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I am told some information about a group $G$ of order $168$. All we are told about G is that: 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 $7$. and later it will be proved to be simple. But for now I am asked:

By looking at the possible orders of subgroups $N$ of $G$, show that if $N$ is a non-trivial proper normal subgroup of $G$, then $|N| ∈ \{2, 4\}$ by using the following fact: For $N$, a normal subgroup of index $n$ of $G$, let $a$ be an element of order $m$. Then if gcd$(m,n)=1$ then $a\in N$.

So at the moment we are not using the fact that $G$ is simple, this will be proved later on.

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So I have tried using the fact that $168=|N||G:N|$. I also know that $|G:N|=|G/N|=n$ but don’t know where to go from here.

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Suppose $N$ is a normal subgroup of order $84$. By the fact given, $N$ must contain every element of order $3$ and $7$. But $104 > 84$.

Similarly, if $|N| = 56$ and $N$ is normal (of index $3$), $N$ must contain every element of order $2,4$ and $7$. They don’t fit.

Can you continue?

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