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

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

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

**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,

We have, therefore established the following result.

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

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.

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.

(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

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.

David Joyner 2007-08-06