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August 13th, 2007, 09:08 PM  #1 
Member Joined: Mar 2007 Posts: 57 Thanks: 0  groups and homo/iso/automorphisms
How many automorphisms are there of Z, and Z_n? (i.e. How many isomorphisms are there from Z to Z, and how many are there from Z_n to Z_n?) How many homomorphisms are there from Z to G, where G is any group of order n? What's the easiest way to describe/list/write out all of them? How many homomorphisms are there from ZxZ to S_n? (Perhaps not in general but juse show a few cases.) Let V = Z/2 x Z/2 be Klein four group. How many homomorphisms are there from V to S_2, S_3, and S_4, in each case? (Z is the set of integers, Z_n = Z mod n, S = "permutations") 
August 14th, 2007, 05:47 AM  #2 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
I'm assuming that an automorphism creates a bijection of a set to itself. If I'm interpreting this correctly, the number of automorphisms of Z should be Z!, which is equivalent to 2^Z=2^aleph_0=c (n! rises faster than 2^n but slower than 2^(2^n)). The number of automorphisms of Z_n should be n!. I'm not going to try to guess at the remaining questions . 
August 14th, 2007, 06:49 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2207 
I suspect aptx4869 meant the group (Z, +) rather than the set Z.

August 14th, 2007, 05:37 PM  #4 
Member Joined: Mar 2007 Posts: 57 Thanks: 0 
Oops. Yes, sorry, all of them are meant to be groups so Z was meant to be (Z, +), Z_n was meant to be (Z_n, +_n), G was meant to be (G, *), S_n was meant to be (S_n, composition), etc. Automorphisms are bijective and needs to satisfy the homomorphism property, so we need the identity element to be mapped to itself in all cases. So far I have found for: Z: Due to the homomorphism property, we need f(1)=1 or f(1)=1 as it is necessary for the map to be surjective since if we have x>y we must also have f(x) > f(y). Therefore there are only two automorphisms: f(x) = x which is the identity mapping, and f(x) = x Z/2: only one automorphism which is the identity mapping Z/3: there are two automorphisms Z/4: there are two automorphisms Z/5: there are four automorphisms ... Z/n for n>2 I think there should be 2*int{ (n1)/2 } automorphisms, where int{.} is the integer function / floor function / removing the noninteger part...... though I haven't proved it yet, and what's more  I could be wrong about that. Correct me if I'm wrong. 
September 19th, 2007, 03:41 AM  #5 
Newbie Joined: Sep 2007 Posts: 3 Thanks: 0 
You are correct about automorphisms of Z, but incorrect about Z/nZ. Since 1 generates Z/nZ a homomorphism is completely determined by the image of 1. Thus there are Z/nZ = n homomorphisms (by the way this answers your question about the number of homomorphisms from Z to G, there is exactly one for each element a in G, given by k \mapsto a^k). Since Z/nZ is finite, a homomorphism is injective iff it is surjective, so it is enough to show that it is surjective. If f(1) = a, the map is surjective iff a is relatively prime to n. Hence the number of automorphisms is phi(n) (where phi is the Euler phifunction). In case you didn't know: phi(n) = \prod_i (p_i  1)p^(e_i  1), where n = \prod_i p_i^(e_i) is the prime number decomposition of n. As you see this is consistent with your claims about Z/nZ, n = 2...5. Hope it helps /d 

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