Is there an isomorphism between the dihedral group and the group defined by the transpose of the dihedral group's Cayley table?
Is the dihedral group isomorphic to its transpose?
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4Perhaps my previous comment was a little silly. I think you are asking about the "opposite group" - see http://en.wikipedia.org/wiki/Opposite_group – 2012-09-14
1 Answers
$\newcommand{\lowast}{*}$ The transpose $B$ of a matrix $A$ has the property that $B_{ij} = A_{ji}$. When these matrices are Cayley tables, then what we are saying is that $i \lowast_B j = j \lowast_A i$. In other words, the multiplication in the “transpose group” is the opposite multiplication in the original group.
It turns out a group is always isomorphic to its transpose group. An isomorphism is given by $g \mapsto g^{-1}$ since $(gh)^{-1} = h^{-1} g^{-1}$ when all multiplications are in the regular group, but $(g \lowast_A h)^{-1} = h^{-1} \lowast_A g^{-1} = g^{-1} \lowast_B h^{-1}$, so that $g\mapsto g^{-1}$ is a homomorphism between the original and the transpose group. It is clearly injective and surjective by definition of group.
One might ask, which group do we take the inverse in? Luckily the answer is the same in both groups (though the answer was just, pick one and be consistent). $g \lowast_A h = 1$ iff $h=g^{-1}$ and $g=h^{-1}$ (under A), but we get the same equation with letters reversed under B, $h \lowast_B g = 1$ iff $h=g^{-1}$ and $g=h^{-1}$ (under B). Hence the property of being an inverse is preserved under transpose.
Gerry's comment makes use of a similar observation: an element has order 2 in the original if and only if it has order 2 under the transpose. This is sufficient to handle the dihedral group of order 8 without handling all groups.
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0I write the post first before previewing it. \lowast wasn't defined in mathjax, but * looked fine. – 2012-09-24