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I want to show that a group of order $42 = 3\times2\times7$ has a normal subgroup, and I'm not completely sure.

I am thinking of injecting into $S_5$, and noting that the subgroup of $G$ of order $7$ cannot be in the image of $\phi$ since that would imply $S_5$ has a subgroup of order 7 contradicting Lagrange's theorem. Thus $Ker(\phi)$ which is a subgroup of G is normal. Is this okay?

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  • $\begingroup$ how are you injecting into $S_5$? Please see below for the solution I believe is intended. $\endgroup$ – qbert Apr 22 '18 at 5:11
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The Sylow theorems do it all for you.

By Sylow 1, the number of Sylow $7$ subgroups $n_7$ satisfies $n_7\cong 1 \mod 7$. It also must divide $6$, leaving you with only the option $n_7=1$. Call it $P_7$.

By Sylow 2, $gP_7g^{-1}=P_7$ for any $g$, and $P_7$ is normal.

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