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define $\mathbb{Z_p^\star}$ as the set of all the non-zero integers $\mod p$ under multiplication, which forms an Abelian group.

My question is this:

if we take the set $\mathbb{Z_7^\star}$, coudl someone explain why for instance, if we take $7\mod 7$ why that'd be $1$ instead of $0$? For instance, $\mathbb{Z_7^\star} = \{1,2,3,4,5,6 \}$ so if we take $2\times 4 = 8$ would $8 = 2$ or $8 = 1$? if we were dealing with the set $\mathbb{Z_7} = \{0,1,2,3,4,5,6\}$ under addition then $4+4 = 8 = 1$, but this is different somehow

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Well since $ 7 \equiv 0$ mod 7, it follows that 7 is not a valid element of your group. It is not 1, it is just 0 in the additive group, and hence not an element of $\mathbb Z_7^*$.

Similarly, $2 \times 4 = 8 \equiv 1$ mod 7.

Does that make sense?

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  • $\begingroup$ yup it does, thank you $\endgroup$
    – Warz
    Feb 8 '14 at 15:59

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