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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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9 divides the order of G – tom Dec 13 '12 at 5:41
Thanks. Where is the problem from? – Jonas Meyer Dec 13 '12 at 5:44
from textbook :basic algebra – tom Dec 13 '12 at 6:02
$G$ is a group of order 27, but it is not necessarily abelian. The three abelian groups of order 27 are straightforward, and according to GP the two nonabelian groups of order 27 are the semidirect product of Z9 by Z3 and the unitriangular matrix group U(3,3), which surely require some sophistication to understand. I think only the former though fits your situation, which is much more palatable. – anon Dec 13 '12 at 6:35
(Um, the previous comment isn't even an attempted proof...) Indeed, your group must be a semidirect product of $H$, a group of order 9, with $Z_3$ the cyclic group of order 3. $H$ can be either $Z_9$ or $Z_3\oplus Z_3$. $Z_3$ has one nontrivial homomorphic image (an embedding) inside $\mathrm{Aut}(Z_9)=U(9)\cong Z_6$ which corresponds to the one nonabelian group $G$, and $Z_3$ has no nontrivial image in $\mathrm{Aut}(Z_3\oplus Z_3)\cong U(3)\oplus U(3)\cong Z_2\oplus Z_2$, so there are no other nonabelian such $G$, whereas the the two (cyclic of order 27 doesn't count) abelian $G$'s are clear. – anon Dec 13 '12 at 6:41

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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You probably mean $H=C_9$ in case (1). – peoplepower Dec 13 '12 at 13:12
Oops! Of course, @peoplepower: thanks. – DonAntonio Dec 13 '12 at 13:16
Your semidirect product sign seems needs reflecting, since $H$ is the normal subgroup. – Derek Holt Dec 13 '12 at 17:16
@DerekHolt, of course. Thanks. – DonAntonio Dec 13 '12 at 17:39

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