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Can anyone please give a clear proof for this statement? Let $G$ be a $p$-group, then, there exists subgroup of $G$ of any order of any power of $p$.

I would appreciate if you don't use 2nd and 3rd Sylow theorems.

Thank you.

MattAllegro
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Bar Dubovski
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1 Answers1

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Let $p$ a prime number. We will prove by induction on $n$ that every group of order $p^n$ has a normal subgroup of order $p^m$ for $0 \le m \le n$. Let $G$ a group of order $p^n$.

If $n=1$, that's clear. Let $n \ge 2$. Clearly the assertion is true for $m=0, m=n$.

Moreover, $G$ has a normal subgroup of order $p$. Indeed, $G$ has an element $a \in Z(G)$ of order $p$ that lies in the center of $G$, which is not trivial (because $|G|=p^n$). Then, the subgroup $H$ generated by $a$ is normal.

Let $\pi : G \to G/H$ the canonical surjective homomorphism. Since $|G/H| = p^{n-1}$, the induction hypothesis applies : there is a normal subgroup $K$ in $G/H$, with $|K|=p^{m-1}$, for $1 \le m \le n$. Therefore, $\pi^{-1}(K)$ is a normal subgroup of $G$ with cardinality $p^m$, as required.

Watson
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  • Thank you for the answer, what is m ? – Bar Dubovski Jan 05 '16 at 22:37
  • We might have $H=G$, but then of course $G$ is cyclic and the claim is clear. Alternatively: It suffices to find any proper nontrivial normal subgroup of $G$, so why not take $Z(G)$ insteadof $\langle a\rangle$? If $Z(G)=G$ then $G$ is the sum of cyclic groups and we can easily find subgroups as required. – Hagen von Eitzen Jan 05 '16 at 22:37
  • Could you please explain why G has a normal subgroup of order p? I know from cauchy theorem that it has subgroup of order p, why is it normal? – Bar Dubovski Jan 05 '16 at 22:41
  • Can you please explain why the subgroup of order p from Cauchy theorem is generated from a∈Z(G)? – Bar Dubovski Jan 05 '16 at 22:50
  • @Bar Dubovski : the center of $G$ is not trivial. Since $Z(G)$ is abelian and finite, you can show (e.g. using this) that is has an element of order $p$. – Watson Jan 05 '16 at 22:53
  • @Watson, can you please clarify this statement : " the induction hypothesis applies : there is a normal subgroup $K$ in $ G/H$, with $ |K|=p^{m−1} $ for $1≤m≤n$. Therefore,$ π^{−1}(K) $is a normal subgroup of $G$ with cardinality $p^m$, as required.". – Bar Dubovski Jan 05 '16 at 23:14
  • @Bar Dubovski : since $G \neq H$ (because $|G|=p^n$ with $n \ge 2$ and $|H|=p$), one has $|G/H|=p^{n-1}<p^n$ is a non-trivial power of $p$, smaller than $p^n$. Then, by induction hypothesis, for each $m \in {0, \dots, n-1}$, there is a normal subgroup $K_m$ in $G/H$. By the correspondence theorem, one can write $K_m = G_m/H$, with $G_m$ normal in $G$. Moreover $|G_m| = |K_m|\cdot|H|=p^m$, as desired. – Watson Jan 06 '16 at 14:14