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