Let p be a prime number which is odd.
(a) Show that $\mathbb{F}^*_p$ has a unique subgroup of order $(p - 1)/2$, namely the subgroup consisting of squares of elements of $\mathbb{F}^*_p$.
My question is basic: what does $\mathbb{F}^*_p$ refers to ?
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6The multiplicative group of the finite field, i.e., without zero (which cannot be invertible). The group $(\mathbb{F}_p^{\times},\cdot)$ has $p-1$ elements and is cyclic, see here. – Dietrich Burde May 13 '15 at 18:58
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This is an answer to the question, give it as such. – bbnkttp May 13 '15 at 19:05
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It is the set of all non-zero elements in the field of $p$ elements. Typically $R^*$ is the set of units in $R$ but since $\mathbb F_p$ is a field, any non-zero element is a unit. The set of units is actually a group with respect to multiplication (assuming $R$ has an identity). In the case of $\mathbb F_p$ this group is just $\mathbb Z/(p-1)\mathbb Z$.
Gregory Grant
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