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Let $A,B,C$ are three subgroups in a way that $1<A \triangleleft B \triangleleft C$. With $B/A$ and $C/B$ are $p$-groups. Then prove that $|C|$ is also a $p$-group. I have been trying to prove it for hours but can't find a way to solve it. So any help will be appreciated.

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  • $\begingroup$ You can't use it because it doesn't say B is the subgroup of C or all it's group is $p-group$. All we can say $C$ is of the form $p^\alpha m$. $\endgroup$ – user631697 Feb 18 at 4:38
  • $\begingroup$ $C$ (not $|C|$ as stated) is a $p$-group here if and only if $A$ is a $p$-group. $\endgroup$ – Nicky Hekster Mar 19 at 9:06
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The result is false. Let $A=\mathbb{Z}_2 \times \{0\} \times \{0\}$, $B=\mathbb{Z}_2 \times \mathbb{Z}_3 \times \{0\}$, and $C=\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3$. $B/A$ and $C/B$ are each isomorphic so $\mathbb{Z}_3$ (so $3$-groups), but $|C|$ is not a power of a prime so $C$ is not a $p$-group for any $p$, let alone $p=3$.

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    $\begingroup$ Yes you are right, I can't believe the question is wrong. I have been trying for hours now. Thank you so much $\endgroup$ – user631697 Feb 18 at 4:51
  • $\begingroup$ Once, I tried for more than a week to prove something. Finally my advisor said "well have you ever considered that it's actually not true?" I had the counterexample within 10 minutes :( $\endgroup$ – Randall Feb 18 at 4:57
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    $\begingroup$ When I was a TA, my professor assigned to his Group Theory students the question: Assume $f(x)$ is irreducible over $\Bbb Q$. Prove the roots of $f(x)$ form a basis over $\Bbb Q$. I got 20 different proofs from 20 students, and none of them were right. I convinced myself of that when I tried for hours to prove it myself. And then I considered $f(x) = x^2-2$. $\endgroup$ – Robert Shore Feb 18 at 5:57

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