Show that every automorphism of $F_{20} = \langle a, b \mid a^5 = b^4 = 1, bab^{-1} = a^3\rangle $ is an inner automorphism

Question

Show that every automorphism of $F_{20} = \langle a, b \mid a^5 = b^4 = 1, bab^{-1} = a^3\rangle $ is an inner automorphism.

I found $Z(F_{20})={1}$ and it follows $F_{20} \cong \text{Inn}(F_{20})$. I know $\text{Inn}(F_{20}) \unlhd \text{Aut}(F_{20})$ but I can't go further!

Can you give me a hint?

Answer 1

Every element of $F_{20}$ can be expressed uniquely in the form $a^ib^j$ for some $0 \le i \le 4$ and $0 \le j \le 3$. Also, from $bab^{-1} = a^3$ it follows easily that $b^r a^s b^{-r} = a^{3^r s}$ whenever $r,s$ are non-negative. Let $f \in \text{Aut}(F_{20})$. Then $f(a) = a^k$ for some $1 \le k \le 4$ since $\langle a \rangle$ is the unique Sylow $5$-subgroup. Now write $f(b) = a^i b^j$ with $0 \le i \le 4$ and $0 \le j \le 3$. We have $$a^ib^ja^kb^{-j}a^{-i} = f(b)f(a)f(b)^{-1} = f(a)^3 = a^{3k}$$ so $b^j a^k b^{-j} = a^{3k}$. Since $b^j a^k b^{-j} = a^{3^j k}$, we must have $3^j k \equiv 3k \pmod{5}$ and thus $j = 1$. So $f(b) = a^ib$ for some $0 \le i \le 4$. Since $f$ is determined by $f(a) = a^k$ and $f(b) = a^ib$, we see that there are at most $4 \cdot 5 = 20$ possible maps that $f$ can be. Thus $|\text{Aut}(F_{20})| \le 20$. You have already shown that $|\text{Inn}(F_{20})| = 20$, so we must have $\text{Aut}(F_{20}) = \text{Inn}(F_{20})$.