$p$-group problem

Let $$A,B,C$$ are three subgroups in a way that $$1. With $$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.

• 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 $p^\alpha m$. – user631697 Feb 18 at 4:38
• $C$ (not $|C|$ as stated) is a $p$-group here if and only if $A$ is a $p$-group. – Nicky Hekster Mar 19 at 9:06

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 $$p=3$$.
• 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 $f(x) = x^2-2$. – Robert Shore Feb 18 at 5:57