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$G \cong \mathbb{Z}/3^6\mathbb{Z} \oplus\mathbb{Z}/3^5\mathbb{Z}\oplus\mathbb{Z}/3^2\mathbb{Z}$

$H\leq G$ so that $G/H \cong \mathbb{Z}/3^2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z} $

Find all possible $H$ up to isomorphism.

This is what I did:

(1) $H$ must be of the form $\mathbb{Z}/3^{k_{1}}\mathbb{Z} \oplus\mathbb{Z}/3^{k_{2}}\mathbb{Z}\oplus\mathbb{Z}/3^{k_{3}}\mathbb{Z}$

(2) WLOG, $0 \leq k_{1} \leq 6$, $0 \leq k_{2} \leq 5$, $0 \leq k_{3} \leq 2$, (The order obiously doesn't matter because a change of order will be isomorphic)

Also, $|G/H|=3^3=3^{13}/(3^{{k_{1}+k_{2}+k_{3}}}) \Rightarrow k_{1}+k_{2}+k_{3}=10$

Thus, by choosing $k_{i}$ we can find all possibilities.

However, I have a few questions:

  1. Is (1) justified? Why $\mathbb{Z}/3^{k_{1}}\mathbb{Z} \oplus\mathbb{Z}/3^{k_{2}}\mathbb{Z}\oplus\mathbb{Z}/3^{k_{3}}\mathbb{Z}\oplus\mathbb{Z}/3^{k_{4}}\mathbb{Z}$ isn't possible?
  2. Is (2) justified? Why $k_{1}=7$ is not possible?

Thank you!

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(1) is justified by the general result that any subgroup of an abelian group with $n$ generators can be generated by at most $n$ elements. (2) is justified by the fact that all elements of $G$, and hence also of $H$, have order dividing $3^6$. – Derek Holt Jun 25 '12 at 11:54
WOLG? Not WLOG? – Gerry Myerson Jun 25 '12 at 12:43
@ Gerry - you're right (: @ Derek - perhaps it was a too easy example. Why $k_{3}=3$ is not possible? – catch22 Jun 25 '12 at 12:56
$k_3=3$ is not possible because ${\mathbb Z}/3^3{\mathbb Z} \oplus {\mathbb Z}/3^3{\mathbb Z} \oplus {\mathbb Z}/3^3{\mathbb Z}$ has more elements of order dividing 27 than $G$ does. – Derek Holt Jun 25 '12 at 16:24

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