Let G=HK, where H, K are characteristic subgroups of G with trivial intersection, i.e. . Then, .

Proof:

Now suppose , where and are characteristic subgroups of with . Define by

is a homomorphism, and bijective since . Thus and similarly, so that is well-defined.

Note that so is a homomorphism.

Suppose . Then . Then for , so that . Thus is injective.

For any , define . Then

So is a homomorphism.

If , then , so that . Then since , so , so that . Thus and is injective.

Any can be written as since is bijective. Similarly, any can be written as . Then so is surjective.

Thus . Note that since . Similarly, . So and is surjective.

Hence is an isomorphism.

### Like this:

Like Loading...

*Related*