Prove that a group of order $56$ is not simple.
For this question my strategy was to look at the prime factorization of the order $$56 = 2^3 \cdot 7$$Notice then that $ 2 \nmid 7$ so we could use the $1\text{st}$ Sylow Theorem to determine there exists a subgroup of order $8 = 2^3$, call this subgroup $H$. This is necessarily a $2$-subgroup since $2 \mid 8$. Furthermore, this is a Sylow $2$-subgroup since $\lvert H \rvert = 8 = 2^3$ is the maximal power that divides $\lvert G \rvert = 56 = 2^3 \cdot 7$. Perfect, we can now apply the $3\text{rd}$ Sylow Theorem and we notice then that $$n_2(G) \mid \frac{\lvert G \rvert}{2^3} = \frac{2^3 \cdot 7}{2^3} = 7,\ \text{hence},\ n_2(G) \mid 7$$ Also that $$n_2(G) \equiv 1 \pmod{2}$$ Great, then since $7$ is prime then $n_2(G)$ can either be $1$ or $7$. But we notice that $$7 \equiv 1 \pmod{2}$$ Which means I am unable to conclude that $H$ is normal by the proposition that states that any Sylow $p$-subgroup is normal if and only if $n_p(G) = 1$. I tried this tactic again with $7$ but found that the subgroup of order $7$, call it $K$, fell into the same trap. Where, $n_7(G)$ can be either $1,2,4, \text{or}\ 8$ but we notice that $$8 \equiv 1 \pmod{7}$$ Hence I am unable to conclude that $K$ is normal in $G$.
I emailed my professor asking for a tip on how to proceed here and he suggested that "It's correct to want to show that G contains either a normal sylow $7$-subgroup or a normal sylow $2$-subgroup. Do this by assuming otherwise and counting elements. Be careful with the intersection when doing your count, though."
I need a little help parsing this tip. By "assume otherwise" I suppose he means 'assume that $G$ does not contain a normal sylow $2$ or $7$-subgroup'?
Additionally, I'm not sure exactly what is meant by counting elements and how that can prove normality of $H$ or $K$. Initially I thought we could use Lagranges Theorem to calculate $$[G : H] = \frac{\lvert G \rvert}{\lvert H \rvert}$$ And show that $[G : H] = 2$ then use the fact that any subgroup with index $2$ is normal to prove normality but that clearly doesn't work, neither does it work for $K$. So, any guidance in how to proceed is greatly appreciated.
Best Answer
Good idea, but you were looking at the wrong Sylow.
Consider the Sylow $7$-subgroups. By Sylow's Third Theorem, there is either $1$ or $8$ of them. If there is one, then it is normal and you are done.
If there are $8$ of them, then note that any two must intersect trivially; this accounts for $8\times 6 = 48$ nontrivial elements. This leaves only $8$ elements that are not of order $7$; that means that there can only be one Sylow $2$-subgroup (there aren't enough elements left for there to be more), and so the Sylow $2$-subgroup is normal.