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Let M be minimal normal in finite solvable group G, and assume that M=C_G (M). Show that G splits over M and that all complements for M in G are conjugate.

Hint. Let L/M be minimal normal in G/M, and note that L/M is a q-group for some prime q. Show that q does not divide |M|, and consider a Sylow q subgroup of L.

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  • $\begingroup$ what have you done? $\endgroup$ – user87543 Dec 25 '13 at 6:17
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In brief: if $q$ divides $|M|$, then $L$ is a $q$-group, so $M \cap Z(L) \ne 1$, hence by minimality of $M$ $M \le Z(L)$ contradicting $M=C_G(M)$. Let $P$ be a Sylow $q$-subgroup of $L$, $N = N_G(P)$. Show $N \cap M = 1$ (otherwise we get $M \cap Z(L) \ne 1$ again), hence $N$ is the required complement, and all complements are normalizers of Sylow $q$-subgroups of $L$, so they are all conjugate in $G$.

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