# No cyclic subgroups of orders 8 and 9 in $G/H$

Consider the group $$G = \mathbb{Z}_{48} \oplus \mathbb{Z}_{36} \oplus \mathbb{Z}_{30}$$, with subgroup $$H = \langle (30, 16, 18) \rangle$$. I wish to show that in $$G/H$$, there are no cyclic subgroups of order $$8$$ and $$9$$.

Some incomplete ideas which I have thought of follow––

• Define the natural homomorphism $$\psi: G \rightarrow G/H, \psi(g) = g+H.$$
• By construction, $$\ker\psi = H,$$ so $$|\ker \psi| = |H| = \text{lcm}(|30|, |16|, |18|) = \text{lcm}(8, 9, 5) = 360.$$
• Thus, $$\psi$$ is a $$360$$-to-$$1$$ homomorphism from $$G$$ onto $$\psi(G)$$.
• Upon the contrary, suppose that in $$G/H$$ there is a cyclic subgroup $$\psi(g') = g' + H$$ of order $$8$$. Since $$8=|g'+H|||g'|$$, then $$|g'| = 8n,$$ for some $$n \in \mathbb{N}.$$ Am I permitted to say that because $$\psi$$ is $$360$$-to-$$1$$, $$|g'| =8 \cdot 360?$$
• Hi, does my solution answer your question? Mar 30 at 8:02
• @MathFail Yes, it does, thank you. I've accepted your answer. :) Mar 30 at 21:38
• Great, you are welcome! Mar 31 at 7:34

"Since $$8=|g'+H|||g'|$$, then $$|g'| = 8n$$", how do you get this?

If such a cyclic subgroup exsits, assume the generator is $$g'+H$$, then we have $$(g'+H)^8=H$$ and $$(g'+H)^k\neq H$$, for $$k=1,2,...,7$$, which is equivalent to $$8g'\in H$$ and $$kg'\notin H$$ for $$k=1,2,...,7$$.

If set $$g'=(a,b,c)$$, then we have $$6|8a, 4|8b, 6|8c$$, hence $$b$$ is arbitray, $$a=3n, c=3m$$, so for $$g'=(3n,b,3m)$$, at least $$4g'\in H$$, hence the order for this cyclic subgroup is $$\le 4$$, which gives contradictions.

Similar argument for the non-existence of the cyclic subgroup with order $$9$$.

• And as to your question, I had used the theorem that the order of $gH$ is $G/H$ divides the order of $g$ in $G$. Mar 28 at 15:10
• Also, how have you obtained that if $g' = (a, b, c)$, then we have $6|8a, 4|8b, 6|8c$? Mar 28 at 16:05
• Can you send the link for this theorem? Mar 28 at 17:51
• Because $8g'\in H$, and $H=\langle 6, 4, 6\rangle$, so $8a=6n_1\Rightarrow 6|8a$ Mar 28 at 17:54
• A link to this theorem follows: math.stackexchange.com/questions/2929919/… Mar 28 at 18:22