# Difference between revisions of "Normal subgroup"

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'''Corollary 3.''' Let <math>\rm G</math> and <math>{\rm H}</math> be groups; let <math>{\rm G'}</math> be a subgroup of <math>{\rm G}</math> and <math>{\rm L}</math> a normal subgroup of <math>{\rm G'}</math>. Let <math>f</math> be a group homomorphism from <math>{\rm G}</math> to <math>{\rm H}</math>, with <math>{\rm N}</math> the kernel of <math>f</math>. Then <math>{\rm LN}</math> is a normal subgroup of <math>{\rm G'N}</math>, <math>\rm L \cdot (G' \cap N)</math> is a normal subgroup of <math>\rm G'</math>, and <math>f({\rm L})</math> is a normal subgroup of <math>f({\rm G'})</math>; furthermore, the quotient groups <math>\rm G'N/LN</math>, <math>\rm G'/L\cdot (G' \cap N)</math>, and <math>f({\rm G'})/f({\rm L})</math> are isomorphic. | '''Corollary 3.''' Let <math>\rm G</math> and <math>{\rm H}</math> be groups; let <math>{\rm G'}</math> be a subgroup of <math>{\rm G}</math> and <math>{\rm L}</math> a normal subgroup of <math>{\rm G'}</math>. Let <math>f</math> be a group homomorphism from <math>{\rm G}</math> to <math>{\rm H}</math>, with <math>{\rm N}</math> the kernel of <math>f</math>. Then <math>{\rm LN}</math> is a normal subgroup of <math>{\rm G'N}</math>, <math>\rm L \cdot (G' \cap N)</math> is a normal subgroup of <math>\rm G'</math>, and <math>f({\rm L})</math> is a normal subgroup of <math>f({\rm G'})</math>; furthermore, the quotient groups <math>\rm G'N/LN</math>, <math>\rm G'/L\cdot (G' \cap N)</math>, and <math>f({\rm G'})/f({\rm L})</math> are isomorphic. | ||

− | ''Proof.'' By theorem 2, <math>f({\rm L})</math> is normal in <math>f({\rm G'})</math>. Let <math>\iota</math> be the canonical homomorphism of <math>f({\rm G'})</math> into <math>f({\rm G'})/f({\rm L})</math>; let <math>f'</math> be the restriction of <math>f</math> to <math>\rm G'N</math>, and let <math>g = \iota \circ f'</math>. Then <math>g</math> is a surjective homomorphism from <math>{\rm G'N}</math> to <math>f({\rm G'})/f({\rm L})</math>, and its kernel is <math>{\rm LN}</math>. Furthermore, <math>g</math> induces a surjective homomorphism from <math>{\rm G'}</math> to <math>f({\rm G'})/f( | + | ''Proof.'' By theorem 2, <math>f({\rm L})</math> is normal in <math>f({\rm G'})</math>. Let <math>\iota</math> be the canonical homomorphism of <math>f({\rm G'})</math> into <math>f({\rm G'})/f({\rm L})</math>; let <math>f'</math> be the restriction of <math>f</math> to <math>\rm G'N</math>, and let <math>g = \iota \circ f'</math>. Then <math>g</math> is a surjective homomorphism from <math>{\rm G'N}</math> to <math>f({\rm G'})/f({\rm L})</math>, and its kernel is <math>{\rm LN}</math>. Furthermore, <math>g</math> induces a surjective homomorphism from <math>{\rm G'}</math> to <math>f({\rm G'})/f({\rm L})</math>; the kernel of this homomorphism is <math>{\rm L \cdot (G' \cap N)}</math>. The corollary then follows from theorem 2. <math>\blacksquare</math> |

For the following three corollaries, <math>{\rm G}</math> will denote a group, and <math>{\rm N}</math> a normal subgroup of <math>\rm G</math>, and <math>\lambda</math> the canonical homomorphism from <math>\rm G</math> to <math>\rm G/N</math>. | For the following three corollaries, <math>{\rm G}</math> will denote a group, and <math>{\rm N}</math> a normal subgroup of <math>\rm G</math>, and <math>\lambda</math> the canonical homomorphism from <math>\rm G</math> to <math>\rm G/N</math>. | ||

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'''Corollary 5.''' Let <math>\rm G'</math> be a subgroup of <math>\rm G</math> containing <math>\rm N</math>. Then <math>\rm G'/N</math> is normal in <math>\rm G/N</math> if and only if <math>\rm G'</math> is normal in <math>\rm G</math>; in this case, the groups <math>\rm G/G'</math> and <math>\rm (G/N)/(G'/N)</math> are isomorphic. | '''Corollary 5.''' Let <math>\rm G'</math> be a subgroup of <math>\rm G</math> containing <math>\rm N</math>. Then <math>\rm G'/N</math> is normal in <math>\rm G/N</math> if and only if <math>\rm G'</math> is normal in <math>\rm G</math>; in this case, the groups <math>\rm G/G'</math> and <math>\rm (G/N)/(G'/N)</math> are isomorphic. | ||

− | ''Proof.'' Note that <math>(\lambda^{-1} \circ \lambda)(\rm G') = G'N = G'</math>. Then by theorem 2, if <math>\lambda(\rm G')=G'/N</math> is normal in <math>\ | + | ''Proof.'' Note that <math>(\lambda^{-1} \circ \lambda)(\rm G') = G'N = G'</math>. Then by theorem 2, if <math>\lambda (\rm G')=G'/ N</math> is normal in <math>\lambda(\rm G) = G / N</math>, then <math>\rm G'</math> is normal in <math>\rm G</math>. Conversely, since <math>\lambda</math> is surjective, if <math>\rm G'</math> is normal in <math>\rm G</math>, then <math>\lambda(\rm G') = G'/N</math> is normal in <math>\rm G'/N</math>. Now, suppose that <math>\rm G'</math> is normal in <math>\rm G</math>. Let <math>f</math> be the canonical homomorphism of <math>\rm G/N</math> onto <math>\rm (G/N)/(G'/N)</math>. Evidently <math>f \circ \lambda</math> is a surjective homomorphism from <math>\rm G</math> to <math>\rm (G/N)/(G'/N)</math>, and the kernel of <math>f \circ \lambda</math> is <math>\rm G'</math>. Then by theorem 2, <math>\rm G/G</math> and <math>\rm (G/N)/(G'/N)</math> are isomorphic. <math>\blacksquare</math> |

'''Corollary 6.''' Let <math>\rm G'</math> be a subgroup of <math>\rm G</math>. Then <math>\rm G'N</math> is a subgroup of <math>\rm G</math> of which <math>\rm N</math> is a normal subgroup, and the groups <math>\rm G'/(G' \cap N)</math> and <math>\rm G'N/N</math> are isomorphic. | '''Corollary 6.''' Let <math>\rm G'</math> be a subgroup of <math>\rm G</math>. Then <math>\rm G'N</math> is a subgroup of <math>\rm G</math> of which <math>\rm N</math> is a normal subgroup, and the groups <math>\rm G'/(G' \cap N)</math> and <math>\rm G'N/N</math> are isomorphic. |

## Latest revision as of 12:22, 9 April 2019

A **normal subgroup** of a group is a subgroup of for which the relation "" of and is compatible with the law of composition on , which in this article is written multiplicatively. The quotient group of under this relation is often denoted (said, " mod "). (Hence the notation for the integers mod .)

An equivalent definition of normal subgroups is this-

is said to be a normal subgroup of a group if .Note that this means but it does not imply that for every , .

## Description

From the characterizations of relations compatible with left and right translation (see the article on cosets), a subgroup is normal if and only if is equivalent to , which is in turn true if and only if implies , which is in turn equivalent to its converse (by replacing , with , ).

Note that the relation is compatible with right multiplication for any subgroup : for any , On the other hand, if is normal, then the relation must be compatible with left multiplication by any . This is true if and only implies Since any element of can be expressed as , the statement " is normal in " is equivalent to the following statement:

- For all and , ,

which is equivalent to both of the following statements:

- For all , ;
- For all , .

By symmetry, the last condition can be rewritten thus:

- For all , .

Equivalently, one can say that a normal subgroup is one that is stable under all inner automorphisms.

The intersection of a family of normal subgroups of a group is a normal subgroup of . For (for each ) implies (for each ); hence implies .

## Examples

In an Abelian group, every subgroup is a normal subgroup. More generally, the center of every group is a normal subgroup of that group.

Every group is a normal subgroup of itself. Similarly, the trivial group is a subgroup of every group.

Consider the smallest nonabelian group, (the symmetric group on three elements); call its generators and , with , the identity. It has two nontrivial subgroups, the one generated by (isomorphic to and the one generated by (isomorphic to ). Of these, the second is normal but the first is not.

If and are groups, and is a homomorphism of groups, then the inverse image of the identity of under , called the *kernel* of and denoted , is a normal subgroup of (see the proof of theorem 1 below). In fact, this is a characterization of normal subgroups, for if is a normal subgroup of , the kernel of the canonical homomorphism is .

Note that if is a normal subgroup of and is a normal subgroup of , is not necessarily a normal subgroup of .

Every characteristic subgroup of is a normal subgroup of .

## Group homomorphism theorems

**Theorem 1.** An equivalence relation on elements of a group is compatible with the group law on if and only if it is equivalent to a relation of the form , for some normal subgroup of .

*Proof.* One direction of the theorem follows from our definition, so we prove the other, namely, that any relation compatible with the group law on is of the form , for a normal subgroup .

To this end, let be the set of elements equivalent to the identity, , under . Evidently, if , then , so ; the converse holds as well, so is equivalent to the statement "". Also, for any , so . Thus is closed under the group law on , so is a subgroup of . Then by definition, is a normal subgroup of .

**Theorem 2.** Let and be two groups; let be a group homomorphism from to , and let be the kernel of .

- If is a subgroup of , then the inverse image of under is a subgroup of ; if is normal in , then its inverse image is normal in . Consequently, is a normal subgroup of , and of this inverse image. If is surjective, then , and induces an isomorphism from to .
- If is a subgroup of , then is a subgroup of ; if is normal in , then is normal in . In particular, if is surjective, then is normal in . The inverse image of under is .

*Proof.* For the first part, suppose are elements of . Then , so is an element of . Hence is a subgroup of . If is a normal in , then for all in and all in ,
so
thus is normal in . Applying this result to the trivial subgroup of , we prove that is normal in ; since the trivial subgroup of is also a subgroup of , is also a normal subgroup of . If is surjective, then by definition . Also, if and are elements of which are congruent mod , then , so . Thus induces an isomorphism from to which is evidently a homomorphism; hence, an isomorphism. This proves the first part of the theorem.

For the second part, suppose that are elements of . Then so is a subgroup of and of . Suppose is normal in . If is any element of , then so is normal in . If is surjective, then , so is normal in .

Finally, suppose that is an element of such that is an element of . Then for some , . Hence Then , for some . Then . This finishes the proof of the second part of the theorem.

**Corollary 3.** Let and be groups; let be a subgroup of and a normal subgroup of . Let be a group homomorphism from to , with the kernel of . Then is a normal subgroup of , is a normal subgroup of , and is a normal subgroup of ; furthermore, the quotient groups , , and are isomorphic.

*Proof.* By theorem 2, is normal in . Let be the canonical homomorphism of into ; let be the restriction of to , and let . Then is a surjective homomorphism from to , and its kernel is . Furthermore, induces a surjective homomorphism from to ; the kernel of this homomorphism is . The corollary then follows from theorem 2.

For the following three corollaries, will denote a group, and a normal subgroup of , and the canonical homomorphism from to .

**Corollary 4.** The mapping is a bijection from the set of subgroups of that contain to the set of subgroups of .

*Proof.* Evidently, if is a subgroup of containing , then is a subgroup of . If is a subgroup of , then is a subgroup of containing , so is surjective. Finally, since is the kernel of , , so is injective.

**Corollary 5.** Let be a subgroup of containing . Then is normal in if and only if is normal in ; in this case, the groups and are isomorphic.

*Proof.* Note that . Then by theorem 2, if is normal in , then is normal in . Conversely, since is surjective, if is normal in , then is normal in . Now, suppose that is normal in . Let be the canonical homomorphism of onto . Evidently is a surjective homomorphism from to , and the kernel of is . Then by theorem 2, and are isomorphic.

**Corollary 6.** Let be a subgroup of . Then is a subgroup of of which is a normal subgroup, and the groups and are isomorphic.

*Proof.* By theorem 2, is the inverse image of the image of under ; hence it is a subgroup of in which is evidently normal. Let be the canonical injection of into , and let be the restriction of to . Then is a surjective homomorphism from to , and its kernel is . Hence and are isomorphic, as desired.