I want to find a non-trivial isomorphism between the dihedral group $D_n$ and itself. Non-trivial means that the isomorphism won't be the identity.

I looked at the group $D_n$ as the set of the following elements:

$$\{e, \sigma, \sigma^2, ...., \sigma^{n-1} , \rho, \rho \sigma, \rho \sigma^2 ..., \rho \sigma^{n-1}\},$$

where $\sigma$ is a $\frac{2\pi}{n}$ rotation (clockwise) and $\rho$ is reflection through the vertical line.

What I thought about is the following function: $f(\sigma^k) = \sigma^{-k}$ and $f(\rho\sigma^{-k}) = \rho\sigma^{-k}$

It seems to that it is legit.

Am I correct here?

  • 1
    $\begingroup$ You should be able to check if your function is a group morphism. I suggest you check that $f(ab)=f(a)f(b)$ for all elements $a$ and $b$ in $D$, and that $f(e)=e$. It would be easier if you have an economical way to keep track of your function. Note that since the group is generated by $\rho$ and $\sigma$, the function is determined by its images $f(\rho)$ and $f(\sigma)$. $\endgroup$ – Leaning Mar 9 at 2:12

An isomorphism from a group $G$ to the same group $G$ is called an automorphism.

See this YouTube video describing the group of automorphisms of the dihedral group. (Yes, they form a group! The operation is composition of functions.)


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