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November 6th, 2010, 08:21 PM   #1
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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?
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November 7th, 2010, 02:17 AM   #2
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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.
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