Proof of a relation of Braid groups Let $B_n$ be the braid group on $n$-strings, generated by $\alpha_1,\ldots, \alpha_{n-1}$ with relations $\alpha_i \alpha_j = \alpha_j \alpha_i$ for $|i-j|>1$ and $\alpha_i \alpha_{i+1} \alpha_{i} = \alpha_{i+1} \alpha_{i} \alpha_{i+1}$ for $1 \leq i \leq n-2$.
How to prove the equations $\alpha_i \alpha^{-1}_{i+1} \alpha_{i} =\alpha_{i+1} \alpha^{-1}_i \alpha_{i+1}$? Thank you. 
 A: The proposition is actually false.
Indeed: the proposition implies $(\alpha_1\alpha_2^{-1})^3=e$, but if you draw the closure of such a 3-string braid, you'll find that it's not tricolorable, unlike 3 unlinked unknots.
A: The proposition is $not$ true. It is known that the braid group has a faithful representation as automorphisms of the free group.
Then, let's show that their automorphisms are not the same. Take $\beta_1=\alpha_i\alpha_{i+1}^{-1}\alpha_i$. I'll denote the automorphism $\rho_{\beta_1}$, which is equal to $\rho_{\alpha_i}\rho^{-1}_{\alpha_{i+1}}\rho_{\alpha_i}$. Take the generator $x_i$ of the free group.
$\rho_{\beta_1}(x_i)=\rho_{\alpha_i}(x_i)\rho^{-1}_{\alpha_{i+1}}(x_i)\rho_{\alpha_i}(x_i)=
x_{i+1}x_ix_{i+1}$
Similarly, let's call $\beta_2=\alpha_{i+1}\alpha_i^{-1}\alpha_{i+1}$. Then,
$\rho_{\beta_2}(x_i)=\rho_{\alpha_{i+1}}(x_i)\rho_{\alpha_i}^{-1}(x_i)\rho_{\alpha_{i+1}}(x_i)=
x_i (x_i x_{i+1} x_i^{-1})x_i=x_i^2x_{i+1}$
Which is clearly not the same thing in the free group.
