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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-$1$, $|g'| =8 \cdot 360?$
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  • $\begingroup$ Hi, does my solution answer your question? $\endgroup$
    – MathFail
    Mar 30 at 8:02
  • $\begingroup$ @MathFail Yes, it does, thank you. I've accepted your answer. :) $\endgroup$ Mar 30 at 21:38
  • $\begingroup$ Great, you are welcome! $\endgroup$
    – MathFail
    Mar 31 at 7:34

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"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 $9$.

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

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