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Is $D_{2n}$ always a subset of $D_{2k}$ where $n,k \in N$ and $k>n$

I though this was true as every element of $D_{2n}$ must surely be in $D_{2k}$.

For example, consider:

$D_{4}=[1,r,s,sr^{-1}]$, and $D_{6}=[1,r,r^{2},s,sr^{-1},sr^{-2}]$.

Note that every element of $D_{4}$ is in $D_{6}$.

Clarification: I am using the $D_{2n}$ notation.

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$D_{2k}$ contains elements with the same names as the ones in $D_{2n}$, but they are not really the same elements. In $D_{2n}$, the element $r$ is rotation by $2\pi / n$ radians, whereas in $D_{2k}$, it is rotation by $2\pi / k$.

The elements satisfy different multiplication rules in the two groups. For example, in $D_4$, the element $r$ satisfies $r^2 = 1$ and $r^3 = r$, whereas in $D_6$, the element $r$ satisfies $r^2 = r^{-1}$ and $r^3 = 1$.

So I don't think it's useful to think of $D_{2n}$ as a subset of $D_{2k}$ by comparing the names of their elements.

On the other hand, $D_{2k}$ may contain subgroups which are isomorphic to $D_{2n}$ for certain choices of $k$ and $n$. For example, $D_{12}$ contains an isomorphic copy of $D_{6}$. To see it, draw a triangle connecting the even-numbered vertices of the hexagon, and consider the six operations which fix the triangle: rotations by 0, 120, and 240 degrees, and three of the six reflections.

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  • $\begingroup$ Thank you, that was such a relief.. This was my initial thought by the names really did throw me off. $\endgroup$ – J.Gudal Mar 24 '16 at 19:35

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