Homomorphisms/Isomorphisms

• October 22nd 2011, 05:58 PM
jzellt
Homomorphisms/Isomorphisms
a) The inclusion Z -> R is a homomorphisms of additive groups.

b) The subgroup {0} of Z is isomorphic to the subgroup {(1)} of S5.

These are both true statements, but I must know how to prove them. Can someone show me please? Thanks.
• October 22nd 2011, 06:15 PM
Drexel28
Re: Homomorphisms/Isomorphisms
Quote:

Originally Posted by jzellt
a) The inclusion Z -> R is a homomorphisms of additive groups.

b) The subgroup {0} of Z is isomorphic to the subgroup {(1)} of S5.

These are both true statements, but I must know how to prove them. Can someone show me please? Thanks.

Let me ask you a question, if $\iota:\mathbb{Z}\hookrightarrow \mathbb{R}$ is our inclusion what is $\iota(x+y)$? What about $\iota(x)+\iota(y)$?
• October 22nd 2011, 06:20 PM
jzellt
Re: Homomorphisms/Isomorphisms
I'm not sure. If f is our function f: Z -> R, doesn't it have to be defined for me to continue?
• October 22nd 2011, 06:21 PM
Drexel28
Re: Homomorphisms/Isomorphisms
Quote:

Originally Posted by jzellt
I'm not sure. If f is our function f: Z -> R, doesn't it have to be defined for me to continue?

We have defined the function, it's the inclusion map. Let's pause for a second, did you take the time to fully understand all the words in the problem statement? Do you know what an inclusion map is?
• October 22nd 2011, 06:25 PM
jzellt
Re: Homomorphisms/Isomorphisms
nevermind see next post...
• October 22nd 2011, 06:32 PM
jzellt
Re: Homomorphisms/Isomorphisms
So I googled it and it looks like the inclusion map of function f is defined by f(x) = x. So to show that f: Z->R is an homomorphism, I must show that f(x+y) = f(x) + f(y). But f(x+y) = x+y and f(x) + f(y) = x+y. So were done. COrrect?
• October 22nd 2011, 07:40 PM
Drexel28
Re: Homomorphisms/Isomorphisms
Quote:

Originally Posted by jzellt
So I googled it and it looks like the inclusion map of function f is defined by f(x) = x. So to show that f: Z->R is an homomorphism, I must show that f(x+y) = f(x) + f(y). But f(x+y) = x+y and f(x) + f(y) = x+y. So were done. COrrect?

Exactly.