I was provided this difficulty for homework and I to be not certain where to start. I know a systems using Lagrange"s theorem, yet we have actually not proven Lagrange"s theorem yet, in reality our teacher hasn"t also mentioned it, so i am guessing there should be another solution. The just thing I could think that was showing that a team of element order \$p\$ is isomorphic to \$ asiilaq.netbbZ/p asiilaq.netbbZ\$. Would this work?

Any guidance would be appreciated.

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As video camer McLeman comments, Lagranges theorem is substantially simpler for groups of element order than for general groups: it states that the team (of prime order) has no non-trivial ideal subgroups.

I"ll usage the complying with

## Lemma

Let \$G\$ be a group, \$xin G\$, \$a,bin asiilaq.netbb Z\$ and \$aperp b\$. If \$x^a = x^b\$, then \$x=1\$.

Proof: by Bezout"s lemma, part \$k,ellin asiilaq.netbb Z\$ exist, such that \$ak+bell=1\$. Then \$\$ x = x^ak+bell = (x^a)^k cdot (x^b)^ell = 1^k cdot 1^ell = 1 \$\$

(If you know a tiny ring theory, you can prefer to an alert that the set \$ x^i=1\subseteq asiilaq.netbb Z\$ forms perfect which need to contain \$(a,b)=1\$ if it consists of \$a\$ and \$b\$.)

## The question

Now permit \$P\$ it is in an arbitrary group of element order \$p\$. Consider any type of \$xin P\$ such the \$x eq 1\$ and consider the set\$\$ S = 1, x, x^2 , dots , x^p-1 \subseteq P.\$\$First assume two of these aspects are equal, say \$x^u=x^v\$ and \$u

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