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Problem

Find all the semidirect products $G=\mathbb Z_3 \rtimes_{\phi} \mathbb Z_4$. Show that one of these is non abelian and is not isomorphic to $\mathbb A_4$

I am a bit lost with this problem. First I've tried to find all the possible morphisms $$\phi:\mathbb Z_4 \to Aut(\mathbb Z_3)$$

The only two possible automorphisms in $\mathbb Z_3$ are $Id$ and $-Id$. $\phi$ is determine by the value it takes at $1$, so I have $\phi_1(1)=Id$ and $\phi_2(1)=-Id$.

For the case $\phi_1=Id$, we can think the external semidirect product as a cartesian product $\mathbb Z_4 \oplus \mathbb Z_3$ with $(z_1,z_2)+(w_1,w_2)=(z_1+\phi_1(z_2)(w_1),z_2+w_2)=(z_1+w_1,z_2+w_2)$. I don't know to which group $\mathbb Z_4 \oplus \mathbb Z_3$ is isomorphic to (it is clear it has to be an abelian group of order $12$).

For the case $\phi_2=-Id$, I don't know what the external semidirect product looks like and to which "known" group is isomorphic to.

Any suggestions would be appreciated.

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    $\begingroup$ For the nonabelian example, all you are asked to do is to show that it is not isomorphic to $A_4$, so I should concentrate on that. Does $A_4$ have an element of order $4$? $\endgroup$ – Derek Holt Sep 24 '14 at 8:50
  • $\begingroup$ No, it doesn't, just of order 1,2 and 3, so I suppose you are suggesting to find an element of order 4 in the semidirect product, I'll do that but a part of the exercise is to find all the semidirect products, I could find one and I am having some difficulty to realize what the other "looks" like, I would appreciate some help with that. $\endgroup$ – user156441 Sep 24 '14 at 13:06
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    $\begingroup$ You have found all of the semidirect products. As soon as you have accurately defined the homomorphisms from ${\mathbb Z}_4 \to {\rm Aut}({\mathbb Z}_3$ that give rise to the possible semidirect products, you are done. Most semidirect products don't have any easier description. If you ask the group theory program GAP to identify this group, it says "C3 : C4" which just that it is a semidirect (but not direct) product of C3 with C4. For the abelian example, it says "C12". $\endgroup$ – Derek Holt Sep 24 '14 at 13:53
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$\mathbf Z_4\oplus\mathbf Z_3$ can be given another name. For a hint, what's the order of $(1,1)$? You aren't supposed to show that the other semidirect product is isomorphic to some known group, just that it's not isomorphic to $A_4$. The easiest way to do this is to remember that elements of $A_4$ all have order $2$ and $3$ and find an element of the semidirect product of higher order.

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  • $\begingroup$ Oh, $Z_4⊕Z_3=Z_{12}$ (the order of $(1,1)$ is $12$), but I have no idea what $Z_3⋊ϕ_2Z_4$ "looks" like (I have to find all the semidirect products first). $\endgroup$ – user156441 Sep 24 '14 at 13:00
  • $\begingroup$ I don't real know what you mean by "looks like." It has underlying set the cartesian product and $(a,b)(c,d)=(a+\phi(b)(c),b+d),$ as in any semidirect product-if you haven't learned that maybe this is the issue. $\endgroup$ – Kevin Carlson Sep 24 '14 at 21:19
  • $\begingroup$ Forget about that, I got mixed up. $\endgroup$ – user156441 Sep 26 '14 at 16:30

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