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I want to prove that there is no simple group of order $1452$. We have $1452 = 2^2\cdot 3\cdot 11^2$, and the Sylow theorems give:

\begin{align} n_2 &\in \{1,3,11,33,121,363\} \\ n_3 &\in \{1,4,22,121,484\} \\ n_{11} & \in \{ 1,12 \} \end{align}

If any of the $n_{p}$ if $1$, done. If $n_2 = 3$ or $n_3 = 4$, action by conjugation in the collection of $2$-Sylows or $3-$Sylows of $G$ induces a morphism of $G$ to a group of order $3!$ or $4!$, respectively, which can't be injective since $1452 \not\mid 6$ and $1452 \not\mid 24$, so the kernel of said morphism is a non-trivial subgroup of $G$, done.

Then we can suppose from now on that $n_{11}=12$, $n_{2}\geq 11$ and $n_3 \geq 22$. I'm stuck now. Help?

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    $\begingroup$ Actually $1452$ does not divide $12!$ either if your prime decomposition is correct. $\endgroup$ – Captain Lama Apr 10 '16 at 20:52
  • $\begingroup$ The decomposition is correct. And you're right, thanks! Then the same argument I gave for $n_2=3$ and $n_3 = 4$ solves the problem. $\endgroup$ – Ivo Terek Apr 10 '16 at 20:56
  • $\begingroup$ @Captain If you want to convert the comment into an answer I'll accept it :P $\endgroup$ – Ivo Terek Apr 10 '16 at 20:57
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Following your reasoning, since $11$ appears twice in $1452$ and only once in $12!$, $1452$ cannot divide $12!$ so you can conclude with $n_{11}=1$.

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