Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.

Question

This time I have no approach. I haven't found any relation between the divisors of $n$ and the order of the subgroups of $G$. How would approach this?

It may not help you to think this way but there is a clear-cut way to deal with cyclic groups. A cyclic group is one which has a surjection from the additive group $\mathbb{Z}\to G$. In this case the kernel would be $n\mathbb{Z}$. Now the subgroups of $G$ and the subgroups containing $n\mathbb{Z}$ are in bijection. Now this should be obvious. — Grobber, Aug 06 '16 at 06:50
See also https://math.stackexchange.com/questions/410389/subgroups-of-a-cyclic-group-and-their-order — lhf, Dec 23 '19 at 01:03

score 1 · Accepted Answer · answered Aug 06 '16 at 06:28

1

If you realize $G$ as the set $\{0,1,\ldots, n-1\}$ with addition modulo $n=de$, then $\{0,e,2e,\ldots, (d-1)e\}$ is a subgroup of order $d$.

answered Aug 06 '16 at 06:28

Peter Franek

11,522

I like your idea. Would it be enough if I prove that this subgroup is a subgroup of G? – TheMathNoob Aug 06 '16 at 06:55
Isn't it obvious? $ke + le = (k+l)e$. – Peter Franek Aug 06 '16 at 07:00
for the inverse can we say $ke+x=de$, so $x=e(d-k)$ where $k<d$? – TheMathNoob Aug 06 '16 at 07:06
Yes, the inverse of $ke$ is $(d-k)e$. – Peter Franek Aug 06 '16 at 07:07
why? can you elaborate on that?, it's a little bit hard for me to see – TheMathNoob Aug 06 '16 at 07:11
Because $ke + (d-k)e = de = n$ which is zero modulo $n$. The point is that both the group operation as well as inversion are the same as i $G$ (that's part of the definition of a subgroup). – Peter Franek Aug 06 '16 at 07:12
Oh didn't realize that you changed your answer XD – TheMathNoob Aug 06 '16 at 07:14

score 0 · Answer 2 · answered Aug 06 '16 at 06:34

‎ ‎$ d‎ ‎\vert n ‎‎‎‎‎‎ ‎\Rightarrow ‎‎‎‎\exists k‎ ‎\in ‎‎\mathbb{N} ‎‎\qquad‎ st ‎‎\qquad‎ n=kd ‎‎$‎‎‎‎

‎$ 1=‎ ‎‎a^n=a^kd=(a^k)^{d} ‎\Rightarrow ‎o(<a^k>)=d ‎‎$‎‎‎

then ‎$<a^k> \leq G ‎$ ‎and ‎‎$‎\vert ‎<a^k‎> \vert = d ‎$‎‎

score 0 · Answer 3 · answered Aug 06 '16 at 06:42

0

For any divisor d of n,we shall get an integer k such that dk=n. Now consider the element $a^k$ and show that the subgroup generated by it is the required one.

answered Aug 06 '16 at 06:42

Shanghaikid

185

Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.

3 Answers3

Linked

Related