I am studying some exercises about semi-direct product and facing this solved one:

Show that the order of group $G=\langle a,b| a^6=1,a^3=b^2,aba=b\rangle$ is $12$.

Our aim is to show that $|G|\leq 12$ and then $G=\mathbb Z_3 \rtimes\mathbb Z_4=\langle x\rangle\rtimes\langle y\rangle$. So we need a homomorphism from $\mathbb Z_4$ to $\mathrm{Aut}(\mathbb Z_3)\cong\mathbb Z_2=\langle t\rangle$ to construct the semi-direct product as we wish: $$\phi:=\begin{cases} 1\longrightarrow \mathrm{id},& \\\\y^2\longrightarrow \mathrm{id},& \\\\y\longrightarrow t,& \\\\y^3\longrightarrow t, \end{cases} $$ Here, I do know how to construct $\mathbb Z_3 \rtimes_{\phi}\mathbb Z_4$ by using $\phi$ and according to the definition. My question starts from this point: The solution suddenly goes another way instead of doing $(a,b)(a',b')=(a\phi_b(a'),bb')$. It takes $$\alpha=(x,y^2), \beta=(1,y)$$ and note that these elements satisfy the relations in group $G$. All is right and understandable but how could I find such these later element $\alpha, \beta$?? Is really the formal way for this kind problems finding some generators like $\alpha, \beta$? Thanks for your help.

  • $\begingroup$ Once you have exhibited elements (such as $\alpha$ and $\beta$ here) that satisfy the given relations, then you are guaranteed the existence of a homomorphism $f$ from $G$ to $\mathbb{Z}_3\rtimes\mathbb{Z}_4$ such that $f(a)=\alpha$, $f(b)=\beta$. As $\alpha,\beta$ generate the semidirect product, you can then deduce that $|G|\ge 12$, and that $G$ has that semidirect product as a quotient group. Combining this with the other fact that $|G|\le12$ then completes the proof. $\endgroup$ – Jyrki Lahtonen Sep 12 '12 at 12:15
  • $\begingroup$ @JyrkiLahtonen: Sorry for asking this but finding $\alpha$ and $\beta$ is really itself a new small problem. Is there any way for guessing them ? At least in this group? $\endgroup$ – mrs Sep 12 '12 at 12:25

The subgroup $A$ generated by $a^2$ is normal and order 3. The subgroup $B$ generated by $b$ is of order 4. The intersection of these is trivial so the product $AB$ has order 12. So $G$ has order at least 12. To show it has order 12, we need to see that $a\in AB$; but $b^2=a^3=a^2a$ so $$a=a^{-2}b^2\in AB.$$ Thus the group is the semidirect product of $A$ by $B$ where $$ba^2b^{-1}=(ba)(ab^{-1})=a^{-1}(ba)b^{-1}=a^{-1}(a^{-1}b)b^{-1}=a^{-2}$$

  • $\begingroup$ May I ask you why we need to have $a$ in $AB$? $\endgroup$ – mrs Sep 12 '12 at 15:55
  • $\begingroup$ @BabakSorouh We want $G=AB$, not just $AB\subset G$. Certainly $b\in B\subset AB$, so we want to also show $a\in AB$ $\endgroup$ – i. m. soloveichik Sep 12 '12 at 16:28
  • $\begingroup$ Thanks for what you noted here for my problem. I learnt some good ideas about your approach. Thank you. But I'd like to know how the generators $\alpha, \beta$ are chosen among $11$ elements of $G$. Indeed, some elements couldn't generate the whole group and they couldn't satisfy the homomorphism which Jyrki noted. Do you think we should search among the elements of $\mathbb Z_3\rtimes\mathbb Z_4$ to find the best selection for $\alpha, \beta$ ? $\endgroup$ – mrs Sep 12 '12 at 18:42
  • $\begingroup$ Let $x=a^2$, $y=b$ as in the statement of the problem. Then $\alpha=(x,y^2)=a^2b^2$ and $\beta=(1,y)=b$ $\endgroup$ – i. m. soloveichik Sep 12 '12 at 18:49

$aba=b$ implies that $bab^{-1}=a^{-1}$, hence $bab^{-1}a^{-1}=a^{-2}$; it follows that $G'\supseteq \langle a^{-2}\rangle\cong \mathbb{Z}_3$. Also, $b^{-1}a^2b=(b^{-1}ab)^2 = (b^{-1}aba.a^{-1})^2=(b^{-1}.ba^{-1})^2=a^{-2}\in \langle a^2\rangle$. Hence $\langle a^2\rangle$ is normalised by both $b$, and $a$; $\langle a^2\rangle\triangleleft G$. Then $G/\langle a^2\rangle=\langle \bar{a},\bar{b}\colon \bar{a}^2=1, \bar{b}^2=\bar{a}, \bar{b}\bar{a}\bar{b}^{-1}\bar{a}^{-1}=1\rangle = \langle \bar{b}\colon \bar{b}^4=1\rangle \cong \mathbb{Z}_4$.

Hence $G'=\langle a^2\rangle\cong \mathbb{Z}_3$, and $G/G'\cong \mathbb{Z}_4$.

Further $aba=b$ implies $abab=b^2$, i.e. $(ab)^2=b^2=a^3$. Therefore, $(ab)^4=(a^3)^2=1$, and $(ab)^2\neq 1$; hence $\langle ab\rangle$ is cyclic subgroup of $G$ of order $4$, and clearly, it should intersect trivially with $G'$ (compare orders). Hence $G=G'\rtimes \langle ab\rangle \cong \mathbb{Z}_3\rtimes \mathbb{Z}_4$.


To construct semidirect product of $\mathbb{Z}_3=\langle x\rangle$ by $\mathbb{Z}_4=\langle y\rangle$, note that $y$ acts on $\langle x\rangle$ by conjugation, and hence $y^{-1}xy\in \{x,x^2\}$. If $y^{-1}xy=x$, then $G$ will be abelian, and if $y^{-1}xy=x^2$, then $G$ will be the non-abelian group of order 12 (it is different from $D_{12}$, and $A_4$; why?).


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