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 \(\displaystyle K \lhd H \lhd G \not\Rightarrow K \lhd G\).