# Normality is strongly intersection-closed

This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., strongly intersection-closed subgroup property)
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## Statement

### Verbal statement

An arbitrary (possibly empty) Intersection of subgroups (?) of Normal subgroup (?)s of a group is normal.

Note: The use of the word strongly is to allow the empty intersection as well. We can also say that normality is intersection-closed and also identity-true.

### Symbolic statement

Let $I$ be an indexing set and $H_i$ be a family of normal subgroups of $G$ indexed by $I$. Then, the intersection, over all $i$ in $I$, of the normal subgroups $H_i$, is also a normal subgroup of $G$. In symbols:

$\bigcap_{i \in I} H_i \triangleleft G$

## Definitions used

### Normal subgroup

A subgroup $N$ of a group $G$ is said to be normal, if given any inner automorphism $\sigma$ of $G$ (viz a map sending $x$ to $gxg^{-1}$), we have $\sigma(N)$$N$.

### Strongly intersection-closed

A subgroup property is termed strongly intersection-closed if given any family of subgroups having the property, their intersection also has the property. Note that just saying that a subgroup property is intersection-closed simply means that given any nonempty family of subgroups with the property, the intersection also has the property.

Thus, the property of being strongly intersection-closed is the conjunction of the properties of being intersection-closed and identity-true, viz satisfied by the whole group as a subgroup of itself.

## Generalizations

The general result (of which this can be viewed as a special case) is that any invariance property is strongly intersection-closed.

Here, an invariance property is the property of being invariant with respect to a certain collection of functions on the whole group. For normal subgroups, the collection of functions is the inner automorphisms.

## Proof

### Hands-on proof using invariance under inner automorphisms definition

This proof method generalizes to the following results: invariance implies strongly intersection-closed Other particular cases of the generalization include characteristicity is strongly intersection-closed, full invariance is strongly intersection-closed]], and strict characteristicity is strongly intersection-closed.

Given: $H_i$ is a family of normal subgroups of $G$ indexed by $i \in I$. $H = \bigcap_{i \in I} H_i$ is the intersection.

To prove: $H \triangleleft G$. In other words, for any $x$ in $H$ and any inner automorphism $\sigma$ of $G$, we need to show that $\sigma(x) \in H$.

Proof: Since $x$ is in $H$, $x \in H_i \forall i \in I$. By the normality of $H_i$, $\sigma(x) \in H_i \forall i \in I$. Hence, $\sigma(x) \in \bigcap_{i \in I} H_i = H$. This completes the proof.

### Proof using kernel of homomorphism definition

Given: $H_i$ is a family of normal subgroups of $G$ indexed by $i \in I$. $H = \bigcap_{i \in I} H_i$ is the intersection.

To prove: $H$ is normal in $G$, i.e., $H$ occurs as the kernel of some homomorphism originating from $G$.

Proof: For each $i \in I$, $H_i$ is the kernel of some homomorphism $f:G \to L_i$, where $L_i$ ca nbe taken as the quotient group $G/H_i$.

Let $L$ be the external direct product (unrestricted) of the $L_i$s and define $f:G \to L$ as the unique map such that for any $g \in G$, the $L_i$-coordinate of $f(g)$ equals $f_i(g)$. Since each $f_i$ is a homomorphism, $f$ is also a homomorphism. Further, the kernel of $f$ is the set of those $g \in G$ for which each $f_i(g)$ is the identity element, which is the intersection of the $H_i$s, which is $H$. Thus, $H$ is the kernel of a homomorphism originating from $G$.

### Proof using commutator definition

Other results proved in a very similar way include commutator-in-center is intersection-closed.

Given: $H_i$ is a family of normal subgroups of $G$ indexed by $i \in I$. $H = \bigcap_{i \in I} H_i$ is the intersection.

To prove: $H$ is normal in $G$, i.e., $[G,H]$ is contained in $H$.

Proof: Since $H \le H_i$ for each $i \in I$, $[G,H] \le [G,H_i]$ for each $i \in I$. Since each $H_i$ is normal in $G$, $[G,H_i] \le H_i$. Thus, $[G,H] \le H_i$ for each $i$, forcing $[G,H] \le \bigcap_{i \in I} H_i = H$. This completes the proof.

## Consequences

A consequence of normality being strongly intersection-closed is the fact that given any subgroup we can talk of the smallest normal subgroup containing that subgroup. This smallest normal subgroup is termed the normal closure.

## References

### Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 88, Exercises 22(a) and (b) (part (a) asks for the case where we're intersecting only two subgroups)
• Topics in Algebra by I. N. Herstein, More info, Page 53, Problem 4 (stated only for intersection of two subgroups)
• An Introduction to Abstract Algebra by Derek J. S. Robinson, ISBN 3110175444, More info, Page 45, Exercise 2