Prove or disapprove: Group of order $135$ must be abelian

Question

The question is as follows: prove or disapprove: every group of order $135$ must be abelian.

I started like this: $G = H \times\ K$ when $H$ is a normal 5-sylow subgroup and $K$ is a normal 3-sylow subgroup (both normal from sylow theory).

$H$ is cyclic and therefor abelian, but what about $K$? if it's cyclic then it proves the statement. if not, I'm not sure how to continue...

I'm not sure how to continue from here.

So the question reduces to: does there exist a nonabelian group of order $27$? — Derek Holt, Sep 20 '17 at 11:45
See https://groupprops.subwiki.org/wiki/Groups_of_order_27, https://math.stackexchange.com/questions/279076/classify-groups-of-order-27, https://math.stackexchange.com/questions/135072/prove-or-disprove-a-group-with-order-p3-is-abelian-if-its-has-a-normal-subgro — lhf, Sep 20 '17 at 11:47
From the linked answers: take $G = \Bbb Z_5 \times H_3(\Bbb Z_3)$, where $H_3(\Bbb Z_3)$ is the Heisenberg group modulo 3. — Ben Grossmann, Sep 20 '17 at 11:55
The title should be "prove or disprove". Or may be "Approve or disapprove". — Dietrich Burde, Sep 20 '17 at 12:22

score 3 · Accepted Answer · answered Sep 20 '17 at 12:04

3

It is enough to exhibit a nonabelian group of order $135$ of the form $C_5 \times K$, where $K$ is a nonabelian group of order $27$.

The set of all matrices of the form $$\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}$$ with entries in $\mathbb Z_3$ is a nonabelian multiplicative group of order $27$, called the Heisenberg group over $\Bbb{Z}_3$.

answered Sep 20 '17 at 12:04

lhf

216,483

Adapted from https://math.stackexchange.com/a/2192101/589. – lhf Sep 20 '17 at 12:04

Prove or disapprove: Group of order $135$ must be abelian

1 Answers1