Determine is endomorphism, kernel, epimorphism, monomorphism

[Shackleford]

I'm reasonably certain I did 1b correctly. I'm not sure about 1h. In both cases, since phi is the transformation from the multiplicative group G to the multiplicative group G, the identity is 1.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110729_200311.jpg?t=1311988303

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110729_200408.jpg?t=1311988318

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110729_200704.jpg?t=1311988336

 

[micromass]

Seems to be entirely correct!

One small detail, though. You checked that 1b was a monomorphism directly with the definition. This is unnecessary work as you also calculated the kernel. Knowing that the kernel is {1} is good enough for showing it's a mono.

 

[SammyS]

Yup, looks good.

 

[Shackleford]

Seems to be entirely correct!

One small detail, though. You checked that 1b was a monomorphism directly with the definition. This is unnecessary work as you also calculated the kernel. Knowing that the kernel is {1} is good enough for showing it's a mono.

Thanks for checking my work.

How does the kernel = {1} show one-to-one?

 

[Shackleford]

Yup, looks good.

Thanks.

According to the definition, a homomorphism is an endomorphism if the mapping is to itself, i.e. G = G'. So, in this problem, since G = G (R - {0}), you just have to show it's a homomorphism or not.

 

[stringy]

How does the kernel = {1} show one-to-one?

You could try showing this! It only requires the definition of a group homomorphism. And it's actually both a necessary and sufficient condition (it'll be an "if and only if" proof).