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Show that there does not exist a simple group of order $126$.

By the Sylow's theorem, I already know that $N_7 | 24$ and $N_7 \equiv 1 \pmod 7$; I found that $N_7 \in \{1,8\}$

I think I can use the generalized Cayley's theorem and the theorem which states:

If $H < G$ such that his index $p := (G : H)$ is the least divisor of O(G), then $H \unlhd G$.

I am a bit locked onto the problem now. Does someone could detail me how to resolve this problem? Otherwise is it possible to give me some steps to understanding and solving this problem?

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  • $\begingroup$ I made a mistake in the factors, the problem becomes trivial in this case. $\endgroup$ – user230283 Dec 15 '15 at 0:26
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$126 = 2\cdot 3^2\cdot 7$, so $n_{7}\equiv 1\bmod 7$ and $n_{7}|126$ (both of which are given by Sylow's theorem) implies $n_{7} = 1$. $n_{7} = 8$ is not a possibility since it doesn't divide the group order. Thus there is only one Sylow 7 subgroup, and so it is normal.

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Since 7 divides 126, Sylow shows that there exists a subgroup H of order 7, Sylow again says that the number of group of order 7 $n_7=1\mod 7$ and divides 18 since 126 is the product of 7 and 18, so $n_7=1$ and $H$ is normal. done.

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  • $\begingroup$ The group cardinality is $126$, not $42$. $\endgroup$ – user230283 Dec 15 '15 at 0:21
  • $\begingroup$ I answered when it was 42 in the question $\endgroup$ – Tsemo Aristide Dec 15 '15 at 0:22

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