Proving Isomorphism of Z4 / (2Z4) to Z2

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SUMMARY

The factor group Z4 / (2Z4) is isomorphic to Z2, as established by analyzing the elements of these groups. Z4 consists of the elements {0, 1, 2, 3}, while (2Z4) includes the subgroup {0, 2}. The resulting factor group contains two cosets: {0, 2} and {1, 3}, confirming the isomorphism to Z2, which has elements {0, 1}. Understanding finite cyclic groups is crucial for this proof.

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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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Dawson64 said:
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?

The groups you're considering are of very small order so in this case, just write out the elements and remember the number of groups of order 2 is ______ . In general, based off your definition, you should be able to identify the group pretty easily by what you know about finite cyclic groups.
 

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