# Calculate $G/Z(G)$ where $G=(\mathbb Z_3 \oplus \mathbb Z_3)\rtimes_{\phi} \mathbb Z_2$

Problem

Let $f \in \operatorname{Aut}(\mathbb Z_3 \oplus \mathbb Z_3)$ be defined as $f(x,y)=(2x,x+y)$ and let $\phi:\mathbb Z_2 \to \mathbb Z_3 \oplus \mathbb Z_3$ be the morphism given by $\phi(0)=Id$, $\phi(1)=f$. Let $G=(\mathbb Z_3 \oplus \mathbb Z_3)\rtimes_{\phi} \mathbb Z_2$ Characterize $G/Z(G)$.

I know that $G/Z(G) \cong \operatorname{Int}(G)$, where $\operatorname{Int}(G)=\{I_g:G \to G, I_g(x)=gxg^{-1}\}$, but I am not so sure if this is of any help. There are $18$ elements in this group so I wouldn't want to calculate by hand how the product given by $\phi$ works between any two elements of the group, I would appreciate any hints or suggestions to work on a solution.

• Is that $\;\cdot_\phi\;$ thing supposed to be a semidirect product? – Timbuc Dec 1 '14 at 4:43
• @Timbuc, yes it is a semidirect product given by the morphism $\phi$. – user156441 Dec 1 '14 at 4:45
• You may then want to use the usual, international notation for semidirect product in order to avoid confusion. Do it in LaTeX with \rtimes: $$(\Bbb Z_3\oplus\Bbb Z_2)\rtimes_\phi\Bbb Z_2$$ – Timbuc Dec 1 '14 at 4:48
• First, that your group isn't abelian. Second, that any non-abelian group of order $\;18\;$ is not nilpotent. Third, that since $\;G/Z(G)\;$ cannot be cyclic non-trivial ever, the only options left are $\;|Z(G)|=1\,,\,2\,,\,3\;$...but $\;|Z(G)|=2\;$ cannot be as any group of order nine is abelian and this would contradict "second" above... – Timbuc Dec 1 '14 at 5:34
• Note that $f(0,y) = (0,y)$. – Derek Holt Dec 1 '14 at 7:33

If we take $\;((a,b)\,,\,r)\;,\;\;((c,d)\,,\,s)\in G\;$ , then

$$((a,b)\,,\,r)\cdot((c,d)\,,\,s):=\left((a,b)+(c,d)^r\,,\,r+s\right)=$$

$$\begin{cases}\left((a,b)+(c,d)\,,\,s\right)=\left((a+c,b+d)\,,\,s\right)&,\;\;r=0\\{}\\\left((a,b)+(2c,c+d)\,,\,1+s\right)=\left((a+2c,b+c+d)\,,\,1+s\right)&,\;\;r=1\end{cases}$$

The excellent, as usual, comment by Derek helps you now to check that: $${}$$ \begin{align*}&\bullet\;\;((a,b),r)\cdot((0,y),0)=((a,b+y), r)\\{}\\ &\bullet\;\;((0,y),0)\cdot((a,b),r)=((a,b+y),r)\end{align*} $${}$$ and the above means $\;Z(G)=\{\;((0,0),0)\,,\;((0,1),0)\,,\;((0,2),0)\;\}\cong\Bbb Z_3\;$ , what gives you $\;\left|G/Z(G)\right|=6\;$

Now, unless you do the calculations by hand, I can't see how to prove that in fact $\;G/Z(G)\cong S_3\;$ since the other possibility is the cyclic group, which would render $\;G\;$ nilpotent.

Another hint that could possible be helpful (but I guess would also require some calculations, too): prove that in fact $\;G\cong S_3\times\Bbb Z_3\;$ , and now directly you can show that

$$Z(G)=Z(S_3)\times Z(\Bbb Z_3)=1\times\Bbb Z_3\cong\Bbb Z_3\;$$

and etc.