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I'm looking for a proof of this that does not rely on the Chinese remainder theorem, since the exercise is from a book that has not reached such a point (Aluffi, Algebra).

Show that for $p$ prime, the multiplicative group $G=(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic. You may use that for $p$ prime, the equation $x^d=1$ has at most $d$ solutions in $\mathbb{Z}/p\mathbb{Z}$. (hint: for $g\in G$ of maximal order, $\forall h\in G, |h|\;|\;|g|$, and in particular, $h^{|g|}=1$.)

My only idea for now is constructing an element $a\in G$ such that $|a|=|G|$, using the fact that a group of order $n$ is cyclic $\iff\exists$ an element of order $n$.

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If you know the structure theorem for finite Abelian groups, you will be able to prove that either a finite Abelian group $G$ is cyclic, or that its exponent is $<|G|$, that is there is $m<|G|$ with $g^m=e$ for for all $g\in G$.

An alternative attack: show that for $d\mid (p-1)$ the group $G=(\Bbb Z/\Bbb Z)^*$ has exactly $d$ solutions of $x^d\equiv1$. If you compare $G$ to $H=\Bbb Z/(p-1)\Bbb Z$, you find that the number of elements of order dividing $d$ is the same in each, therefore the number of elements of order exactly $d$ is the same in each, and now take $d=p-1$.

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    $\begingroup$ The question specifically asked for a proof that did not use the structure theorem for f.g. abelian groups (but it has been edited out after G. Sassatelli's comment). $\endgroup$ – Arnaud D. Jul 7 '17 at 12:30
  • $\begingroup$ Note that the alternative attack should be approached by induction on $d$. $\endgroup$ – Dustan Levenstein Jul 7 '17 at 12:32
  • $\begingroup$ Could you explain a bit more your alternative attack? How can I find the number of elements of order dividing $d$? $\endgroup$ – George Jul 7 '17 at 14:31
  • $\begingroup$ @George There are $\le (p-1)/d$ solutions of $x^{(p-1)/d}\equiv1$, so $\ge d$ different values of $y^{(p-1)/d}$ modulo $p$, each of which solves $y^d\equiv1$. $\endgroup$ – Lord Shark the Unknown Jul 7 '17 at 16:08
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Hint: Use the hint.

Among all elements if $(\Bbb Z/p\Bbb Z)^\times$, let $g$ have maximal order, $d$. If we assume that $d<p-1$, then there exists $h\in (\Bbb Z/p\Bbb Z)^\times$ that is not among the $d$ solutions of $x^d=1$. The order $m$ of $h$ is thus not a divisor of $d$. Let $h'=h^{m/\gcd(m,d)}$. Then $h'$ is of order $m'=\frac m{\gcd(m,d)}$, which is coprime to $d$. Conclude that $hg$ has order $m'd>d$, contradiction.

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  • $\begingroup$ Shouldn't $hg$ be $h'g$? $\endgroup$ – Pierre-Yves Gaillard Jul 7 '17 at 13:14
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    $\begingroup$ $m'$ might not be coprime with $d$, for instance $m=8$, $d=6$. $\endgroup$ – lhf Jul 7 '17 at 13:16
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    $\begingroup$ Isn't it enough that the order of $h$ does not divide $d$? ie: We know clearly that $(\mathbb{Z}/p\mathbb{Z})^*$ is a finite group, so the orders of all its elements is finite. If $g$ has maximal order $d$, and $d<p-1$, then $\exists h$ such that $h^d\neq 1$. But then $|h|$ does not divide $d$, and hence since it is a finite group, we get a contradiction. $\endgroup$ – George Jul 7 '17 at 18:13

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