This exercise will compute the automorphism group of $D_{10}$. The presentation of the group is $D_{10 }= <r, s |r^5 = s^2 = 1, s^{−1}rs = r^{−1}>$
(a) (i) Show that under any automorphism $σ : D_{10} → D_{10},$ $r$ can have a maximum of $4$ images, and $s$ can have a maximum of $5$ images.
(ii) Show that any combination of $φ(r)$ and $φ(s)$ above give an automorphism of$ D_{10}$. Hence prove that$ |Aut(D_{10})| ≤ 20.$
(b) Let $σ$ be the homomorphism defined by $r→ r$ and $ s → sr,$ and $τ$ be the homomrphism defined by $r→ r^2$ and $s → s$. Show that $Aut(D_{10})$ is generated by $σ$ and $τ$.
(c) Show that $Aut(D_{10})\cong F_5.$
I tried part (a). for part (b) and (c) , by given $σ$ and $τ$, I found associated permutation representation by labeling element in $D_{10}$ from $1$ to $10$ then I got
$σ=(6 \space 7 \space 8 \space9\space 10)$ and $\tau=(2 \space 3 \space 5 \space4)(7\space 8 \space 10\space9)$
and I know presentation of $F_5=<a,b|a^5=b^4=1,bab^{-1}=a^3>$
But $\tau\sigma\tau^{-1}$ is not equal to $\sigma ^3$. Is there anything wrong with my work? Please help me.
I think, If I define $σ$ by $r→ r$ , $ s → sr,$ and $τ$ by $r→ r^3$ , $s → s$ , I can find a isomorphism to $F_5$
Actually we can have this presentation for $F_{5}$, too. That is $F_5=\langle a,b|a^5=b^4=1,b^{-1}ab=a^3\rangle$. Now we can get the result, with those two mentioned generators.
Also the new generators $\tau$ and $\sigma$, that you defined, work with the presentation $F_5=\langle a,b|a^5=b^4=1,bab^{-1}=a^3\rangle$. So it depends on how we view $F_{5}$.