3

I think there is not.

In order to proof this I've written all the element's order and I've seen that the possible orders in A4 were only 1,2 and 3 (3 elements whit order 2 and 8 elements whit order 3). The orders in D4 were 1,6, 3 and 2 but there were 7 elements with order 2. So because an isomorphism is first of all an homomorphism, my map have to respect this condition: ord(f(x)) divides ord(x), but in my exemple I have 8 elements whit order 3 in the domine and 3 elements whit order 3 or 1 in the codomine, so I'm forced to send seven elements in only three, so the map can't be injective, so can't be an isomorphism.

I think it is correct, but I would like to know if there is a smarter way to proof this involving generators.

Thanks!

pter26
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    How many elements of order 6 in $A_4$? In $D_6$? – Jyrki Lahtonen Jan 31 '17 at 14:31
  • In A4 there aren't element whit order 6, in D6 there are 2 – pter26 Jan 31 '17 at 14:37
  • @JyrkiLahtonen I think notation differs as to what $D_6$ is. – Asinomás Jan 31 '17 at 14:45
  • D6 is the dihedral group – pter26 Jan 31 '17 at 17:21
  • Good job, user411485. So if $f:A_4\to D_6$, $f(x)$ has order six and $ord(f(x))$ divides $ord(x)$ what can you say about $x$? Because an isomorphism has an inverse mapping that is also a homomorphism, you actually have $ord(x)=ord(f(x))$ whenever $f$ is an isomorphism. – Jyrki Lahtonen Feb 01 '17 at 06:59
  • @Jorge There are many clues here that to OP $D_6$ is a group of order twelve. 1) Otherwise the question of whether it can be isomorphic to $A_4$ is moot. 2) It has elements of order six. I am aware that there are people who call the dihedral group of order twelve $D_{12}$, but that is not the case here. – Jyrki Lahtonen Feb 01 '17 at 07:02
  • @JyrkiLahtonen sorry, I missread your comment, I thought it only asked to look at the number of elements in each group. – Asinomás Feb 01 '17 at 07:03

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