TITLE: proving a group is soluble (solvable if you are american)
QUESTION [5 upvotes]: (7 marks) (a) Show that every group $G$ of order 56 has a normal Sylow subgroup. (Hint: If $n_7 \neq 1$ then show that $n_7 = 8$ deduce from this that $G$ has 48 elements of order 7, leaving just enough elements for one Sylow 2-subgroup). 

If $n_7=1$, the the Sylow 7-subgroup of $G$ is normal. Suppose $n_7=8$. The eight distinct Sylow 7-subgroups are cyclic so that $G$ has at least $6\cdot8 = 48$ elements of order 7. Now $n_2$ divides 7 and is odd, so that $n_2\in \{1,7\}$. If $n_2=7$, then let $P_1$ and $P_2$ be distinct Sylow 2-subgroups of $G$. There must exist a non-identity element $x\in P_1$ and $x \notin P_1$, but then $G$ has at least $48+8+1=57$ elements, which is nonsense. Thus $n_2=1$, so that the unique Sylow 2-subgroup of G is normal.  
b) Explain why (a) means that every group of order 56 is soluble.

REPLY [5 votes]: The theorem you need is

Every finite $p$-group is solvable.  

and 

If $N\lhd G$, $N$ and $G/N$ are solvable, then $N$ is solvable. 

So from part (a), you have proven that $n_7=1$ or $n_2=1$.
Let $P\in Syl_7(G)$.
Then $P\lhd G$. Also note that $P$ is $7$-group and $G/P$ is $2$-group.
Hence $P$ and $G/P$ are solvable, so is $G$.
If you let $Q\in Syl_2(G)$, the same argument follows as well.