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This is taken from Keith Conrad's expository paper on subgroups of cyclic groups.

Let $G$ be a cyclic group, with a generator $g$. For a subgroup $H \subset G$, we will show $H=\langle g^n\rangle$ for some $n\geq0$, so $H$ is cyclic. (...) The idea is that every element in a subgroup of the form $\langle g^n\rangle$ looks like $g^{nk}$, so $n$ is the smallest positive exponent among the elements of the subgroup.

I'm not sure how to think about this... If we take the cyclic group with ten elements $(e,g^1,g^{-1},\ldots, g^5)$ and build the subgroup $\langle g^4\rangle=\{g^4,g^{-2},g^2,g^{-4},e\}$ so $g^{4*3}=g^2$ and $2<4$. Am I making some mistake?

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    $\begingroup$ How are you getting a subgroup of $4$ elements in a group of $10$? Lagrange would be very disappointed in you! $\endgroup$ – Theo Bendit Mar 12 '18 at 16:39
  • $\begingroup$ Ouch! Thanks about that. $\endgroup$ – nek28 Mar 12 '18 at 16:42
  • $\begingroup$ So $g^4$ and $g^2$ generate the same subgroup. Why do you think this is a problem? $\endgroup$ – ancientmathematician Mar 12 '18 at 16:42
  • $\begingroup$ The way it is stated in the proof seems to imply that 4 should be the minimal power in a subgroup generated by $g^4$. $\endgroup$ – nek28 Mar 12 '18 at 16:44
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    $\begingroup$ The important thing that the paper says is that we can get a generator for $H$ by looking at $g^n$ where $n$ is minimal in $\{i\in\mathbb{N} : g^i\in H\}$ $\endgroup$ – ancientmathematician Mar 12 '18 at 16:47
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Note that it says it will show $H=\langle g^n\rangle$ for some $n > 0$. That is, it is only necessary to find a single $n$, not every $n$ that will work. The claim is that if $H \ne \{e\}$, it is sufficient to choose $$n = \min \{k \in \Bbb N^+\mid g^k \in H\}$$

Since this set cannot be empty (I'll leave it to you to figure out why), it always has a least element. For any $g^k \in H$, we can write $k = qn + r$ for some $q$ and $0 \le r < n$. But $g^r = (g^n)^{-q}g^k \in H$, and since $n$ is the smallest positive exponent of $g$ in $H$, it must be that $r = 0$. I.e., $g^k = (g^n)^q \in \langle g^n \rangle$.

It doesn't matter if there are other values $n_1$ such that $\langle g^{n_1} \rangle = H$. (In your example, $n = 2$ and $n_1 = 4$.) All that matters is that the $n$ chosen can be proven to work.

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