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1. How many homomorphism are there of [tex]\mathbb{Z}[/tex] onto [tex]\mathbb{Z}[/tex]?

Answer from the book: 2

I don't know

Anything else won't work, for example: if

[tex]\phi(x)=2x[/tex]

then [tex]\phi(xy) = 2xy[/tex]

and [tex]\phi(x)\phi(y) = 2x2y[/tex]

[tex]2xy \neq 2x2y.[/tex] ?

Or should I be using additive notation because [tex]\mathbb{Z}[/tex] is the set of integers?

Okay next question:

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

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 ...

Answer from the book: 2

I don't know

*why*this is the answer! Is it because the only homomorphism possible are [tex]\phi(x) = x[/tex] and [tex]\phi(x) = e [/tex] (where e is the identity) ?Anything else won't work, for example: if

[tex]\phi(x)=2x[/tex]

then [tex]\phi(xy) = 2xy[/tex]

and [tex]\phi(x)\phi(y) = 2x2y[/tex]

[tex]2xy \neq 2x2y.[/tex] ?

Or should I be using additive notation because [tex]\mathbb{Z}[/tex] is the set of integers?

Okay next question:

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

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 ...

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