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I'm trying to understand the group $2 \cdot A_5$, i.e. the double cover of the alternating group on 5 elements. I'd like to think of elements of this group as members of the set $$\{+1,-1\} \times A_5$$ When viewing elements as members of this set, is there a (relatively) intuitive definition of the group operation?

xyz
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  • Is this not the product group $\Bbb Z_2\times A_5$? – Andrew Oct 19 '12 at 20:30
  • No, I believe that it is not. In particular, I don't believe that the group operation operates independently on each part of the pairs in ${+1,-1} \times A_5$. – xyz Oct 19 '12 at 20:32
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    @Andrew It is apparently $SL_2(\Bbb F_5)$. – anon Oct 19 '12 at 20:41
  • It is indeed, but I'd specifically like to understand the group operation when viewing elements in the way that I mentioned. – xyz Oct 19 '12 at 20:43
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    Writing the group elements in the way you suggest seems misleading and not really helpful to me. The point is that as has been pointed out, the right double cover is ${\rm SL}(2,5)$. The centre os this group consists of scalar matrices of determinant $1$, so is ${ \pm I }$. Factoring out the center leads to ${\rm PSL}(2,5) \cong A_{5}.$ – Geoff Robinson Oct 19 '12 at 20:52
  • @GeoffRobinson it may not be helpful to you, but it is to me ;-) Though if the answer when viewing elements this way is "there is no shorter description of the group operation other than writing out the multiplication table", then I can live with that. – xyz Oct 19 '12 at 20:53
  • Related. – anon Oct 19 '12 at 21:09

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