Every group of order $56$ has a proper non - trivial normal subgroup? I tried getting $n_{2}$ and $n_{7}$ , denote the number of Sylow-2-Subgroups and Sylow-7-Subgroups respectively.
I got two cases for $n_{2} = 1 , 7 $ and for  $n_{7} = 1 , 8$ , i noticed that if $n_{2} = 1$ and $n_{7}= 1$ ,then we are done since they are unique subgroups and hence will be normal in $G$.
Next how to proceed with the cases of $n_{2} = 7$ and $n_7 = 8$.
Please help?
 A: We have $o(G)=56=2^3.7$
Possible no. of sylow 2 subgroup $n_2=1+2k|7$ i.e either $1$ or $7$.
Similarly possible no. of sylow $7$ subgroup $n_7=1+7k|8$ i.e either $1$ or $8$.If there are $1$ sylow $7$ subgroup,then we are done because unique sylow-$p$ subgroups are normal.Suppose there are $8$ distinct sylow-$7$ subgroup say, $p_1,p_2\cdots p_8$.
Now $$|p_i\cap p_j||_{|p_i|or|p_j|}$$
$$|p_i\cap p_j|=1;i\neq j$$
$$|p_1\cup p_2\cdots\cup p_8|=56-7=49$$ i.e there are $48 $ elements of order $7$.We know that order of sylow- $2$ subgroup of $G$ is $8$.If there are $8$ sylow-$7$ subgroup then, there can only $1$ sylow-$2$ subgroup is possible which will be normal.
A: Count the total number of elements. Suppose neither $n_2$ nor $n_7$ is equal to $1$. Then $n_7=8$, which makes $8\cdot 6$ elements which belong to a $7$-Sylow subgroup, and to no other Sylow subgroup.
On the other hand, since $n_2=7$, each $2$-Sylow subgroup contains at least $4$ elements which belong to no other Sylow subgroup. In all $7\cdot 4$ elements.
Summing up, the group would contain at least $48+28+1$ elements. This is more than $56$, if I'm not mistaken.
