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If $G$ be a cyclic group of prime order $p$, prove that every non-identity element of $G$ is a generator of $G$.

WE know that no of generator= $\phi{(p)}=p-1$.

We have, if $a$ is a generator of $G$ then $a^k$ is a generator iff $gcd(p,k)=1$. Here as $p$ is prime, then $k=1, 2,3,\cdots, p-1$ as gcd of each of $k$ with $p$ is $1$. Therefore, generator of $G$ are $a, a^2, a^3, \cdots a^{p-1}$. They are $p-1$ in number.

I am not satisfied with my above argument. I need more significant proof. Please help me.

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  • $\begingroup$ Look at whether a power $a^i$ will be equal to $a^j$. If not, then all powers of distinct and <a> becomes a generator. $\endgroup$ – Shailesh Sep 25 '15 at 4:34
  • $\begingroup$ @Shailesh How to show that " power $a^i$ will be equal to $a^j$ ..." $\endgroup$ – rama_ran Sep 25 '15 at 4:42
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    $\begingroup$ Ok, First of all every group of prime order is cyclic. Let $a \ne e$ be an element of the group. Now consider all powers of $a$. If, for some $i,j$, $a^i = a^j$, then $a^{i-j} = e$, so the order of this element $a$ is $i-j \lt p$, but the order of an element divides the order of the group. Use this to show that all the powers of $a$ are distinct, which means any $a$ is a generator. Please fill in the gaps. $\endgroup$ – Shailesh Sep 25 '15 at 4:48
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Hint: Recall that by Lagrange's thereom, $\vert g \vert$ divides $\vert G \vert $ $\forall$ $g$ $\in $ $G$.So what can you about $\vert g\vert $ ?

Edit: let $g \in G$,consider the subgroup $H=<g>$ and by Lagrange's theorem $\vert H \vert$ divides $\vert G \vert $,hence $\vert g \vert$ divides $\vert G \vert$.

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  • $\begingroup$ Lagrange's thereom tells that order of every subgroup of a finite group G is a divisor of $o(G)$. But If d divides $o(G)$ it is not necessarily true that there is a subgroup of $o(d)$. You are telling about order of element divides $o(G)$ but Lagrange's theorem tells $o(subgroup)$ divides $o(G)$. Also I am not understand how your argument proves the desired result? $\endgroup$ – rama_ran Sep 25 '15 at 4:39
  • $\begingroup$ I've edited my answe to make it more clear. $\endgroup$ – Arpit Kansal Sep 25 '15 at 4:48
  • $\begingroup$ Does the downvoter care to explain? $\endgroup$ – Arpit Kansal Sep 25 '15 at 4:49
  • $\begingroup$ I understand your point of "order of element divides o(G)". But how this argument proves the desired result? $\endgroup$ – rama_ran Sep 25 '15 at 4:55
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    $\begingroup$ you're right,Bravo! $\endgroup$ – Arpit Kansal Sep 25 '15 at 5:05
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We know that for all g in G, <(g)> is a subgroup of G, so by Lagrange's Theorem, |<(g)>| = |g| divides |G|. Since |G| = p, a prime, for all g in G, |g| is 1 or p. But, if |g| = 1, then g=e, thus for all non-identity g in G, |g|=p, so |<(g)>|=|G|, and <(g)> is a subset of G, and <(g)> and G are finite, so <(g)>=G, for all non-identity g in G.

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