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Assume that:

$G$ contains a normal subgroup $H$ of order $9$, and $G$ is generated by $H$ and an element $x\in G-H$ of order $3$.

How to classify all such groups $G$?

I think $9$ divides the order of $G$, and G is isomorphic to an abelian group of order 27.

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    See http://www.math.columbia.edu/~bayer/S09/ModernAlgebra/semidirect.pdf2012-12-13

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First, it must be clear that in any group of order $\,27\,$ , a subgroup of order $\,9\,$ is normal (why? what is this subgroup's index?)

Second, as was already pointed out above, you already know that

$G=C_3\ltimes H\,\,\,,\,\,C_3:=\langle x\;\;;\;\;x^3=1\rangle$

We now have two possible cases:

$(1)\;\;H= C_9\;\;\Longrightarrow\,\operatorname{Aut}H\cong C_6\;\;(\text{Hint:}\,\,\phi(3^2)=3\cdot 2=6)$

The only non-trivial homomorphisms $\,C_3\to C_6=\langle y\rangle\,$ are $\,x\to y^2\,\,,\,\,x\to y^4\,$ , which give us non-trivial groups of order $\,27\,$ . These two though are isomorphic, as you can read in page $\,49-50\,$ in the PDF here

$(2)\;\;H=C_3\times C_3\;\;\Longrightarrow\,\operatorname{Aut}(H)\cong\,C_2\times C_2$

The only homomorphism possible here is the trivial one (why?), giving us a direct product and thus an abelian group.

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    @DerekHolt, of course. Thanks.2012-12-13