Same as above. I'm trying to show that for any n being odd, $D_n$ has exactly n elements of order 2 where $D_n$ is non-abelian. I know that for $n\ge3$ this is true, but what about for $n=1$.
2026-03-26 11:00:20.1774522820
The Dihedral group $D_1$ is non-abelian?
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Here is an answer to your group theory question. The dihedral group $D n$ has 2*n elements. There is only one group with two elements. So D1 is isomorphic to the group with 2 elements. That group is like odd and even numbers with addition. The even numbers are the identity element. Here is the Cayley table - e for even and o for odd.
$+$ | e o
_____
e | e o
o | o e
Regards, Matt