November 6th, 2010, 08:21 PM  #1 
Newbie Joined: Nov 2010 Posts: 7 Thanks: 0  Integers mod 4
Why does it make sense (when considering Z4)to form the factor group Z4 / (2Z4) where kZn = {0, k mod n, 2k mod n, ..., nk mod n}? I believe that this above factor group is isomorphic to Z2, but how can I prove this? 
November 7th, 2010, 02:17 AM  #2 
Newbie Joined: Nov 2010 Posts: 3 Thanks: 0  Re: Integers mod 4
Why it makes sense? That question is a bit vague. But every group of order two is isomorphic to Z/2Z (which is sometimes denoted Z2). It's not hard to see why: There is always the identity, and the other element must be its own inverse. Formally: Let G=2. An isomorphism f:G>Z/2Z must always map the identity of G to the identity of Z/2Z. Hence the nonidentity of G gets mapped to the nonidentity of Z2. So we choose e>e and g>1. So f is obviously a bijection. You can easily check that f(x y) = f(x)f(y) for every x and y. 

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