0
$\begingroup$

The Fundamental Theorem of Cyclic Groups states the following:

Let $G$ be a cyclic group generated by $g$ with $|G| = n$ (order of $G$ is $n$). Then:

  1. For any subgroup $H\le G$, $H$ is cyclic as well
  2. For any subgroup $H\le G$, $|H|$ (order of $H$) divides $n$
  3. For each positive divisor $t$ of $n$ there is exactly one subgroup $H\le G$ such that $|H| = t$, and $H = \langle g^{n/t} \rangle$.

I was able to follow the proof given here: http://site.iugaza.edu.ps/mabhouh/files/2011/01/alg1-ch4.pdf

However, I don't quite understand what points (2) and (3) mean. Could you share any intuition you have for these?

$\endgroup$
7
  • 2
    $\begingroup$ (2) is Lagrange's theorem. $\endgroup$
    – Shahab
    Mar 19 at 10:34
  • 3
    $\begingroup$ There's an isomorphism from any cyclic group to $\mathbb{Z}/n\mathbb{Z}$ for some $n$. Have you tried interpreting the results there? It might be more intuitive in that setting. $\endgroup$
    – user816709
    Mar 19 at 10:57
  • $\begingroup$ @user816709 what would be a subgroup in $\mathbb{Z}_n$? $\endgroup$
    – cbakos
    Mar 19 at 11:11
  • $\begingroup$ All subgroups of $\mathbb{Z}/n\mathbb{Z}$ are cyclic and are thus of the form $\langle m \rangle$ for some $0\leq m \leq n-1$. $\endgroup$
    – user816709
    Mar 19 at 11:58
  • 1
    $\begingroup$ Cyclic group is defined as a group generated by powers of an element (where powers mean repeated group actions). So, for a size $n$ cyclic group, we can write: $$ \langle g \rangle = \{ g ,g^2...,g^{n-1}, g^n \}$$ But consider a divisor of the order of group, say $t$, it is clear that $\frac{n}{t} = p$ where $p$ is a positive integer. So, it is clear that the following is also a group: $$ \langle g^p \rangle= \{ 1, g^p ,g^{2p}..., g^{n-p},g^{n} \}$$ $\endgroup$
    – Ethakka appam with Chai
    Mar 19 at 13:12

1 Answer 1

Reset to default
0
$\begingroup$

To understand this, perhaps it is useful to try to "build" a subgroup yourself. Let $G = \mathbb Z_n$, the integers modulo $n$ --- we can do this since all cyclic groups are the same as $\mathbb Z_n$ (can you see why?). OK, let's try to make a subgroup $H<G$. We first need to take $0$ since all groups need their identity element. Now, let's say we take the element $2$ into the group $H$. We are then forced to also include $4$, and $6$, and so on, in order to make sure that our group $H$ is closed under the group operation. This is the smallest subgroup generated by the element $2$, and so denoted $\langle 2 \rangle$. If $n=10$, can you write down explicitly all its elements?

Now think about the difference if $n=11$. Do you see a difference if $n$ is even or $n$ is odd? What if, instead of picking the element $2$, we picked some other number? What subgroup would it generate then? How does this depend on the choice of $n$? Does the prime factorisation of $n$ somehow matter?

Once you play around with enough examples, and you read the proof again, all the steps should then feel natural and easy to follow. There's no way to convey the intuition for this easily except to ask you to think about these examples yourself, IMO.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.