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As the title reads: In a cyclic group with $n$ elements, how many elements $x$ have the following property: $x^n=1$ ? where $1$ is the identity

I tried searching for an answer to this in my book and here, but I couldn't find a direct answer, so I apologize if this has been answered before in another post.

I feel like there's a theorem for this, I just can't seem to find it. I'm thinking the answer has to do with the number of generators, since the generators guarantee that we find the identity, so multiplying a generator element n times should give the identity, or am I completely off track?

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It's a property of every finite group $G$ that the order of any element $g \in G$ divides $|G|$. That is, if the order of $g$ is $m$, then $m $ divides $|G|$. A consequence of this is that $g^{|G|}=1$ for every $g \in G$.

In particular, if $G$ is cyclic of order $|G|=n$, then $g^n=1$ for every $g \in G$.

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  • $\begingroup$ Ah of course! The specification of a cyclic group got me confused, but this is actually related to Lagrange's theorem? $\endgroup$ – asdwasd18 Oct 14 '19 at 13:43
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    $\begingroup$ That's right, nothing special about cyclic groups is used here. Although as the other answer shows, for cyclic groups this can be proved more simply (without invoking Lagrange's Theorem). $\endgroup$ – kccu Oct 14 '19 at 13:46
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If $G$ is cyclic, it's generated by some $g\in G$, so $x=g^t$ for some integer $t$ and $g^n=1$. Then $x^n=(g^t)^n=(g^n)^t=1^t=1$.

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