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I'm trying to solve the following exercise but I appear to be stuck.

Prove that every finite solvable group $G$ contains a fully invariant abelian $p$-subgroup for some prime $p$.

I know that the next-to-last derived subgroup is abelian and fully invariant plus I know that every finite solvable group has a synthetic series in which every quotient group is of prime rank, but I cannot combine these two to get what I want. Any help?

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All terms in the derived series of a solvable group are fully invariant. In particular, the last nontrivial term, $H$ say, in the derived series is abelian and fully invariant.

Now, if $H$ is finite, then it is easy to see that any Sylow $p$-subgroup of $H$ is fully invariant in $G$, and is an abelian $p$-group.

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  • $\begingroup$ Thank you very much. It didn't occur to me to look for the Sylow groups of the last derived subgroup, but it makes sense now. $\endgroup$ – Amontillado Jan 5 '18 at 12:53
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This solution is for "characteristic" in stead of "fully-invariant", which is slightly different - see the remark by prof D. Holt.
The trick is to look at a minimal normal subgroups of $G$.
If $M$ is minimal normal and non-trivial, then because a subgroup of $G$ it is also solvable and we must have $M' \lt M$. Since $M'$ char $M \lhd G$, we have $M' \lhd G$, and by the minimality of $M$, we get $M'=1$, that is, $M$ is abelian.
Now let $p$ be a prime dividing $|M|$, then $H=\{m \in M: m^p=1 \}$ is a non-trivial (Cauchy!) characteristic subgroup of $M$ (use that $M$ is abelian!). Again we can conclude that $H=M$, so $M$ must be elementary abelian. Finally, the group you are looking for is the product of all such $M$ for a fixed $p$. This subgroup is clearly characteristic.

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    $\begingroup$ But fully invariant is not the same as characteristic. $H$ fully invariant in $G$ means $\phi(H) \le H$ for all homomorphisms $\phi:G \to G$, whereas $H$ characteristic means $\phi(H)=H$ for all automorphisms of $G$. Since all terms in the derived series are fully invaiaint, you could take a Sylow $p$-subgroup of the last nontrivial term in the derived series to answer the question. $\endgroup$ – Derek Holt Jan 5 '18 at 8:46
  • $\begingroup$ Ah Derek, thanks for correcting my omission. Of course you are right. $\endgroup$ – Nicky Hekster Jan 5 '18 at 9:18

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