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

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?

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 $$g \in G$$.

• Ah of course! The specification of a cyclic group got me confused, but this is actually related to Lagrange's theorem? – asdwasd18 Oct 14 '19 at 13:43
• 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). – kccu Oct 14 '19 at 13:46

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 $$x^n=(g^t)^n=(g^n)^t=1^t=1$$.