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I want to understand the proof that $g^x = g^{x\bmod m}$ for $g\in G$ for $G$ such that $\vert G\vert = m$

The proof goes as follows

say that $x = qm+r$

then $$g^x = g^{qm+r} = g^{qm}g^r = (g^m)^qg^r = 1^qg^r = g^r = g^{x\bmod m}$$

How is $g^m = 1$? Or is there an hypothesis missing? Such as $g$ is a generator.

Moreover, if I take $\mathbb{Z}^*_5$, then $2^5 = 2 \ne 1 = 2^{5\bmod 5}$

Michael Hardy
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user405156
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    This is a fundamental property that if $|G| = m$, then for all $g \in G$, $g^m = 1$ . This is a corollary of Lagrange's theorem. https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory) – JWM Jan 23 '18 at 16:24
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    The order of the multiplicative group $\mathbb Z^*_5$ is not $5$. – Cheerful Parsnip Jan 23 '18 at 16:26
  • Note the use of \bmod in my edits. The "b" stands for "binary", meaning the spacing to the left and right of "mod" is what is appropriate for a binary operation symbol. – Michael Hardy Jan 23 '18 at 16:27
  • I always make the mistake that the order of $\mathbb{Z}^*_p$ is $p$ and not $\phi(p)$. – user405156 Jan 23 '18 at 16:29

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In a finite group $G$ of order $m$ we have $g^m=e$ for all $g\in G$. This follows from Lagrange's Theorem. A special case is for $G=(\mathbb{Z}/n\mathbb{Z})^{\times}$, of order $m=\phi(n)$, which is called Euler's Theorem.

Dietrich Burde
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The order of the group generated by $g$, $<g>$, must divide the order of the group, by Lagrange's theorem. But the order of $<g>$ is the order of $g$. Set $d=\lvert <g >\rvert $. Thus $m=k\cdot d$ for $k\in \mathbb Z$... So $g^m=g^{kd}=(g^d)^k=(e)^k=e$...

Secondly, $\phi (5)= 5-1=4$.