Let and be groups. Suppose and are subgroups and that there exists an isomorphism . Define by . Note that is normal in since , using a previous theorem. We denote by the quotient . (In particular, depends on .) Think of as the direct product “collapsed” by identifying each element with its -image in .
- Prove that the images of and in are isomorphic to and , respectively, and that these images intersect in a central subgroup isomorphic to . Find .
- Let . Let and as usual. Let be the central product of and which identifies and . (I.e. , , and .) Let be the central product of and which identifies and . (I.e. , , and .) Prove that .
- Let denote the canonical projection, and identify with and with in .
Let , and suppose . Then , so that . Thus , and hence is injective. Similarly, is injective.
Note that the restriction is also injective. Suppose ; then for some and we have . Thus ; by definition, then, and . Note that for all we have , so that is in the center of . Moreover, . Conversely, if , we have , so that . Then , and by the First Isomorphism Theorem, . In addition, is a central subgroup of .
Finally, by Lagrange write and . Note that . Now . We may also write this equation in the form .
- Define as follows: , , and . Because generates and these images satisfy the relations , , , and , extends to a homomorphism .
If we let denote the natural projection, is a mapping . We claim that is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of , namely . In fact, = . Thus . By the remarks on page 100 in the text, there exists a unique group homomorphism such that , and in particular, .
We now show that . Suppose . Then ; note that for some integers , and that . Then .
If , then mod 2 and mod 4. Now if mod 4, then mod 4 and . If mod 4, then mod 2 and .
If , then mod 2 and mod 4. If mod 4, then mod 4 and . If mod 4, then mod 4 and .
Thus , hence is injective.
Now by part (a), we see that both and have order 16. Thus is an isomorphism.