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Is $A_{4}\times Z_2\simeq \langle g,h \mid g^{12},h^2,(gh)^{12}, gh=hg\rangle$? In addition, is $\operatorname{Aut}(A_{4}\times Z_2)= \operatorname{Aut}(A_{4})\times \operatorname{Aut}(Z_2$)=$S_{4}\times Z_2$? Also, assume G=$\langle g,h \mid g^{12},h^2,(gh)^{12},gh=hg \rangle$, what's the automorphism group of G?

As J.P said, the isomorphism doesn't exist.

6666
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  • By $A_{12}$ do you mean the alternating group of even permutations on ${1, \ldots, 12}$? – Sammy Black Mar 30 '15 at 02:48
  • @ Sammy Black Yes! – 6666 Mar 30 '15 at 02:49
  • Do you have a group presentation for the alternating group? I would start there. – Sammy Black Mar 30 '15 at 02:51
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    Also, if $h^2 = e$ (your second relation), then writing $gh^{12}$ in the third relation is unnecessary: $gh^{12} = g(h^2)^6 = ge^6 = ge = g$. – Sammy Black Mar 30 '15 at 02:52
  • @Sammy Black Sorry I made a mistake, here I mean $A_4$,not $A_{12}$ – 6666 Mar 30 '15 at 02:52
  • @Sammy Black Also the relation I mean here is ${{gh}}^{12}$= ${{g}}^{12}h^{12}$ – 6666 Mar 30 '15 at 02:55
  • @Joseph: Are you assuming that $g$ and $h$ commute? $(gh)^{12}$ (which I suppose is what you mean), is not the same as $g^{12}h^{12}$. – bzc Mar 30 '15 at 02:58
  • @ Brandon Carter, yes I think you are right, they can be not commutative – 6666 Mar 30 '15 at 02:59
  • If you map $g$ to a generator of the cyclic group $Z_{12}$ with $12$ elements, and $h$ to the identity, you see that $Z_{12}$ is a quotient of the group given by the presentation. $A_4\times Z_2$ does not have an element of order $12$... – j.p. Mar 30 '15 at 06:43
  • @j.p thank you very much, and do you have idea about the automorphism group of G? – 6666 Mar 30 '15 at 07:00
  • According to the theorem in a link you had in an earlier version of your question the statement about the automorphism group of $A_4\times Z_2$ looked correct. But as $G$ is some other group, you first have to identify it (from the presentation I see only an automorphism exchanging $g$ and $gh$ fixing $h$). – j.p. Mar 30 '15 at 07:38
  • @j.p, yes, I know the aut of A4xZ2 is right, but I am still struggling to find the G, do you know some tricks to find the group by presentation? – 6666 Mar 30 '15 at 07:42
  • The group $G$ is an infinite hyperbolic triangle group. – Derek Holt Mar 30 '15 at 08:10
  • It is not even clear what a question like "what's the automorphism group of $G$?" means. – Derek Holt Mar 30 '15 at 08:16
  • @Derek Holt, I mean Aut(G) – 6666 Mar 30 '15 at 08:58
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    To answer your latest question, yes, ${\rm Aut}(A_4 \times C_2) \cong S_4 \times C_2$, since both direct factors are characteristic. – Derek Holt Mar 30 '15 at 09:25

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In general, the triangle group $\langle x,y \mid x^k=x^l=(xy)^m=1 \rangle$ is infinite if and only if $1/k + 1/l + 1/m \le 1$. It is virtually abelian when $1/k + 1/l + 1/m = 1$ and hyperbolic when $1/k + 1/l + 1/m < 1$, which is true in your example. So it is certainly not isomorphic to $A_4 \times C_2$ (which doesn't even have any elements of order $12$).

Derek Holt
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  • Thank you, but I think to the 2 genus surface made by (12,2,3) triangles with 24 edges, it's $\Gamma^{+}$ group is just that. – 6666 Mar 30 '15 at 09:03
  • Also here maybe I forget to say g and h are commutative – 6666 Mar 30 '15 at 09:19
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    So you are not asking about the group $\langle g,h \mid g^{12},h^2,(gh)^{12} \rangle$ but about the group $\langle g,h \mid g^{12}=h^2=(gh)^{12}=1, gh=hg \rangle$ ? That's a commutative group of order $24$ isomorphic to $C_{12} \times C_2$. – Derek Holt Mar 30 '15 at 09:23
  • On the question of ${\rm Aut}(G)$, I have found assertions that the outer automorphism groups of triangle groups are finite, but I haven't found any more details or references. – Derek Holt Mar 30 '15 at 09:27
  • Thank you very much, now I wonder what's the automorphism group of $Z_{12}\times Z_2$? – 6666 Mar 30 '15 at 09:39
  • How to determine if Z4, Z6 are characteristic subgroup of Z4xZ6? – 6666 Mar 30 '15 at 10:09
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    That's easy - they are not. But I think you are asking too many new questions in comments. – Derek Holt Mar 30 '15 at 10:17
  • Could you tell me how do determine is or not? I really want to compute Aut(Z4xZ6) – 6666 Mar 30 '15 at 10:19
  • @Joseph: If you want to know the automorphism group of $Z_4\times Z_6$ then you can take a look at http://math.stackexchange.com/questions/102895/how-can-i-compute-mathrmaut-mathbbz-4-mathbbz-times-mathbbz-6-mathbb – j.p. Mar 30 '15 at 10:30
  • @j.p.Thank you very much! – 6666 Mar 30 '15 at 15:05