Let $G$ be an abelian group. Assume that $|G|=p^km$ where $p$ is a prime and $p\nmid m$. We know from Sylow's theorem that $G$ has a subgroup of order $p^k$. But is it true that $G$ has a subgroup of order $p^j$ for all $1\leq j\leq k$ ?
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Yes, this follows easily by induction on $j$. – Tobias Kildetoft Nov 13 '16 at 20:05
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Yes...in particular any finite $p$ group has this property. So just take subgroups of the $p$ subgroup you already found. By the way, $G$ need not be abelian. – Eric Auld Nov 13 '16 at 20:05
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Duplicate of https://math.stackexchange.com/questions/1603023 – Watson Nov 13 '16 at 20:23
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Yes, this is true! Even for non-abelian group.
Proposition. Let $G$ be a finite group and $p$ be a prime divisor of the order of $G$, then for all $j\in\mathbb{N}$ such that $p^j$ divides the order of $G$, there exists $H\leqslant G$ such that $|H|=p^j$.
The proof relies on the following lemma:
Lemma. Let $G$ be a finite group and $H\triangleleft G$. If $B$ is a subgroup of $G/H$, let $A:=p^{-1}(B)$, one has: $$[G:A]=[G/H:B].$$
Proof. Let $p:G\twoheadrightarrow G/H$ be the canonical surjection, $p_{\vert A}:A\twoheadrightarrow B$ is a surjective group homomorphism whose kernel is $A\cap H=H$. The result follows from first isomorphism theorem. $\Box$
Then the proof goes like this:
Step 1. If $G$ is abelian of order $p^m$, $G$ admits a subgroup of index $p$.
If $G$ is cyclic, this is trivial, otherwise proceed by induction on $m$.
Step 2. If $G$ is non-abelian of order $p^m$ with $m\geqslant 1$, then $G$ has a normal subgroup of index $p$.
Proceed by induction on $m$ and use the lemma with $H$ being the center of $G$.
Step 3. If $G$ has order $p^m$ with $m\geqslant 1$, then for all $j\in\{0,\cdots,m\}$, there exists $H\leqslant G$ such that $|H|=p^j$.
Step 4. Conclude.
C. Falcon
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For step 1, it is a lot easier to find a normal subgroup of order $p$ than one of index $p$. – Tobias Kildetoft Nov 13 '16 at 20:16