# Proving that a subgroup is normal

This survey article is about proof techniques for or related to satisfaction of the following property: normal subgroup
Find other survey articles about normal subgroup | Find fact articles that prove satisfaction of this property

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

This method is existential, or demonstrative -- exhibiting one homomorphism suffices.

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:

Statement Statement with symbols Type Quick explanation of proof using this approach
Every group is normal in itself (proof section) For any group $G$, $G$ is normal in $G$ Subgroup metaproperty satisfaction for identity-true subgroup property Consider the trivial homomorphism from $G$ to the trivial group, i.e., the homomorphism that sends every element of $G$ to the identity element of the trivial group. The kernel of this is $G$ itself.
Trivial subgroup is normal In any group $G$, the trivial subgroup (the subgroup comprising only the identity element) is normal Subgroup metaproperty satisfaction for trivially true subgroup property Consider the identity map from $G$ to itself. This is a homomorphism, and its kernel is the trivial subgroup.
Normality is strongly intersection-closed If $H_i, i \in I$ are all normal in $G$, so is $\bigcap_{i \in I} H_i$ Subgroup metaproperty satisfaction for strongly intersection-closed subgroup property Consider the homomorphism from $G$ to the external direct product of all the quotient groups $G/H_i$, which in each coordinate is the corresponding quotient map. The kernel of this is the intersection $\bigcap H_i$.
Normality satisfies intermediate subgroup condition $H \le K \le G$, $H$ normal in $G$, then $H$ normal in $K$ Subgroup metaproperty satisfaction for intermediate subgroup condition Compose the inclusion of $K$ in $G$ with the quotient map from $G$ to $G/H$. This is a homomorphism from $K$ with kernel $H$.
Center is normal The center $Z(G)$ is normal in $G$ Subgroup-defining function property satisfaction for center Center is the kernel of the homomorphism from the group to its automorphism group via the conjugation action.
Direct factor implies normal $G = G_1 \times G_2$, then $G_1 \times \{ e \}$ is normal in $G$. Subgroup property implication from direct factor $G_1 \times \{ e \}$ is the kernel of the projection to the second coordinate: $(g_1,g_2) \mapsto g_2$.

### Verify invariance under inner automorphisms

This method requires a universal check.

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 $H$.

The generating set approach is also useful in computational situations, both when dealing with groups described by presentations and with groups described by means of generating sets inside larger ambient groups. For more on this, see normality testing problem.

Here are some applications of the generic element approach:

Statement Statement with symbols Type Quick explanation of proof using this approach
Every group is normal in itself (proof section) For any group $G$, $G$ is normal in $G$ Subgroup metaproperty satisfaction for identity-true subgroup property Boils down to closure of the group under multiplication, inverse map
Trivial subgroup is normal In any group $G$, the trivial subgroup (the subgroup comprising only the identity element) is normal Subgroup metaproperty satisfaction for trivially true subgroup property Boils down to using the property of identity element and inverses to show $geg^{-1} = g$
Normality is strongly intersection-closed If $H_i, i \in I$ are all normal in $G$, so is $\bigcap_{i \in I} H_i$ Subgroup metaproperty satisfaction for strongly intersection-closed subgroup property For $h \in \bigcap H_i$, $ghg^{-1} \in H_i$ for each $i$, hence in intersection.
Normality satisfies intermediate subgroup condition $H \le K \le G$, $H$ normal in $G$, then $H$ normal in $K$ Subgroup metaproperty satisfaction for intermediate subgroup condition use the fact that any element of $K$ is also in $G$.
Center is normal The center $Z(G)$ is normal in $G$ Subgroup-defining function property satisfaction for center In fact, for $g \in G$, $h \in Z(G)$, $ghg^{-1} = h$.

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:

Statement Statement with symbols Type Quick explanation of proof using this approach
Normality is strongly join-closed $H_i, i \in I$ normal in $G$, then so is $\langle H_i \rangle_{i \in I}$ Subgroup metaproperty satisfaction Any element of the join is a product of elements in the $H_i$s. Use the fact that conjugation is an automorphism and invariance of each letter in the product. See also endo-invariance implies strongly join-closed
Normal subset generates normal subgroup If $A$ is a normal subset of $G$, $\langle A \rangle$ is a normal subgroup Random fact Write arbitrary element of $H$ as a product of elements of $A$ and their inverses, now use that conjugation is an automorphism.

### Determine its left and right cosets

This method requires a universal check.

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:

Statement Statement with symbols Type Quick explanation of proof using this approach
Every group is normal in itself (proof section) For any group $G$, $G$ is normal in $G$ Subgroup metaproperty satisfaction for identity-true subgroup property For any $g \in G$, $gG = G$ and $Gg = G$. So $gG = Gg$.
Trivial subgroup is normal In any group $G$, the trivial subgroup (the subgroup comprising only the identity element) is normal Subgroup metaproperty satisfaction for trivially true subgroup property For any $g \in G$, $g\{e \} = \{ g \}$ and $\{ e \} g = \{ g \}$. So $g \{ e \} = \{ e \} g$.
Subgroup of index two is normal If the index of a subgroup $H$ in $G$ is 2, then $H$ is normal in $G$ Subgroup property implication from subgroup of index two For any $g \in H$, $gH = Hg = H$. For any $g \notin H$, $gH$ is the complement of $H$ in $G$, and so is $Hg$.
Center is normal The center $Z(G)$ is normal in $G$ Subgroup-defining function property satisfaction for center In fact, for $g \in G$, $h \in Z(G)$, $gh = hg$. Thus, $gZ(G) = Z(G)g$ element-wise and hence also as whole sets.

### 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 derived 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 $H \le K \le G$ 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.

### Centralizers

Further information: Normality is centralizer-closed

The centralizer of a normal subgroup is normal. Thus, a group obtained by starting from a normal subgroup and taking the centralizer is normal. Moreover, the taking the centralizer operation can be combined with joins, intersections, and many other operations.

### Commutators

The commutator of two normal subgroups is also a normal subgroup. Also, the commutator of the whole group and any subset is normal. These facts can be useful in establishing that certain subgroups are normal.

## 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 derived subgroup, and the Frattini subgroup are normal.

## Starting from normal subgroups and using deterministic processes

An even more general idea than that of subgroup-defining functions is the following: any subgroup that is obtained by starting from a collection of normal subgroups and using deterministic processes, which may involve joins, intersections, centralizers, and commutators, or other processes, still yields a normal subgroup. Here, the term deterministic means invariant under automorphisms of the entire system, which implies, in particular, invariance under inner automorphisms.

## 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:

### 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. Further information: Abelian implies every subgroup is normal