Let $G = G_1 \ast G_2$ where $G_1$ and $G_2$ are nontrivial groups and $\ast$ is the free product. Show that if $w \in G$ has odd length (at least $3$), then $w$ is conjugate to an element of shorter length.
This is a problem from Munkres, page 421. I've tried starting with an element $w=xyz \in G$ and conjugating it by a few words, but haven't been able to find a way to reduce it to a word of length less than 3, let alone find a way to do this for all words of odd length.
Just looking for a hint to start with, not a full solution. Thanks.
Edit: fixed notation.
If we let $x_i\in G_1$ and $y_i\in G_2$ for suitable indices $i$, then a word of odd length $2n+1$ is of either of the two forms $$ x_1y_1x_2y_2\cdots x_{n-1}y_{n-1}x_n\quad\text{or}\quad y_1x_1y_2x_2\cdots y_{n-1}x_{n-1}y_n $$ Now take, for instance, the first word, and conjugate by $x_1^{-1}$. We have $$ x_1^{-1}x_1y_1x_2y_2\cdots x_{n-1}y_{n-1}x_nx_1 = y_1x_2y_2\cdots x_{n-1}y_{n-1}(x_nx_1) $$ where the first "letter" of the word has disappeared, and the last two "letters" go together to become one, so this new word has length $2n$.