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I need help to prove that if $(G,\cdot)$ is a cyclic group generated by $g^n$, ($g\in G$, $n\in \mathbb{N}$), then $G$ is also generated by $g$.

I have the definition that $(G,\cdot)$ is a cyclic group if $G=\langle g \rangle$ for some $g\in G$ $\iff$ we can find $g\in G$ such that $\forall h \in G, \exists n \in \mathbb{Z}$, such that $h=g^n$. We then say that $g$ is a generator of $G$.

It follows that if $n \in \mathbb{N}$, then $ord_G(g)=\min\{n\mid g^n=e\}$. Then the following are equivalent:

  • $ord_G(g)=n$.
  • $\langle g \rangle = \{e,g,g^2,...,g^{n-1}\}$, with all $n$ distinct.
  • $\left|\langle g \rangle\right|$ = n.
  • $g^n=e$ and if $g^m=e$, then $n\mid m$.

Additionally, if $d=g^m$, then $ord(d)=\dfrac{n}{\gcd(n,m)}$

For my proof, I'm not sure if I need to prove that the cyclic subgroup generated by $\langle g \rangle$ is equivalent to $G$ itself, so that $g^n$ is then a generator of $G$. But then I'm using what I want to prove as an assumption, so, I believe my argument is for the converse. Also, I don't know if that is a correct way to go about this.

Alternatively, I'm wondering if I'm supposed to consider $\langle g \rangle$ as $\langle g^n \rangle$ instead. But I don't know the implications of that. Does this mean that $\langle g^n \rangle$ = $\{g^{n^n}\mid n \in \mathbb{Z}\}$? Or does that not follow from that?
Am I on the right track with either of those ideas? If not, I'd appreciate any sort of hint or starting point that could help me develop this further. Would going about this by method of contradiction be a better starting point?

I appreciate any feedback. I'm still very new to group theory and so these concepts haven't quite sunk in yet.

shp
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sadmathbean
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    First of all, $(g^n)^k$ is equal to $g^{kn}$, and not $g^{n^k}$. Also, note that a cyclic group might also be infinite, in which case the equivalent conditions you wrote are not really relevant. Finally, what you need to prove is that every element of $G$ is equal to some power of $g$. Can you prove it, given that every element is a power of $g^n$? – Mark Feb 20 '23 at 18:13
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    If every element of $G$ is a power of $g^n$, then every element of $G$ is also a power of $g$: $(g^n)^k = g^{nk}$. – Randall Feb 20 '23 at 18:13
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    $G=\langle g^n \rangle \subseteq \langle g \rangle \subseteq G$ – lhf Feb 20 '23 at 18:20
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    Hi @lhf : would you consider converting your comment-answer into a solution answer? Regards – rschwieb Feb 20 '23 at 18:40
  • @Mark Thank you for that clarification. That makes more sense looking at it like that. – sadmathbean Feb 20 '23 at 19:29
  • @Randall That makes more sense. Thank you for that equality! – sadmathbean Feb 20 '23 at 19:31
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    @rschwieb, done, thanks for the nudge. – lhf Feb 24 '23 at 23:17

3 Answers3

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You're trying to confuse yourself by using notation that conflicts with the given data.

The exercise asks you to prove that if $g^n$ is a generator of the cyclic group $G$, then also $g$ is a generator thereof.

Now ask yourself how could this fail: there should be an element $x\in G$ that's not a power of $g$. On the other hand, $x=(g^n)^m$ for some integer $m$, because $g^n$ is a generator of $G$ by assumption. But…

More directly, saying that $G$ is generated by $g^n$ means that every element in $G$ is a power of $g^n$. Hence every element of $G$ is also a power of $g$, because $$ (g^n)^m=g^{nm} $$ Hence $$ G=\langle g^n\rangle\subseteq\langle g\rangle\subseteq G $$ and we have proved that $G=\langle g\rangle$.

egreg
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If $G$ is finite, the order of $g^n$ is always less than or equal to that of $g$, since $$\lvert g^n\rvert =\dfrac {\lvert g\rvert}{(\lvert g\rvert, n)}.$$

Here is a proof i gave.

If $G$ is infinite, then $G\cong \Bbb Z.$ The only generators are $\pm 1.$ And, if $g^n=\pm 1,$ then $g=\pm 1.$

calc ll
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Here is a one-line proof: $$ G=\langle g^n \rangle \subseteq \langle g \rangle \subseteq G $$ The first equality is the hypothesis.

The second inclusion follows from $g^n \in \langle g \rangle$.

The third inclusion follows from $g \in G$.

These two inclusions follow from $h \in H \implies \langle h \rangle \subseteq H$, for $H$ a subgroup of $G$.

lhf
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