Every group of order $440$ is solvable

Question

Given a group $G$ of order 440, it has a unique subgroup of order 11 which is normal in $G$. Let it be denoted $H$.

$H$ is clearly solvable, if $G/H$ was solvable, so it would be $G$. However I cannot seem to be able to show this last bit. Any idea on how to show that $G/H$ is solvable? Also, any other idea on how to prove that $G$ is solvable would be greatly appreciated.

What did you try? Do you know how to prove that a group is solvable? What ways do you know? — Fareed Abi Farraj, May 04 '19 at 15:30
@FareedAF Basically in my algebra course we have two basic methods so far, either you find a radical tower, or either you find an subgroup H, which is solvable and so that G/H is solvable — miraunpajaro, May 04 '19 at 15:54
@DietrichBurde Thank you very much, that solves my problem :) — miraunpajaro, May 04 '19 at 15:57

Ehsaan · Accepted Answer · 2019-05-05T02:19:59.330

0

$G/H$ is a group of order $40$, so your same Sylow technique shows that it has a unique normal subgroup of order $5$, and once again reduces to a smaller case: now we can assume $|G|=8$. This is a $p$-group for $p=2$, therefore it is nilpotent and solvable.

edited May 05 '19 at 02:19

answered May 04 '19 at 15:34

Ehsaan

3,227

If $H$ and $G/H$ are nilpotent, $G$ need not be nilpotent. in fact, not every group of order $440$ is nilpotent, see here. – Dietrich Burde May 04 '19 at 15:37
The dihedral group $D_{220}$ of order $440$ is not nilpotent, since $220$ is not a power of $2$. – Dietrich Burde May 04 '19 at 15:54
OK, but it still shows OP's group is solvable. – Ehsaan May 05 '19 at 02:19
Yes, this is true. – Dietrich Burde May 05 '19 at 08:16

Every group of order $440$ is solvable

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