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**Swlabr** Not quite...you are taking a sylow subgroup of M, not of G. There is no reason to say these coincide.

Let $\displaystyle p | |M|$. As $\displaystyle M$ is abelian, the sylow $\displaystyle p$-subgroups of $\displaystyle M$ are normal and unique, so let $\displaystyle P$ be the maximal subgroup of $\displaystyle M$ of order a power of $\displaystyle p$. Clearly, if $\displaystyle |M| = p^nq$, $\displaystyle p \not\div q$ then $\displaystyle |P| = p^n$.

Now, we know that $\displaystyle P \lhd M$, but we wish to prove that $\displaystyle P \lhd G$. This is true because if $\displaystyle \phi: M \rightarrow M$ is an automorphism, then $\displaystyle P\phi = P$ because every element has order a power of $\displaystyle p$ and no other element in the group has this order, and automorphisms preserve order (does that make sense)? Therefore, $\displaystyle P \text{ char } M$.

Next, notice that for $\displaystyle g \in G$, $\displaystyle ^g:M \rightarrow M$, $\displaystyle h \mapsto h^g$ is an automorphism of $\displaystyle M$. This is bacause $\displaystyle M$ is normal in $\displaystyle G$. As this is an automorphism of $\displaystyle M$ it must fix $\displaystyle P$. Therefore, $\displaystyle P^g = P$ for all $\displaystyle g \in G$, and we thus conclude that $\displaystyle P \lhd G$, as required to get our contradiction.

Note that $\displaystyle K \lhd H \lhd G \not\Rightarrow K \lhd G$.