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If $G$ is a group and $m\geqslant 1$ is an integer put $G[m] := ${$x^m = 1 : x ∈ G$}
a) If $G$ is cyclic of order $n$ and if $m|n$, show that $|G[m]| = m$.
(b) Suppose that $G = \langle x \rangle$ has order $n$, and that $m|n$. Prove that $G[m] = \langle x^{\frac{n}{m}} \rangle $

Not sure how to start this. Do we use $\sum_{d|n} φ(d) = n$?

Gauri Sharma
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1 Answers1

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HINT: If $G$ is cyclic of order $n$ and $x$ is a generator, then $$G=\{x^0,x^1,\ldots,x^{n-1}\}.$$ Which of these elements satisfy $y^m=1$?

Servaes
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