Normality is strongly intersection-closed

Verbal statement
An arbitrary (possibly empty) fact about::intersection of subgroups of fact about::normal subgroups 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:

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

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 $$\sigma(N)$$ &sube; $$N$$.

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.

Results following from the same generalization

 * Characteristicity is strongly intersection-closed
 * Fully characteristic satisfies strongly intersection-closed

Similar results

 * Normality is strongly join-closed
 * Characteristicity is strongly join-closed

Hands-on proof using invariance under inner automorphisms definition
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.

Textbook references

 * , Page 88, Exercises 22(a) and (b) (part (a) asks for the case where we're intersecting only two subgroups)
 * , Page 53, Problem 4 (stated only for intersection of two subgroups)
 * , Page 45, Exercise 2