroup created by the association).