I am told some information about a group $G$ of order $168$. All we are told about $G$ is: It has one element of order one, $21$ elements of order $2$, $56$ elements of order $3$, $42$ elements of order $4$ and $48$ elements of order $7$.

A previous question, see here found that: For $N$, a proper non trivial normal subgroup of $G$, then the only possible values for $|N|$ are $|N|∈\{2,4\}$

Now using the fact $|N|∈\{2,4\}$, prove $G$ is simple. I know this would mean showing that $|N|$ isn't actually either $2$ or $4$, since a simple group has no non trivial proper normal subgroups. I've tried doing it with Sylow theorems but got stuck, is there a better route to go down?


Assume $|N|=2$. As $N$ has only the identity automorphism, $N$ commutes with all of $G$. In particular, if $a$ is the nontrvial element of $N$ amd $b\in G$ is of order $7$, then $ab$ is of order $14$, which is absurd.

Assume $|N|=4$. Consider $2$-Sylow groups: They have order $8$, their number $n_2$ divides $21$, their union consists of the $42+21+1=64$ elements of orders $1,2,4$. Then the intersection of any two distinct 2-Sylows is precisely $N$ (it must be a proper subgroup of each Sylow, and it must contain $N$). We conclude that $4+n_2\cdot 4= 64$, i.e., $n_2=15$, contradicting $n_2\mid 21$. This excludes the possibility $|N|=4$.

  • $\begingroup$ I can't quite make sense of this line: $4+n_2\cdot 4= 64$ So we have the intersection + $n_2$ $\cdot$ the intersection? Why can we write it like this? $\endgroup$
    – harry55
    Apr 6 '17 at 20:29

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