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Suppose that $n = dm$ where $d$ and $m$ are positive integers with $m\ge 3$. Consider the dihedral group $D_n = \langle \{\mu, \rho\}\rangle,$ where $|\mu| = 2$, $|\rho| = n$ and $\rho\mu = \mu\rho^{−1}$, and the dihedral group $D_m = \langle \{s, r\}\rangle,$ where $|s| = 2$, $|r| = m$ and $rs = sr^{−1}$.

Define $\psi : D_n \to D_m$ by $ψ(\mu^a\rho^b)=s^ar^b$, for any integers $a,b$.

Show that $\psi$ is well-defined.


Here's the stuff I noticed:

  • different values of $a,b$ can give the same group element $\mu^a\rho^b$, and I need to show that they also give the same $\psi(\mu^a\rho^b)$.

  • if $n$ is not a multiple of $m$, then $\psi$ not well-defined.


So here is what I did so far, (tried to make a proof sketch):

from integer division, there exists unique integers $i, j, s, t$ with $0 \le i < 2$ and $0 \le j < n$ and $a = i + 2s$ and $b = j + nt$.

So, the group element $\mu^a\rho^b=\mu^{i+2s}\rho^{j+nt}$ uniquely determined by $i$ and $j$, since changing $s$ and $t$ won't make a difference.

So, I think I need to show that $\psi(\mu^a\rho^b)$ depends only on $i$ and $j$, and not on $s$ or $t$. (this is what I'm having a hard time doing.)

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1 Answer 1

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I notice that the symbol you use for the quotient of $a$ when divided by $2$ is the same as the generator $s$ in $D_m$. To avoid confusion, I replace it by $u$.

The main idea is to show that $\psi(\mu^a\rho^b)=\psi(\mu^i\rho^j)$.

By how the function is defined, $\psi(\mu^a\rho^b)=\psi(\mu^{i+2u}\rho^{j+nt})=s^{i+2u}r^{j+nt}$.
Since $|s|=2$, we have $s^{i+2u}=s^i(s^2)^u=s^i$.
Next, $r^{j+nt}=r^{j+dmt}=r^j(r^m)^{dt}=r^j$.
Therefore, $\psi(\mu^a\rho^b)=s^ir^j=\psi(\mu^i\rho^j)$.

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  • $\begingroup$ I understand how you showed $\psi(\mu^a\rho^b)=\psi(\mu^i\rho^j)$, but I'm having some difficulty understanding what well-definedness means and what I need to show when given problems on well-definedness. I tend to confuse injective with well-definedness usually. Can you explain why we must show $\psi(\mu^a\rho^b)=\psi(\mu^i\rho^j)$? $\endgroup$
    – eddie
    Apr 4, 2022 at 1:09
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    $\begingroup$ @eddie This is because there can be more than one way to write each element in $D_n$. For example, $\rho$ can be also written as $\rho^{1+n}$ in $D_n$. Since the image of each element depends on $a,b$, if the function is not well-defined, then it may happen that $\psi(\rho)\neq \psi(\rho^{1+n})$ which is absurd. Because the quotient and remainder of $a,b$ when divided by $2,n$ respectively are unique, by showing that $\psi(\mu^a\rho^b)=\psi(\mu^i\rho^j)$, the image of each element in $D_n$ is always the same regardless of how we write them. $\endgroup$ Apr 4, 2022 at 6:02
  • $\begingroup$ I tried to find the kernel and image of this function, is this done correctly? math.stackexchange.com/questions/4427919/… $\endgroup$
    – eddie
    Apr 14, 2022 at 22:45

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