The (finite abelian) multiplicative group of the integers modulo a prime $p$ can't have a subgroup isomorphic to $C_q\times C_q$, for some prime $q$, because in that case $x^q=1$ would have $q^2$ solutions, which is not possible for a well-known lemma on the field ${\bf{Z}}/(p)$. Is it correct? Does it follow from this argument that the group is cyclic? (I know that this group is cyclic, so I'm asking whether this is a valid proof of it.)
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2I believe everything you say is correct! (well, $x^q=1$ would have $q^2$ solutions) – Greg Martin Jan 14 '23 at 18:48
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Thank you @GregMartin. Indeed, I forgot the unit. Edited. – citadel Jan 14 '23 at 18:52
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Do you mean the group of units of the ring $\Bbb Z/p$? Some people write $U(p)$ or $(\Bbb Z/p)^{\times}$. See also this post for proofs, or this one. – Dietrich Burde Jan 14 '23 at 19:28
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1Yes, @DietrichBurde, I mean that group. I think this collects them (proofs) all: https://kconrad.math.uconn.edu/blurbs/grouptheory/cyclicmodp.pdf – citadel Jan 14 '23 at 19:38
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Yes, you are right. This is a nice collection. – Dietrich Burde Jan 14 '23 at 19:45
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1Infact writing the group with primitive roots helps and make out a different proof. – Jan 18 '23 at 12:10
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Since p and q are prime – Jan 18 '23 at 12:10