If $G$ is an abelian group of order 187, then every nontrivial subgroup of $G$ has order 11 or 17.

Question

I am trying to show that if $G$ is an abelian group of order 187, then the order of every nontrivial subgroup must be either order 11 or 17. I haven't learned the Sylow Theorems to be able to prove it using that method, but I have learned about Lagrange's Theorem, and I am not sure if it is applicable here.

Attempt: If $G$ is an abelian group of order 187, then by Lagrange's Theorem, every element in $G$ must have orders that are factors of 187, that is, 1, 11, 17, or 187. I am not exactly sure how to approach from here, but some help would be appreciated.

You must use the word "non-trivial". You must also only consider proper subgroups. — John Douma, Nov 19 '23 at 06:07
Try to see what happens when your subgroup has two elements which are no generated one from the other. For instance, you should that if your subgroup has two elements of order 11 and 17. They generate the whole group. — 196884e, Nov 19 '23 at 10:42
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. — Shaun, Nov 19 '23 at 13:01

score 1 · Accepted Answer · answered Nov 19 '23 at 05:13

1

The order of a subgroup divides the order of a group. Since $187=11\cdot 17$, the only possible subgroups have order $1$, $11$, $17$, or $187$. (For an abelian group, all four possibilities will occur.)

answered Nov 19 '23 at 05:13

pancini

20,030

Tri · Answer 2 · 2023-11-19T13:03:26.307

0

Since $187=17\times 11$, if you pick any nontrivial subgroup $H \le G$, then Lagrange's theorem states that the order of $H$ divides 187, so there are 4 possible cases: $1,11,17,187$. Since the problem asks to show any nontrivial subgroup is either of order 11 or 17, you have to eliminate the case 1 and 187. What will happen if subgroup $H$ has orders 1 and 187?

edited Nov 19 '23 at 13:03

answered Nov 19 '23 at 05:32

Tri

1,513

2

This is completely unrelated to Cauchy's Theorem. – Tobias Kildetoft Nov 19 '23 at 08:32
@TobiasKildetoft Oh I thinh I missed the part "every subgroup" when I read this post last night. Thank you. I will edit that. – Tri Nov 19 '23 at 12:50

score 0 · Answer 3 · answered Nov 19 '23 at 12:52

0

Hints:

For all $d\mid |G|$ for finite, abelian $G$, there exists a subgroup $H\le G$ such that $|H|=d.$ (See here.)
Recall Lagrange's Theorem.

answered Nov 19 '23 at 12:52

Shaun

47,747

If $G$ is an abelian group of order 187, then every nontrivial subgroup of $G$ has order 11 or 17.

3 Answers3