It is isomorphic to $\mathbb{Z}_{2^8},$ only difference is the symbols usually identifying the elements of the set are from $\{-128, \ldots, 127 \}$ and not $\{0, \ldots, 256\}.$
What is an elegant way to say this in math?
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It is isomorphic to $\mathbb{Z}_{2^8},$ only difference is the symbols usually identifying the elements of the set are from $\{-128, \ldots, 127 \}$ and not $\{0, \ldots, 256\}.$ What is an elegant way to say this in math? |
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A group needs to satisfy the group axioms, so you need to define the group operation. Presumably you are using addition modulo 256. There is a natural correspondence between the two sets that takes $n \in [-128,127]$ to $\begin {cases} n+256 & n \lt 0 \\n & n \ge 0 \end {cases}$ and you can show that the group structure is maintained. |
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modswhich represents $\mathbb{Z}_m$ using $\{-\lfloor \frac{m - 1}{2} \rfloor, \ldots, +\lfloor \frac{m}{2} \rfloor\}$ rather than $\{0, 1, \ldots, m-1 \}.$ – user2468 Mar 24 '12 at 0:48