My Math Forum Integers mod 4

 Abstract Algebra Abstract Algebra Math Forum

 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 non-identity of G gets mapped to the non-identity 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.

 Tags integers, mod

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post shunya Algebra 2 March 11th, 2014 03:26 AM yo79 Math Events 4 March 17th, 2013 01:50 AM sulonski Number Theory 11 August 31st, 2012 04:49 PM Julie13 Calculus 0 August 18th, 2010 11:08 PM cafegurl Elementary Math 3 January 7th, 2009 08:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top