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Show: all $p$-Sylowgroups are normal subgroups $\implies$ group $G$ is solvable.

I know that all subgroups of the different $p$-Sylowgroups are solvable, but do not know if this helps.

Other idea is to show that all $p$-Sylowgroups are solvable and that the factor groups are solvable as well, then it follows, that $G$ is solvable. But I don't know why $p$-Sylowgroups should be solvable..

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    $\begingroup$ Even stronger: the group's nilpotent and then, obviously, solvable. $\endgroup$ – DonAntonio Jun 4 '13 at 15:26
  • $\begingroup$ I believe the question is reduced to this one: math.stackexchange.com/questions/333977/… -- Ben's answer works well with DonAntonio's. $\endgroup$ – Jack Schmidt Jun 4 '13 at 16:45
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Here's a hint:

If two subgroups $H$ and $K$ are normal and intersect precisely in $\{e\}$, then $HK = KH = H \times K$. What is the intersection of a Sylow $p$-subgroup and a Sylow $q$-subgroup for different $p$ and $q$?

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  • $\begingroup$ I think your hint is a bit misleading. It is possible that $H_i \trianglelefteq G$ and $G = H_1 H_2 \ldots H_k$ and $H_i \cap H_j = 1$ for $i \neq j$, yet $G$ is not isomorphic to $H_1 \times H_2 \times \ldots \times H_k$. For more than two normal subgroups $H_i \trianglelefteq G$ with $G = H_1 H_2 \ldots H_k$, we have that $G$ is a direct product of the $H_i$ when $H_i \cap H_1 H_2 \ldots H_{i-1} H_{i+1} \ldots H_k = 1$ for all $i$. $\endgroup$ – Mikko Korhonen Jun 4 '13 at 16:00
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    $\begingroup$ @m.k.: I assume the next step is $H_1$ is a normal $\{p,q\}$-subgroup and $K_1$ a normal $r$-subgroup, so you again get $H_1 \times K_1 = (H \times K) \times K_1$ etc. $\endgroup$ – Jack Schmidt Jun 4 '13 at 16:37
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Try to prove by induction the following easy

Claim: is $\;G\;$ is a finite $\,p$-group, say $\,|G|=p^n\;$ , then

$$\forall 0\le k\le n\;\;\exists\,H_k\lhd G\;\;s.t.\,\,|H_k|=p^k$$

With the above you're done since then $\,1\lhd H_1\lhd\ldots\lhd H_n=G\;$ is an abelian series for $\,G\,$ ...

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