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Is there an isomorphism between the dihedral group and the group defined by the transpose of the dihedral group's Cayley table?

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What is the transpose of a group? – William Sep 14 '12 at 3:17
@William -- edited. – idonutunderstand Sep 14 '12 at 3:19
The Cayley table is not unique. Is it clear that you even get a group by transposing a group table? If it is a group, it must be the dihedral group, as there's no other non-abelian group of order 8 with so many elements of order 2. – Gerry Myerson Sep 14 '12 at 3:26
Perhaps my previous comment was a little silly. I think you are asking about the "opposite group" - see – Gerry Myerson Sep 14 '12 at 3:27

$\newcommand{\lowast}{*}$ 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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I will read this carefully :) but first I'm just wondering why you made a newcommand mapping \lowast to *? – idonutunderstand Sep 24 '12 at 4:20
I write the post first before previewing it. \lowast wasn't defined in mathjax, but * looked fine. – Jack Schmidt Sep 24 '12 at 11:43

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