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A cyclic group $G := \langle x \rangle$ of finite order $n$ has a unique subgroup of order $d$, namely $$\langle x^{n/d} \rangle = \{g \in G : g^d = 1\}$$ for every divisor $d$ of $n$.

I wanted to show the equality $\langle x^{n/d} \rangle = \{g \in G : g^d = 1\}$. Now for the inclusion $\subseteq $ we take an element of $\langle x^{n/d} \rangle$ which has the form $x^{kn/d}$ for some natural number $k$. Hence $$(x^{kn/d})^d = x^{kn} = (x^n)^k =1^k = 1$$ and so $x^{kn/d} \in \{g \in G : g^d = 1\}$. However I am a bit lost proving the inclusion $\supseteq$.

Shaun
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TheGeekGreek
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3 Answers3

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Assume $g$ satisfies $g^d = 1$. Since $G$ is cyclic, you can find an integer $m$ so that $g = x^m$. But this means $x^{md} = 1$ and since $n$ is the order of $x$, this implies $n | md$. We can now conclude $g = x^m = x^{(n/d) \cdot (dm/n)} \in \langle x^{n/d}\rangle$.

Dominik
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Assume $G$ is a finite cyclic group and $|G|=n$. There exists $g\in G$ such that $G=<g>$ and $|g|=n$.

$\mathbf{Existence}$: Put $e=\dfrac{n}{d}$ and define $H_d=<g^e>$. We will show that $|H_d|=d$.

$\bullet$ Indeed, $(g^e)^d=g^{ed}=g^n=1$. From that, we have that $|g^e|=|H_d|\leq d$.

$\bullet$ Suppose $|g^e|=k$. Therefore, $g^{ek}=1$. Since $|g|=n$, we have that $n\leq ek$. It follows that $\dfrac{n}{e}=d\leq k$. So $|H_d|=|g^e|= k\geq d$.

Conclusion: $|H_d|=d$.

$\mathbf{Unicity}$: Suppose $H<G$ and $|H|=d$. We will show that $H=H_d$, where $H_d$ is defined above. We know that $H$ is cyclic. Therefore, $H=<g^k>$, where $k$ is the smallest positive integer such that $g^k\in H$.

$\bullet$ Since $|H|=d$, we get that $g^{kd}=1.$ But $|g|=n$ and $g^{kd}=1$, it follows that $kd\geq n$.

$\bullet$ By the same way, $d$ is the smallest positive integer such that $(g^k)^d=g^{kd}=1$. And since $g^n=1$, it follows that $kd\leq n$.

From that, we get that $kd=n$ which implies that $k=\dfrac{n}{d}=e$. Finally, we get that $H_d=H$.

Cachorro
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Hint: Since $g \in G = <x>$ then $g=x^k$ for some $k$. Now use $g^d=1$.

N. S.
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