Does Every p-Group Have a Subgroup of Every Order Between 0 and n?

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Homework Statement



If G is a p group, show that it has a subgroup of order p^m for every 0<=m<=n.

The Attempt at a Solution



The only thing I know about p-groups is that they have nontrivial centers.
 
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I take it that the order of G is p^n? And I take it you're not supposed to use Sylow?

How about induction on n then?
 
yes, p-group is by definition a group of order p^n. And it's OK to use Sylow theorems, but how?
 
Well, it has another definition where every element has order p^k for some k. Anyway, I think induction is pretty much the only way to go. As I recall, this proof is rather tricky so don't get discouraged.
 
This is really a restatement of one of the Sylow theorems for p-groups. So it's a good idea to try to study a proof of the appropriate Sylow theorem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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