I am self studying abstract algebra from Thomas Hungerford and I have a question in it in the chapter of free groups.
I have a question on page 65 of the book. I am adding it's image.
Question is in line 6 of sketch of proof.
Question : What is the reasoning behind author reducing $ |x| |x^{-1}| = 1_{F} $ .
Kindly tell.
We have to show that $|x|(|x^{-1}|(w)) = w$ for every reduced word $w$.
If $w = 1$, I leave it for you to show that $|x|(|x^{-1}|(w)) = w$ (it should be easy if you understand what follows). Otherwise, we can write $w = x_1^{\delta_1}\cdots x_n^{\delta_n}$. According to the definition of $|x^{-1}|$, there are two cases:
Case 1) $x^{-1} \neq x_1^{-\delta_1}$: by the definition: $$ |x^{-1}|(x_1^{\delta_1}\cdots x_n^{\delta_n}) = x^{-1}x_1^{\delta_1}\cdots x_n^{\delta_n} $$ But then $|x|(x^{-1}x_1^{\delta_1}\cdots x_n^{\delta_n})$ falls under the second case in the definition so we have: $$ |x|(x^{-1}x_1^{\delta_1}\cdots x_n^{\delta_n}) = x_1^{\delta_1}\cdots x_n^{\delta_n} $$
Case 2): $x^{-1} = x_1^{-\delta_1}$, so that $x_1 = x$ and $\delta_1 = 1$: by the definition: $$ |x^{-1}|(x_1^{\delta_1}\cdots x_n^{\delta_n}) = x_2^{\delta_2}\cdots x_n^{\delta_n} $$ Note that because $x_1^{\delta_1}\cdots x_n^{\delta_n}$ is reduced, either $x_2 \neq x_1$ or $\delta_2 = 1$. Hence $x \neq x_2^{-\delta_2}$ and ${|x|}(x_2^{\delta_2}\cdots x_n^{\delta_n})$ falls under the first case of the definition so we have: $$ |x|(x_2^{\delta_2}\cdots x_n^{\delta_n}) = xx_2^{\delta_2}\cdots x_n^{\delta_n} = x_1^{\delta_1}x_2^{\delta_2}\cdots x_n^{\delta_n} $$
So in both cases $|x|(|x^{-1}|(w)) = w$ and we are done.