Let $F(a,b)$ be the free group on $\{a,b\}$. Show that the set $\{a,b^2\}$ is free, yet cannot be extended to a basis.
How do I show that a set is free. I couldn't manage to do working with the definition of free groups.
Definition as given in class:
suppose we have a reduced word in $a$ and $b^2$ - $$w=a^{k_1}\cdot b^{2l_1}\cdots a^{k_r}.$$ which gives the trivial element. then we get a reduced word on $a, b$ that gives the trivial element, which is a contradiction. we can do the same for a word that ends with a power of $b^2$.
now, if we could complete $\{a,b^2\}$ to a basis of $\mathbb{F}_2$ then we would have $$a^{k_1}\cdot b^{2l_1}\cdots a^{k_r}=b$$ for some coefficients (the same is needed to be done for the case where the word ends with a power of $b^2$) so we have that $a^{k_1}\cdot b^{2l_1}\cdots a^{k_r}b^{-1}$ is the trivial element in $\{a,b\}$, which is again a contradiction