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Find $3$ non-isomorphic groups of order $2012$.

Is the following correct?

First of all, we have the two non-isomorphic abelian groups $\mathbb Z_{2012}$ and $\mathbb Z_{2}\times\mathbb Z_{1006}$.

Then I want to construct a non-abelian group from $\mathbb Z_4$ and $\mathbb Z_{503}$. We have $\text{ Aut }(\mathbb Z_{503})\cong\mathbb Z_{503}^\times\cong\mathbb Z_{502}$. Let $a$ be an element of order $2$ in that group (exists because of Cauchy's theorem), then $\varphi:\mathbb Z_4\to\text{ Aut }(\mathbb Z_{503}), \bar 1\mapsto a$ is a non-trivial group homomorphism and thus $\mathbb Z_4\rtimes_\varphi\mathbb Z_{503}$ is non-abelian.

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You could do it even easier. $2012$ is an even number so there is also the dihedral group $D_{1006}$.

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    $\begingroup$ Yes, but that wasn't my question. $\endgroup$
    – RedLantern
    Commented Feb 15, 2019 at 11:47
  • $\begingroup$ What you did looks fine. A semidirect product can't be abelian unless it comes from a trivial homomorphism. (which means it is actually a direct product). I just wanted to point on the fact that it is actually very easy to find a non abelian group of even order (if the order is at least $6$) without using semidirect products. $\endgroup$
    – Mark
    Commented Feb 15, 2019 at 11:56
  • $\begingroup$ Each dihedral group is a semidirect product. $\endgroup$
    – Shaun ♦
    Commented Feb 15, 2019 at 16:25
  • $\begingroup$ @Shaun, yes, but you don't really need to know about semidirect products to understand these groups. I knew about dihedral groups long time before I learned about semidirect products. $\endgroup$
    – Mark
    Commented Feb 15, 2019 at 16:27

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