# Group Theory/Normal subgroups and the Noether isomorphism theorems

**Definition (normal subgroup)**:

Let be a group. A subgroup is called a **normal subgroup** if and only if for each we have .

**Proposition (elements of disjoint normal subgroups commute)**:

Let be a group, and let so that . Then .

**Proof:** Let and . Since is a normal subgroup, , so that there exists such that . Similarly, since , there exists so that . Then , hence and since we get , so that and since and were arbitrary, .

**Proposition (characterisation of direct products within groups)**:

Let be a group, and let be subgroups of . Consider the group generated by these groups. The following are equivalent:

- The function is an isomorphism
- For we have and
- Each element in can be uniquely written as a product , where for we have

**Proof:** Certainly 1. 2., since for all , the subgroup of corresponds via to the subgroup , which is certainly normal. For 2. 3., observe that any element of *may* be written as a product , where and each is an element of some . But by 2., the are pairwise disjoint, so that any elements in distinct commute. Hence, we may sort the product so that the first few entries are in , the following entries are in and so on. The products of the entries that are contained within then form the element as required by the decomposition in 3. Finally, for 3. 1., observe that 3. implies that the given function is bijective. But it is also a homomorphism, because 3. immediately implies that the are disjoint, and we may use that elements of disjoint normal subgroups commute to obtain that indeed commutes with the respective group laws.

**Definition (subgroup product)**:

Let be a group, and let be subgroups. Then the **subgroup product** of and is defined to be

- .

In general, the subgroup product is not a subgroup. However, if one of the subgroups involved in the product is a normal subgroup, then it is:

**Proposition (subgroup product of subgroup and normal subgroup is subgroup)**:

Let be a group, and let be subgroups such that one of is normal. Then and are subgroups of .

**Proof:** Without loss of generality assume that is normal. Let , where and . Then

for some because is normal (here we applied the subgroup criterion). Similarly, is a subgroup.

**Proposition (product of normal subgroups is normal)**:

Let be a group, and let be normal subgroups of . Then , ie. is a normal subgroup of , and the same holds for .

**Proof:** By symmetry, it suffices to prove that is a normal subgroup. Indeed, let and , where and . Then

for some , , since are normal.

## Exercises

edit- Prove that the intersection of normal subgroups is again normal.
- Let be a group, and let such that are pairwise coprime. Prove that .