# Show the function between the dihedral groups is well defined

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.)

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, $$\psi(\mu^a\rho^b)=s^ir^j=\psi(\mu^i\rho^j)$$.
• 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 $\psi(\mu^a\rho^b)=\psi(\mu^i\rho^j)$? Apr 4, 2022 at 1:09
• @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. Apr 4, 2022 at 6:02