# Isomorphism theorems

Let and be two groups and let be a homomorphism of onto . Furthermore, let . If , then (see Theorem 7.1.4) . If then it is readily seen that : For if , then

so (see Chapter 2, exercise 2). Now since , where designates the identity of , we have and so , since is onto and also using exercise 6(b) from Section 1.1. We therefore have shown that any subgroup of is of the form
 (8.7)

where satisfies the condition that .

Finally let be any subgroup of that contains the kernel, . Then, of course,

(See exercise 6(a) of Section 1.1, where an equality is given if is 1-1, but the above inclusion holds for any . Why?) However since is a homomorphism, we can show equality. For if , then , so , where . Thus , where (note ), but ; whence , and we have
 (8.8)

With these results at our disposal, we are now in a position to prove the following result.

Theorem 8.3.1 (Correspondence Theorem)   Let be a homomorphism of the group onto the group with . Let be the class of all subgroups of which contain . The mapping (or correspondence)

is a 1-1 correspondence between the family and the class of all subgroups of . Moreover, if and only if .

Remark 8.3.2   The condition that is onto in this theorem is really no restriction because if is not onto, we just replace with .

Proof: Part of the theorem has already been established in our remarks preceding the theorem. In particular, we have noted that the mapping defined in the statement of the theorem is onto (see (8.7); i.e., is the mapping of the family to the family of all subgroups of given by

We also note that each subgroup Hà is such that , for all . It is also clear that is 1-1, for suppose , then . But from the remarks preceding the theorem in particular equation (8.8), we get

Thus is 1-1.

Finally, if then since is onto, for arbitrary there exists a such that . Thus

and so . Conversely, if , consider , where . Then

But then and since is 1-1, we have and so . We note that is actually also one of the subgroups of , which contains because is normal in ; for implies .

We apply Theorem 8.3.1 to the particular case of the canonical homomorphism (see Example 7.1.2) of a group onto a factor group , where is, of course, a normal subgroup of ; i.e., . We claim . Indeed, recall the identity in is , so that , as claimed. Thus, by Theorem 8.3.1, any subgroup of is of the form where and . However,

 (8.9)

We have, therefore established the following result.

Corollary 8.3.3   Let be a group and let . Any subgroup of is of the form where is a subgroup of containing . If and are two such subgroups of , then implies . Moreover if and only if .

We continue to assume is a homomorphism of the group onto with kernel . Let be a normal subgroup of that contains and let . Consider the mappings

where is the canonical map of onto . Note , by Theorem 8.3.1, since .

The composite map

is, of course, a homomorphism of onto . Suppose and , i.e., suppose . Then and conversely. Hence, the is (by equation (8.8)). Applying the FHT (Theorem 7.1.8), we have , where the isomorphism of onto is given by . We summarize this in the following theorem, frequently called the first isomorphism theorem.

Theorem 8.3.4 (First Isomorphism Theorem)   Let be a homomorphism of the group onto the group with . Let such that . Then , and

by the mapping .

Again, we consider the special case of a group and the canonical map onto a factor group . If and , then

by equation (8.9). Thus Theorem 8.3.4 gives the following result.

Corollary 8.3.5   Let be a group and let and be normal subgroups of such that . Then

Now assume that and are subgroups of a group , and moreover that . Then, in particular, for all , and so clearly . This implies by Theorem 4.2.1, that . As we will have occasion to use this fact in the future, we state it here as the following result.

Proposition 8.3.6   If , and , then .

(Of course and since , we have .) Next consider the mapping

given by , where . This map is a homomorphism of into (see exercise 1 for this section). We claim tht is, in addition, onto . Indeed, for any coset of in is of the form , where and . The kernel, , consists of those such that , i.e., those elements of or . Thus applying the FHT (Theorem 7.1.8) to yields the second fundamental isomorphism theorem.

Theorem 8.3.7 (Second Isomorphism Theorem)   If and are subgroups of a group and is also normal in , then and

The isomorphism is given by the mapping , where .

The second isomorphism theorem can probably best be remembered by the following mnemonic device:

Label the vertices of the figure as indicated, it being immaterial which side are writes or on. One then reads the isomorphism by reading modulo the opposite sides.'' Should also be normal in , then we obtain, symmetrically, , which may be read by reading modulo'' the other pair of opposite sides of the figure.

There is another fundamental theorem of isomorphism (the third isomorphism theorem) due to Zassenhaus, but we postpone a consideration of this theorem until we reach the section to which it is most relevant.

Subsections

David Joyner 2007-08-06