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I know that in some version of Sylow's 1st theorem, it states that

if $|G|=p^nm$ for some $n\geq 1$ and where $p$, a prime number, does not divide $m$, then every subgroup $H$ of $G$ of order $p^i$ is a normal subgroup of a subgroup of order $p^{i+1}$ for $1 \leq i <n$.

Now we have the series of subgroups: $$\{e\}=H_0 < H_1 < \cdots <H_r = G.$$ My question comes: why are the quotient groups, $\frac{H_{i+1}}{H_i}$, abelian?

Besides, this is the version of Sylow's 1st theorem that I learnt:

If $G$ is a finite group, $p$ is a prime number and $n \geq 1$ is the largest integer such that $p^n | |G|$, then there exists a subgroup of $G$ of order $p^n$.

How can I deduce the equivalence relation between the above two versions? Is it possible that I don't make use of the first version of 1st theorem to prove the statement?

Thanks in advance.

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  • $\begingroup$ What are the $H_i$, please? Are they $p$-groups? $\endgroup$
    – rogerl
    Nov 8 '14 at 14:11
  • $\begingroup$ Yes, they are $p$-groups, but what does that imply? $\endgroup$
    – Nighty
    Nov 8 '14 at 14:13
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    $\begingroup$ So if the order of $H_i$ is $p^i$, then the order of each quotient is... $\endgroup$
    – rogerl
    Nov 8 '14 at 14:13
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    $\begingroup$ Each quotient should be of order $p$, which means that it is cyclic? $\endgroup$
    – Nighty
    Nov 8 '14 at 14:15
  • $\begingroup$ Yes, exactly right. $\endgroup$
    – rogerl
    Nov 8 '14 at 14:59
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Since each $H_i$ is a $p$-group of order $p^i$, the quotients $H_{i+1}/H_i$ are all groups of order $p$ and thus cyclic, hence abelian.

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  • $\begingroup$ They are normal since (trivially) $p$ is the least prime dividing $|H_{i+1}|$ and $[H_{i+1}:H_i]=p$. $\endgroup$
    – Pedro Tamaroff
    Nov 9 '14 at 0:49
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The theorems you mention are NOT equivalent. In general:

if $H$ is a $p$-group and $K$ is a subgroup of index $p$, then $K$ must be normal in $H$.

Why? Let $H$ work by right multiplication on the right cosets of $K$ in $H$. This action has as kernel $C$=core$_H(K)=\cap_{h \in H}K^h$ and $K/C$ embeds injectively in $S_p$ Note that $C$ is normal in $H$ and that $C \subseteq K \subseteq H$. Since the index index$[H:K]=p$, it follows that index$[K:C] \mid (p-1)!$, which implies $K=C$, since index$[K:C]$ is a $p$-power. Hence $K$ is normal.

Also, it is easy to see that every $p$-group $H$ of order $p^n$ has a subgroup $H_i$ of order $p^i$ ($0 \leq i \leq n)$: (sketch - induction on the order of $H$) $Z(H)$ is non-trivial (well-known fact about $p$-groups), and hence has an element of order $p$, say $z$. Then apply induction on $H/\langle z \rangle$ - this group as a subgroup $K/\langle z \rangle$ of order $p^{i-1}$, and hence $|K|=p^i$.

You are now able to construct a series $H_0=1 \lhd H_1 \lhd H_2 \cdots H_{n-1} \lhd H$, where in each step the index is $p$, whence all the normal subgroup signs. Since $H_{i+1}/H_i \cong C_p$, all the quotients are abelian.

So you do not need Sylow theory after all.

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  • $\begingroup$ (More generally, if $p$ is the least prime dividing $|G|$ and $[G:H]=p$ then $H\lhd G$.) $\endgroup$
    – Pedro Tamaroff
    Nov 9 '14 at 0:50
  • $\begingroup$ Yes of course but the OP title talks about $p$-groups ... $\endgroup$ Nov 9 '14 at 0:52
  • $\begingroup$ I am still confused for the abelian part. The group (quotient group) is cyclic implies it is abelian? Why? $\endgroup$
    – Nighty
    Nov 9 '14 at 2:22
  • $\begingroup$ Yes, every cyclic group is abelian! $\endgroup$ Nov 9 '14 at 9:38

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