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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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What have you tried? –  Jonas Meyer Dec 13 '12 at 5:22
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
(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

1 Answer 1

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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