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I'm studying the proof of the theorem that says:

A subgroup of a cyclic group is cyclic

The proof goes as follows.

Let $G = \langle g \rangle$ be a cyclic group. Let $H$ be a subgroup of $G$. For $H = \{e\}$, the statement is trivial. So suppose $H \neq \{e\}$ en let $g^k \in H$ with $k$ minimal in $\mathbb{N_0}$ (so $k$ is the first non zero natural number such that $g^k \in H$). We now show that $H = \langle g^k \rangle$, from which the result follows. Let $g^n \in H$. Write $n = qk+r$ with $0 \leq r <k$. Then, $g^r = g^{n-qk} = g^n(g^k)^{-q} \in H$, from which must follow that $r = 0$ (because we took $k$ minimal and $r < k$). So $g^n \in \langle g^k \rangle$, hence $H = \langle g^k \rangle$.

Now, I'm wondering why the proof doesn't work if $H = \{e\}$ if I would not treat the case $H = \{e\}$ separately.

I think it is because the proof would fail if $G$ is generated by an element of infinite order, so we cannot take $g^k \in H$ with $k$ minimal. Can someone verify whether this is correct?

Thanks in advance.

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    $\begingroup$ You need a non-zero $k$ in order to invoke the division algorithm, writing $n = qk+r$. $\endgroup$ – James Jun 21 '17 at 14:50
  • $\begingroup$ Yes and we cannot always find such a $k$, for example as in the example I mentioned? $\endgroup$ – user370967 Jun 21 '17 at 14:51
  • $\begingroup$ That's correct. If $H$ is not trivial, then you know that there is a non-zero power of the generator $g$ that belongs to $H$, and so you can take that power to be the least positive such power. $\endgroup$ – James Jun 21 '17 at 14:58
  • $\begingroup$ Write this as an answer :) $\endgroup$ – user370967 Jun 21 '17 at 14:59
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The point of treating the case of a non-trivial subgroup $H$ separately is to get a non-zero power $g^k$ (i.e., $k\neq 0$) of the generator $g$ of $G$ in $H$. Then there is a positive such power $g^k$; i.e., with $k>0$ (since $g^k\in H$ implies that $g^{-k}\in H$). Then you can take $k$ to be the least positive power of $g$ that belongs to $H$. That allows you to use the division algorithm to write $n=qk+r$, and the rest of the proof goes through.

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There's no real need to separate off the case $H=\{e\}$ when $G$ is finite; in this case the minimal $k>0$ such that $g^k\in H$ is $k=|G|$.

If $G$ is infinite, you indeed get into trouble, because $g^k\ne e$ for all $k>0$.

I believe you can appreciate a different proof. Consider the map $$ \varphi_g\colon\mathbb{Z}\to G \qquad \varphi_g(n)=g^n $$ Since $G=\langle g\rangle$, this is a surjective homomorphism. Thus the subgroups of $G$ are in bijection with the subgroups $K$ of $\mathbb{Z}$ containing $\ker\varphi_g$ and the bijection sends $K$ to $\varphi_g(K)$.

Since every subgroup of $\mathbb{Z}$ is of the form $k\mathbb{Z}$ and is cyclic, we are done.


The argument with the division algorithm is the key step in proving that the subgroups of $\mathbb{Z}$ are of the form $k\mathbb{Z}$: the statement is obvious for $K=\{0\}$. If $K\ne\{0\}$, then it contains a positive element, hence a minimal positive element $k$. Now it's easy to see that $K=k\mathbb{Z}$.

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  • $\begingroup$ Ah nice proof. Thanks for your answer. Now I'm sure the proof can only fail in the infinite case, as I expected. $\endgroup$ – user370967 Jun 21 '17 at 22:03

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