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

Question

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?

Answer 1

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