Let g be a group and $a \in G$. Define $\phi_a:G\rightarrow G$ by $\phi_a(g)=aga^{-1}.$ Now Let $G=D_4$ and $a=r$, where $r$is the rotation. We must show that $\phi_r: D_4\rightarrow D_4$. So show what it is mapped to.

So far I have, $\phi_r(r)=r$, $\phi_r(r^2)=r^2$, $\phi_r(r^3)=r^3$, $\phi_r(e)=e.$

Now for the reflections, I have defined $h$ as the horz reflection and $v$ as the vert reflection. I also have said $d_1$ is the diagonal reflection around the line y=-x if it were to be graphed. (I dont know of a way to say it better) and $d_2$ as the reflection y=x. So $\phi_r(h)=d_2$, $\phi_r(v)=d_1$, $\phi_r(d_1)=rd_1r=h$ and $\phi_r(d_2)=v$. Is this correct?

Lastly explain how there can be no automorphisms of $D_4$ so that $\phi(r)=f$, where $f$ is a reflection.


With the definitions above, \begin{align*} \phi_r(h) &= rhr^{-1} = v\\ \phi_r(v) &= rvr^{-1} = h\\ \phi_r(d_1) &= rd_1r^{-1} = d_2 \\ \phi_r(d_2) &= rd_2r^{-1} = d_1. \end{align*} (To see this, you can do these operations on a sheet of paper marked in such a way that you can tell what orientation it ends up in.) So what you have written above is in fact not correct.

For your second question, an automorphism of $D_4$ is an isomorphism $\phi$ from $D_4$ to itself. So if $\phi(a) = b$, the orders of $a$ and $b$ must be the same in $D_4$. But $r$ has order $4$ and $f$ has order $2$.

  • $\begingroup$ does that clarify it? $\endgroup$ – Jack Armstrong Oct 16 '14 at 0:14
  • $\begingroup$ so for rhr^-1, do we do r, then h, then r^-1? Same goes for the rest. $\endgroup$ – Jack Armstrong Oct 16 '14 at 15:19
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    $\begingroup$ The answer to that really depends on how your teacher has defined these operations. But in this particular case, it doesn't matter whether you apply them right-to-left or left-to-right; the answers will be the same. $\endgroup$ – rogerl Oct 16 '14 at 16:34

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