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Let $X$ be a non-empty subset and $F(X)$ the free group on $X$. If $z\in F(X)$, it can be expressed in the form $z=x_1^{\epsilon _1} \cdots x_n^{\epsilon _n} $ where $\epsilon _i =\pm 1$, $x_i\in X$ and it contains no pair of consecutive symbols of the form $xx^{-1}$ or $x^{-1}x$. Such an expression is called the reduced form of $z$.

Suppose that $z = x_1^{\epsilon _1}\cdots x_n^{\epsilon _n} = y_1^{\delta _1}\cdots y_n^{\delta _n}$ (both of them are reduced forms of $z$).

Can we conclude that $x_i=y_i$ for each $i$?

Thanks!

user73564
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1 Answers1

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Suppose that this were not the case, so you have $z = x_1^{\epsilon _1}\cdots x_n^{\epsilon _n} = y_1^{\delta _1}\cdots y_n^{\delta _n}$ and suppose that $x_{i}\neq y_{i}$ for some $i$ or that if $x_{i}=y_{i}$ for all $i$, then $\epsilon_{i}\neq\delta_{i}$ for some $i$. Then you have a non-trivial relation in your group, namely

$1=x_1^{\epsilon _1}\cdots x_n^{\epsilon _n}(y_1^{\delta _1}\cdots y_n^{\delta _n})^{-1}=x_1^{\epsilon _1}\cdots x_n^{\epsilon_n}y_{n}^{-\delta_{n}}\cdots y_{1}^{-\delta_{1}}$.

Thus your group is no longer free.

David Ward
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