Show that $D_3\times_\rho\mathbb{Z}_2$ is not isomorphic to $A_4$

This is the third part of a problem. I will list here the first two parts as a reference and then my attempt to solve the third one. I want to verify if the first and second are right and some help for finishing the third part.

(1) Find Aut$(D_3)$

Solution: Let's characterize $D_3$ as the group generated by $r,t$ with $$r^3 = t^2 = (rt)^2 = e$$ $\hspace{10pt}$Suppose that $\phi:D_3 \to D_3$ is an automorphism. Then, $\phi(t)$ must be of order two $\Rightarrow$ $\phi(t) \in \{t,rt, r^2 t\}$. Similarly we can see that $\phi(r)$ must be of order three $\Rightarrow$ $\phi(r) \in \{r,r^2\}$. Since these two choices determine $\phi$ (and any two are compatible), we have $2 \times 3 = 6$ possible choices for $\phi$.
Hence, $$\mbox{Aut}(D_3) = \{\phi_0, \phi_1, \phi_2,\phi_3,\phi_4,\phi_5\}$$ such that $\phi_i :D_3 \to D_3$, $i = 0,...,5$ are the following isomorphisms $$\phi_0(t) = t, \mbox{ } \phi_0(r) = r$$ $$\phi_1(t) = t, \mbox{ } \phi_1(r) = r^2$$ $$\phi_2(t) = rt, \mbox{ } \phi_2(r) = r$$ $$\phi_3(t) = rt, \mbox{ } \phi_3(r) = r^2$$ $$\phi_4(t) = r^2t, \mbox{ } \phi_4(r) = r$$ $$\phi_5(t) = r^2t, \mbox{ } \phi_5(r) = r^2$$ (2) Find an appropriate homomorphism $\rho: \mathbb{Z}_2 \to$ Aut$(D_3)$ to construct $G = D_3\times_{\rho}\mathbb{Z}_2$.

$\textbf{Solution:}$ We must exhibit an homomorphism that takes $\mathbb{Z}_2$ to an automorphism (isomorphic function from $D_3$ to $D_3$). Then $(D_3\times_{\rho}\mathbb{Z}_2,\cdot)$ is a group when the binary operation $\cdot$ is defined as $$(d_i,z_j)\cdot(d_k,z_l) = (d_i\rho(z_j)d_k,z_jz_l)$$ Consider $\rho(z) = \begin{cases} \phi_{0} & \mbox{ if } z = 0 \\ \phi_{1} & \mbox{ if } z = 1 \end{cases}$

$\hspace{10pt}$ By part $(a)$, is clear that $\rho$ sends each element in $\mathbb{Z}_2$ to an element (function) in Aut$(G)$. The function $\rho$ is an homomorphism since for any $z_1,z_2 \in \mathbb{Z}_2$ $$\rho(z_1 + z_2) = \rho(z_1)\circ\rho(z_2)$$

Case 1: $z_1 = 0, z_2 = 0$ $$\rho(0+0) = \rho(0) = \rho(0)\circ \rho(0) = \phi_0\circ\phi_0 = \phi_0$$

Case 2: $z_1 = 0, z_2 = 1$
$$\rho(0+1) = \rho(1) = \phi_0\circ\phi_1 = \phi_1$$ Case 3: $z_1 = 1, z_2 = 1$ $$\rho(1+1) = \rho(0) = \phi_1\circ\phi_1 = \phi_0$$ $$\therefore \rho \mbox{ is an homomorphism }$$ Hence, $(G,\cdot) = (D_3\times_{\rho}\mathbb{Z}_2,\cdot)$ is a group called the external semi-direct product of $D_3$ and $\mathbb{Z}_2$ by $\rho$.

(3) Show that $G = D_3\times_{\rho}\mathbb{Z}_2$ cannot be isomorphic to $A_4$ (Alternating group of 4 symbols).

$\textbf{Solution:}$ $A_4 < S_4$ is the subgroup with $|A_4| = 12 = \frac{4!}{2} = \frac{|S_4|}{2}$. This subgroup is the set of all permutations in $S_4$ which can be written as an even number of transpositions with the binary operation of composition. In cycle notation $A_4$ is the set $$\{e,(12)(34),(13)(24),(14)(23),(123),(132),(124),(142),(134),(143),(234),(243)\}$$ We know that if a permutation is written as a product of disjoint cycles, then its order is the least common multiple of the lengths of the cycles. We have each permutation element in $A_4$ written as a product of disjoint cycles. By Lagrange’s Theorem, the proper subgroups of $A_4$ can only have orders $1, 2, 3, 4$ or $6$. Looking first for cyclic subgroups, we see that $A_4$ has one element of order $1$ (e), three of order $2$ (the pairs of disjoint transpositions), and eight of order $3$.

I just remain to prove that $(D_3\times_{\rho}\mathbb{Z}_2)$ has an element of order 6 right ? I am trying to figure it out how find that element but I had no luck yet. Anyone can verify this and help me to finish my proof? Thanks in advance.

• The claim follows easily from math.stackexchange.com/questions/582658/… – ahulpke Dec 6 '17 at 20:01
• Part (1): see this question. – Dietrich Burde Dec 6 '17 at 20:11
• @DietrichBurde see the answer, it was my own. But it's quite different notation from the original question. – Richard Clare Dec 6 '17 at 21:07
• @ahulpke which claim? this doesn't help me at all, since I already have all the subgroups of $A_4$ and none have order 6. I just have to prove that the other semidirect product have an element of order 6. – Richard Clare Dec 6 '17 at 21:12
• The group $(D_3\times_{\rho}\mathbb{Z}_2)$ obviously has a subgroup of order $6$, namely $D_3$. But $A_4$ hasn't , so they cannot be isomorphic. – Dietrich Burde Dec 6 '17 at 21:13