Algebra questions, (emergency)

[futurebird]

1. How many homomorphism are there of $$\mathbb{Z}$$ onto $$\mathbb{Z}$$?

Answer from the book: 2

I don't know why this is the answer! Is it because the only homomorphism possible are $$\phi(x) = x$$ and $$\phi(x) = e$$ (where e is the identity) ?

Anything else won't work, for example: if
$$\phi(x)=2x$$

then $$\phi(xy) = 2xy$$

and $$\phi(x)\phi(y) = 2x2y$$

$$2xy \neq 2x2y.$$ ?

Or should I be using additive notation because $$\mathbb{Z}$$ is the set of integers?



Okay next question:

2. Let $$\phi: G \rightarrow G'$$ be a group homomorphism. Show that if $$|G|$$ is finite, then $$|\phi[G]|$$ is finite and is a divisor of $$|G|$$.

Homomorphisms don't need to be one-to-one or onto, so I'm havinga hard time seeing why this must be true. Why can't each element in G map to an infinite number of items in G'? My book uses a diagram that suggests that this kind of mapping is like a projection to the x axis, so the set G may have two dimensions, but the set G' has only one. I don't see why this has to be the case, though ...

 

[Mystic998]

1. First, the integers aren't even a group under multiplication. So you should be using addition as your operation. Under addition the integers are an (infinite) cyclic group. What does that tell you?

2. I'm not sure what you mean by mapping to an infinite number of items. But for the first part you can ignore that it's a homomorphism, how big can the image of a finite set under some function be? For the second part, you have to assume G' is finite to make sense of the order of the image being a divisor of the order of G'. As for why it divides, what structure does the image of a homomorphism have in the group in which it's contained?