Proving that a subgroup is normal: Difference between revisions

Revision as of 02:55, 23 December 2009

This survey article is about proof techniques for or related to: normal subgroup
Find other survey articles about normal subgroup

This article explores the various ways in which, given a group and a subgroup (through some kind of description) we can try proving that the subgroup is normal (or that it is not normal). We first discuss the leading general ideas, and then plunge into the specific cases.

Other things instead of proving normality

In some cases, proving that a certain subgroup is normal may be hard or impossible, perhaps because the subgroup is not normal. The following alternative approaches are useful here:

Replacing a subgroup by a normal subgroup: There are many techniques to guarantee, from the existence of a subgroup satisfying certain conditions, the existence of a normal subgroup satisfying similar conditions.

Using the standard definitions

The best way to try proving that a subgroup is normal is to show that it satisfies one of the standard equivalent definitions of normality.

Construct a homomorphism having it as kernel

To prove that $H$ is a normal subgroup of $G$ , we can construct a homomorphism $\varphi :G\to K$ such that the kernel of the homomorphism, i.e., the set of elements that map to the identity, is precisely $H$ . Note that if we do this successfully, it is not even necessary to establish separately that $H$ is a subgroup.

Here are some examples:

Every group is normal in itself: This can be proved by taking the trivial homomorphism from $G$ to the trivial group. The kernel of the trivial homomorphism is precisely the whole group $G$ .
Trivial subgroup is normal: This can be proved by taking the identity homomorphism from $G$ to $G$ . The kernel of this homomorphism is the trivial subgroup.
Center is normal: Although there are other, more direct ways of seeing this, one way of seeing it is that the center is the kernel of the homomorphism from the group to its automorphism group via the conjugation action.

Other examples involve using the fact that some subgroups are normal to prove that other subgroups are normal. For instance:

Normality is strongly intersection-closed: Given a collection of normal subgroups of $G$ , we can construct a homomorphism from $G$ to the external direct product of the corresponding quotient groups. The kernel of this is the intersection of all the normal subgroups. This proves that an arbitrary intersection of normal subgroups is normal.
Normality satisfies intermediate subgroup condition: If $H\leq K\leq G$ , and $H$ is normal in $G$ , then $H$ is normal in $K$ . To prove this, note that $H$ is the kernel of the homomorphism from $K$ to $G/H$ obtained via composition of the inclusion of $K$ in $G$ and the quotient map from $G$ to $G/H$ .

Verify invariance under inner automorphisms

Another way of proving normality is using the inner automorphism, or conjugation, definition. This states that a subgroup $H$ of a group $G$ is normal if for every $h\in H,g\in G$ , we have $ghg^{-1}\in H$ .

This definition could be used in three broad ways:

Try everything: Here, we basically check every element of $G$ and every element of $H$ .
Use generic elements: Here, we don't actually try each element, but rather, argue that for an arbitrary choice of element of $G$ and element of $H$ , the result holds.
Use generating sets: This is a somewhat stronger version. It says that if $A$ is a generating set for $H$ and $B$ is a generating set for $G$ , then $H$ is normal in $G$ if for every $a\in A,b\in B$ , $bab^{-1}$ and $b^{-1}ab$ are both in $H$ .Further information: Normality testing problem

Here are some applications of the generic-element approach:

Every group is normal in itself: It is easy to see, using generic elements, that the condition is satisfied.
Trivial subgroup is normal: This is also easy to see.
Normality is strongly intersection-closed
Normality satisfies intermediate subgroup condition
Center is normal

The generating set approach, or ideas of that kind, are more useful when the subgroup is described by means of generating elements or as a join of subgroups. For instance:

Normality is strongly join-closed

Determine its left and right cosets

A subgroup $H$ is normal in a group $G$ iff, for every $g\in G$ , $gH=Hg$ . This definition is useful for proving normality is some situations:

Compute its commutator with the whole group

A subgroup $H$ of a group $G$ is normal iff the commutator $[G,H]$ is contained in $H$ . This definition is useful for proving, for instance, that the commutator subgroup is normal.

Methods involving metaproperties of normality

Joins and intersections

Further information: Normality is strongly join-closed,Normality is strongly intersection-closed

If the given subgroup can be described using joins and intersections ,starting with normal subgroups, then it is normal.

Upper joins

Further information: Normality is upper join-closed

If the given subgroup is normal in a bunch of intermediate subgroups that together generate the whole group, it is normal in the whole group.

Quotient-transitivity

Further information: Normality is quotient-transitive

If $H\leq K\leq G$ are such that $H$ is normal in $G$ and $K/H$ is normal in $G/H$ , then $K$ is normal in $G$ . This is a frequently used fact.

Images, inverse images, transfer, intermediate subgroups

Further information: Normality satisfies intermediate subgroup condition, Normality satisfies transfer condition, Normality satisfies image condition, Normality satisfies inverse image condition

Subgroup-defining function

One of the simplest ways of showing that a subgroup is normal is to show that it arises from a subgroup-defining function. A subgroup-defining function is a rule that associates a unique subgroup to the group.

Any subgroup obtained via a subgroup-defining function is invariant under any automorphism of the group, and is hence a characteristic subgroup. In particular, it is invariant under inner automorphisms of the group, and is hence normal.

With this approach, for instance, we can show that the center, the commutator subgroup, and the Frattini subgroup are normal.

The example of the center: crude versus refined argument

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

The deviation method of proving normality

In measuring deviation from normality, we see three ways of measuring the extent to which a subgroup deviates from normality: the normalizer, the normal closure and the normal core. Here, we explore each of these as a tool for trying to prove normality.

The normal core method and group actions

Further information: Group acts on left coset space of subgroup by left multiplication

The idea behind using the normal core to prove normality is to show that the given subgroup equals its normal core: the largest normal subgroup contained in it. In other words, we try to establish that the intersection of all conjugates of the subgroup equals the subgroup itself. This method is particularly useful in cases where the subgroup has small index in the whole group.

The normal core method is typically applied along with group actions, as in the setup described below.

Let $H$ be a subgroup of $G$ . Then, $G$ acts on the coset space of $H$ . This gives a homomorphism from $G$ to the symmetric group on the coset space, and the kernel of the homomorphism is the normal core $H_{G}$ . Hence, the quotient group $G/H_{G}$ sits as a subgroup of the symmetric group $\operatorname {Sym} (G/H)$ .

This approach can be used to prove results like the following:

Index two implies normal
Subgroup of least prime index is normal
A related result, that does not prove the subgroup itself is normal, but that it contains another large enough normal subgroup: Poincare's theorem, which states that any subgroup of index $n$ contains a normal subgroup of index dividing $n!$ .

Normal closure

The idea behind using the normal closure in order to prove normality is to prove that the subgroup equals its own normal closure. In other words, we show that the subgroup equals that subgroup generated by all its conjugates. This method is particularly useful when the subgroup is given in terms of a generating set.

Suppose $G$ is a group and $H$ is a subgroup with generating set $S$ . The normal closure of $H$ in $G$ can be obtained as the subgroup generated by all conjugates of elements of $S$ be elements of $G$ . Thus, to show that $H$ is normal in $G$ , it suffices to show that all such conjugates are again in $H$ .

In fact, if we are given a generating set $A$ for $G$ , it suffices to prove that conjugating any element of $S$ by any element in $A\cup A^{-1}$ gives an element of $H$ .

Normalizer

The idea behind using the normalizer to prove normality is to prove that the normalizer of the subgroup equals the whole group. In other words, we show that every element of the group commutes with the subgroup.

Methods suited for particular groups

For abelian groups

If the whole group is abelian, then every subgroup is normal, so there is nothing to prove.

For nilpotent groups

Nilpotent implies every maximal subgroup is normal

@@ Line 14: / Line 14: @@
 ===Construct a homomorphism having it as kernel===
+To prove that <math>H</math> is a normal subgroup of <math>G</math>, we can construct a homomorphism <math>\varphi:G \to K</math> such that the [[kernel]] of the homomorphism, i.e., the set of elements that map to the identity, is precisely <math>H</math>. Note that if we do this successfully, it is not even necessary to establish separately that <math>H</math> ''is'' a subgroup.
+Here are some examples:
+* [[Every group is normal in itself]]: This can be proved by taking the trivial homomorphism from <math>G</math> to the trivial group. The kernel of the trivial homomorphism is precisely the whole group <math>G</math>.
+* [[Trivial subgroup is normal]]: This can be proved by taking the identity homomorphism from <math>G</math> to <math>G</math>. The kernel of this homomorphism is the trivial subgroup.
+* [[Center is normal]]: Although there are other, more direct ways of seeing this, one way of seeing it is that the center is the kernel of the [[group acts on itself as automorphisms by conjugation|homomorphism from the group to its automorphism group via the conjugation action]].
+Other examples involve using the fact that some subgroups are normal to prove that other subgroups are normal. For instance:
+* [[Normality is strongly intersection-closed]]: Given a collection of normal subgroups of <math>G</math>, we can construct a homomorphism from <math>G</math> to the [[external direct product]] of the corresponding [[quotient group]]s. The kernel of this is the intersection of all the normal subgroups. This proves that an arbitrary intersection of normal subgroups is normal.
+* [[Normality satisfies intermediate subgroup condition]]: If <math>H \le K \le G</math>, and <math>H</math> is normal in <math>G</math>, then <math>H</math> is normal in <math>K</math>. To prove this, note that <math>H</math> is the kernel of the homomorphism from <math>K</math> to <math>G/H</math> obtained via composition of the inclusion of <math>K</math> in <math>G</math> and the quotient map from <math>G</math> to <math>G/H</math>.
 ===Verify invariance under inner automorphisms===
+Another way of proving normality is using the [[inner automorphism]], or conjugation, definition. This states that a subgroup <math>H</math> of a group <math>G</math> is normal if for every <math>h \in H, g \in G</math>, we have <math>ghg^{-1} \in H</math>.
+This definition could be used in three broad ways:
+* Try everything: Here, we basically check every element of <math>G</math> and every element of <math>H</math>.
+* Use generic elements: Here, we don't actually try each element, but rather, argue that for an arbitrary choice of element of <math>G</math> and element of <math>H</math>, the result holds.
+* Use generating sets: This is a somewhat stronger version. It says that if <math>A</math> is a generating set for <math>H</math> and <math>B</math> is a generating set for <math>G</math>, then <math>H</math> is normal in <math>G</math> if for every <math>a \in A, b \in B</math>, <math>bab^{-1}</math> and <math>b^{-1}ab</math> are both in <math>H</math>.{{further|[[Normality testing problem]]}}
+Here are some applications of the generic-element approach:
+* [[Every group is normal in itself]]: It is easy to see, using generic elements, that the condition is satisfied.
+* [[Trivial subgroup is normal]]: This is also easy to see.
+* [[Normality is strongly intersection-closed]]
+* [[Normality satisfies intermediate subgroup condition]]
+* [[Center is normal]]
+The ''generating set'' approach, or ideas of that kind, are more useful when the subgroup is described by means of generating elements or as a [[join of subgroups]]. For instance:
+* [[Normality is strongly join-closed]]
 ===Determine its left and right cosets===
+A subgroup <math>H</math> is normal in a group <math>G</math> iff, for every <math>g \in G</math>, <math>gH = Hg</math>. This definition is useful for proving normality is some situations:
+* [[Every group is normal in itself]]
+* [[Trivial subgroup is normal]]
+* [[Subgroup of index two is normal]]
+* [[Center is normal]]
 ===Compute its commutator with the whole group===
+A subgroup <math>H</math> of a group <math>G</math> is normal iff the [[commutator of two subgroups|commutator]] <math>[G,H]</math> is contained in <math>H</math>. This definition is useful for proving, for instance, that the [[commutator subgroup is normal]].
 ==Methods involving metaproperties of normality==