Normal implies permutable

Verbal statement
Any fact about::normal subgroup of a group is a fact about::permutable subgroup.

Symbolic statement
Let $$G$$ be a group and $$H$$ a normal subgroup of $$G$$. Then $$H$$ is a permutable, or quasinormal, subgroup of $$G$$. In other words, $$HK = KH$$ for any subgroup $$K$$ of $$G$$ (or equivalently, the fact about::product of subgroups $$HK$$ is a subgroup for any subgroup $$K \le G$$).

Property-theoretic statement
The subgroup property of being normal is stronger than the subgroup property of being permutable.

Normal subgroup
A subgroup $$H$$ of a group $$G$$ is a normal subgroup if for any $$g \in G$$, $$gH = Hg$$ (viz, the left cosets are the same as the right cosets).

Permutable subgroup
A subgroup $$H$$ of a group $$G$$ is a permutable subgroup if for any subgroup $$K \le G$$, $$HK = KH$$ (or equivalently, $$HK$$ is a group). In other words, $$H$$ and $$K$$ are permuting subgroups for every $$K$$. Here $$HK$$ denotes the product of subgroups:

$$HK = \{ hk : h \in H, k \in K \}, \qquad KH = \{ kh: k \in K, h \in H \}$$

Hands-on proof
Given: Let $$H$$ be a normal subgroup of $$G$$.

To prove: $$H$$ is permutable in $$G$$.

Proof: Let $$K$$ be any subgroup of $$G$$. For every $$g \in K$$, $$Hg = gH$$. Now we have:

$$HK = \bigcup_{g \in K} Hg$$

and

$$KH = \bigcup_{g \in K} gH$$

Since $$Hg = gH \forall g \in K$$, we conclude that $$HK = KH$$.

Notice that the above proof does not anywhere use the fact that $$K$$ is a subgroup.

Permutable subgroups need not be normal
A permutable subgroup need not be normal. There are many counterexamples for finite groups.

Textbook references

 * , Page 53, Problem 3 (the notion of permutable subgroup is not explicitly introduced, but what we're asked to show is implicitly the same)
 * , Page 75, Exercise 5 of Section 5 (Restriction of a homomorphism to a subgroup) (the notion of permutable subgroup is not explicitly introduced, but what we're asked to show is implicitly the same)