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I seek to prove that a group G of order 189 is not simple.

So, for contradiction, I assume G is simple. $|G|=189=3^3 7$. Now, by the Sylow theorems,

$n(7)=1+7k$ divides $3^3=27$. But this is only true when $n(7)=1$, thus G has a normal Sylow 7-subgroup, and so G is not simple.

Is this correct? This is the first time I've encountered this situation where $n(p)=1$ because $n(p)$ couldn't be $m$, which is $3^3=27$ in this case. All other examples I've done, it can usually be $m$ and then I prove it isn't by counting the number of elements.

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    $\begingroup$ This is perfectly correct. $\endgroup$ – Infinity Jan 13 '18 at 23:15

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