Showing $A_4$ is a subgroup of $SU(2)$

Asked Sep 04 '21 at 11:36

Active Sep 04 '21 at 11:38

Viewed 63 times

I am trying to show that $A_4$ is a subgroup of $SU(2).$

We know that $A_4$ is a group of even permutations of four elements, and that $SU(2)$ is the Lie group of $2\times 2$ unitary matrices with determinant $1$. So how can we show that?

I'll be thankfull for having answers.

edited Sep 04 '21 at 11:38

Shaun

44,997

asked Sep 04 '21 at 11:36

Julie

1

Up to conjugation it is enough to realise $A_4$ as subgroup of $GL_2(\Bbb C)$. This is explained, for example, here. – Dietrich Burde Sep 04 '21 at 12:11
@DietrichBurde I am missing something. The irreducible representations of $A_4$ have dimensions $1,1,1,3$. And the sum of two 1-dimensional representations is not faithful, because the commutator subgroup is in the kernel. The usual averaging trick does lead to unitary matrices, but the other problem needs to be solved first. – Jyrki Lahtonen Sep 04 '21 at 14:19
1

@JyrkiLahtonen ...and your comment shows that $A_4$ is not a subgroup of $SU(2)$. I think there is probably confusion here between $SU(2)$ and its quotient $SO(3)$. Of course $A_4$ (the "tetrahedral group") is a subgroup of $SO(3)$. – Sean Eberhard Sep 04 '21 at 15:01
See also this post. – Dietrich Burde Sep 04 '21 at 19:25

Showing $A_4$ is a subgroup of $SU(2)$

0 Answers0