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Let $F$ be the free group on two elements $x$, $y$. To make our lives easier, we denote $x^{-1}$ by $X$ and $y^{-1}$ by $Y$. Let $g_i$ for $i = 1, \ldots, 6$ be the following elements:$$g_1 = xyxY,$$$$g_2 = XYXyxYxyXyxYXYXyxYxyxyXYY,$$$$g_3 = XYXyx,$$$$g_4 = YxyXyxYXYXyyx,$$$$g_5 = YxyXYxyxyX,$$$$g_6 = xxYXXy.$$Let $G$ be the subgroup of $F$ generated by the $g_i$.

Question. How do I see that $G$ is free of rank $6$?

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    $\begingroup$ Are you aware of Nielson reduction? You could also use Stalling folding? Did you just write down 6 random words or do these come from somewhere? $\endgroup$ – Paul Plummer Oct 19 '16 at 18:06
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    $\begingroup$ When you wrote "...to make our lives easier..." you actually meant "...to make my life easier...", right? Anwyay, (1) you can check all the elements given are in reduced normal form, and (2) multiplying two of those elements doesn't result in trivial reduction. Not that it will be terribly hlepful, but for what it's wroth $\;g_6\in F'\;$ as the final exponent sums in all its letters ($\;x,y\;$ and its inverses ) is zero. $\endgroup$ – DonAntonio Oct 19 '16 at 21:17
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    $\begingroup$ I can see two immediate simplifcations. $g_3$ is a prefix of $g_2$, so you could remove that prefix. After doing that, $g_2$ and $g_4$ have a long common prefix, which you could use to further shorten $g_2$. $\endgroup$ – Derek Holt Oct 20 '16 at 12:17
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I'm not an expert in free goups, but I'll give it a go: simplifying the generators like @Derek suggested:

$$g_1 = x\color{red}{yx}Y,$$ $$g_2 = XYx\color{red}{YxyxyXY}Y,$$ $$g_3 = XY\color{red}{X}yx,$$ $$g_4 = YxyX\color{red}{yxYXYXy}yx,$$ $$g_5 = YxyX\color{red}{Yxyxy}X,$$ $$g_6 = x\color{red}{xYX}Xy.$$ $$G_1 = y\color{red}{XY}X,$$ $$G_2 = y\color{red}{yxYXYXy}Xyx,$$ $$G_3 = XY\color{red}{x}yx,$$ $$G_4 = XY\color{red}{YxyxyXY}xYXy,$$ $$G_5 = x\color{red}{YXYXy}xYXy,$$ $$G_6 = Yx\color{red}{xyX}X$$ Now consider all possible concatenations $g_ig_j, G_ig_j, g_iG_j, G_iG_j$ (with $i\neq j$). You will see that the red part of the words never cancel (provided I didn't make a mistake). This implies that in any concatenation of arbitrary length of the $G$'s and $g$'s, the red part will never cancel. This means that there are no relations possible between them, and they indeed form a free group of rank $6$.

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