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Show that every group of order $4125=3\cdot 5^3\cdot 11$ is solvable.

Proof: Suppose $G$ is a group of order $4125$. By Sylow's Theorems,

  • $n_3 \equiv 1 \mod 3$ and $n_3 | 5^3\cdot 11$ $\Rightarrow$ $n_3 = 1,25$.
  • $n_5 \equiv 1 \mod 5$ and $n_5 | 33$ $\Rightarrow$ $n_5 = 1,11$.
  • $n_{11}\equiv 1 \mod 11$ and $n_{11} | 5^3\cdot 3$ $\Rightarrow$ $n_{11} = 1,375$.

Sadly none of my $n_p$ are only $1$. How would I go about this problem?

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  • $\begingroup$ $n_5=11$ will force there to be a normal subgroup whose order is divisible by $5$ (a closer look suggests $15$) and index divisible by $11$. From there getting solubility should b easy. $\endgroup$ Commented Mar 12, 2021 at 16:18
  • $\begingroup$ There are several solutions in the web, e.g., see here. Also compare with this post. $\endgroup$ Commented Mar 12, 2021 at 16:20

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If you’re looking for a quick answer, every group of odd order is solvable.

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