View Full Version : Properties of Homomorphism
sleventh
Dec14-11, 11:43 PM
I am wondering what are the possible homomorphisms
\tau : Z\overline{+} -> Z\overline{+}
From this it should be possible to determine which is injective, surjective, and which are isomorphic.
Homomorphisms between Z plus to Z plus will all be of the form \tau(x) = nx
since \tau(x) = \tau(1)\underline{1} + \tau(1)\underline{2} + ... + \tau(1)\underline{x}
since we have a homomorphism and x is one summed x times.
all are injective
now im not sure how to tell which are surjective
DonAntonio
Dec28-11, 09:19 AM
I am wondering what are the possible homomorphisms
\tau : Z\overline{+} -> Z\overline{+}
From this it should be possible to determine which is injective, surjective, and which are isomorphic.
Homomorphisms between Z plus to Z plus will all be of the form \tau(x) = nx
since \tau(x) = \tau(1)\underline{1} + \tau(1)\underline{2} + ... + \tau(1)\underline{x}
since we have a homomorphism and x is one summed x times.
all are injective
now im not sure how to tell which are surjective
What kind of algebraic structure and under what operation(s) you think "Z plus" (the natural numbers, I presume?) is for you to talk about "homomorphisms"? Perhaps a monoid?
Tonio
spamiam
Dec28-11, 12:21 PM
What kind of algebraic structure and under what operation(s) you think "Z plus" (the natural numbers, I presume?) is for you to talk about "homomorphisms"? Perhaps a monoid?
Tonio
I'm guessing he means the group ##(\mathbb{Z},+)##, the group of integers under addition.
sleventh, you're right that each homomorphism can be written ##\tau_n(x) = nx## for an integer n. (Since ##\mathbb{Z}## is generated by 1, everything is determined by ##\tau_n(1)##.) Can you write out the range of the homomorphism for each n? There shouldn't be very many that are surjective.
Also, not all are injective. I can think of one homomorphism (a boring one, admittedly) that isn't.
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