# To find a non trivial proper cyclic subgroup.

Does every non prime order finite group have at least one non trivial proper cyclic subgroup?

Let $$a\neq e$$ be an element of $$G$$. Then $$\langle a \rangle$$ is always a subgroup of $$G$$, rather it is a cyclic subgroup of $$G$$. But my doubt arises, what will guarantee us that we shall always have at least one such proper subgroups generated by the elements of $$G.$$

## 2 Answers

By Cauchy theorem, if $$p\mid n$$ and order of $$G$$ is $$n$$ then there exists an element of order $$p$$ in $$G$$ call it $$g$$ then $$\langle g\rangle$$ a cyclic subgroup of order $$p$$. Examples: $$\langle r\rangle \leq D_8$$ and $$\langle 2\rangle \leq \mathbb{Z}/12\mathbb{Z}$$

• Cauchy's theorem is way overkill here. – Tobias Kildetoft Feb 12 '20 at 18:57
• Thanks, actually Cauchy's theorem together with the fundamental theorem of arithmetic makes the whole thing clear. – sukanta das Feb 12 '20 at 19:08

Hint Pick any $$a \neq e$$.

Look at $$$$. If this is strictly smaller than $$G$$ you are done.

If $$=G$$ then $$ord(a)=|G|$$ which is composite. Therefore $$ord(a)=mn$$ for some $$m,n >1$$. Show that $$$$ is a proper cyclic subgroup.

• Yes, nice argument. Thank you. – sukanta das Feb 12 '20 at 19:11