TITLE: There is no simple group of order 56
QUESTION [7 upvotes]: I'm working on question 16 of http://www.austinmohr.com/Work_files/hw4.pdf
I'm confused by why there are 48 elements of order 7. I understand that $n_7 = 8$, but no further.

REPLY [2 votes]: We have that $56=2^3\cdot 7$. Then the number of Sylow $7$-subgroups is $1+7k$ , for some $k \geq 0$ and divides $8$. So the number of Sylow $7$-subgroups is either $1$ or $8$. If it is $1$, we are done. So let us assume that it is $8$. Each Sylow $7$-subgroup has order $7$. In a group of prime order every nonidentity element has order $7$. Thus each Sylow $7$-subgroups has $6$ elements of order $7$. Also, they intersec trivially. It follows that there are $8\cdot 6 =48$ elements of order $7$. This leaves only $56-48=8$ elements, wich is the size of one Sylow $2$-subgroup. Thus, there is only room for one o such subgroups. That is, either the Sylow $7$-subgroup or the Sylow $2$-subgroup  must be normal.