[proofplan] We construct the loop operation directly on the same underlying set $Q$. Choose two elements $p,q \in Q$ and use the left and right translations determined by $p$ and $q$ to re-label the row and column inputs of the quasigroup multiplication. The resulting operation has identity element $e = p \circ q$, remains a quasigroup because translations and the re-labeling maps are bijections, and therefore is a loop. The same re-labeling maps give the required isotopy. [/proofplan] [step:Choose base elements and define the row and column relabeling maps] Since $(Q,\circ)$ is a finite [quasigroup](/page/Quasigroup), $Q$ is nonempty. Choose elements $p,q \in Q$, and define \begin{align*} e := p \circ q \in Q. \end{align*} Define the right translation by $q$ to be the map $R_q: Q \to Q$ given by \begin{align*} R_q(z)=z \circ q \quad \text{for every } z \in Q. \end{align*} Define the left translation by $p$ to be the map $L_p: Q \to Q$ given by \begin{align*} L_p(z)=p \circ z \quad \text{for every } z \in Q. \end{align*} By the quasigroup property, for every $x \in Q$ there is a unique $z \in Q$ satisfying $z \circ q = x$, and for every $y \in Q$ there is a unique $z \in Q$ satisfying $p \circ z = y$. Hence $R_q$ and $L_p$ are bijections. Define the bijection $\alpha: Q \to Q$ by \begin{align*} \alpha=R_q^{-1}. \end{align*} Define the bijection $\beta: Q \to Q$ by \begin{align*} \beta=L_p^{-1}. \end{align*} Thus, for every $x,y \in Q$, \begin{align*} \alpha(x) \circ q = x. \end{align*} Also, \begin{align*} p \circ \beta(y) = y. \end{align*} In particular, \begin{align*} \alpha(e)=p. \end{align*} Also, \begin{align*} \beta(e)=q. \end{align*} These identities hold because $R_q(p)=p\circ q=e$ and $L_p(q)=p\circ q=e$. [/step] [step:Transport the quasigroup operation through the relabeling maps] Define the binary operation $*: Q \times Q \to Q$ by \begin{align*} x*y=\alpha(x) \circ \beta(y) \quad \text{for every } (x,y) \in Q \times Q. \end{align*} For each fixed $a \in Q$, define the left translation for $*$ by $a$ to be the map $L_a^*: Q \to Q$ given by \begin{align*} L_a^*(y)=a*y=\alpha(a)\circ \beta(y) \quad \text{for every } y \in Q. \end{align*} This is the composition of the bijection $\beta:Q\to Q$ with the left translation $L_{\alpha(a)}^\circ: Q \to Q$ in the quasigroup $(Q,\circ)$, where \begin{align*} L_{\alpha(a)}^\circ(z)=\alpha(a)\circ z \quad \text{for every } z \in Q. \end{align*} Therefore $L_a^*$ is bijective. For each fixed $b \in Q$, define the right translation for $*$ by $b$ to be the map $R_b^*: Q \to Q$ given by \begin{align*} R_b^*(x)=x*b=\alpha(x)\circ \beta(b) \quad \text{for every } x \in Q. \end{align*} This is the composition of the bijection $\alpha:Q\to Q$ with the right translation $R_{\beta(b)}^\circ: Q \to Q$ in the quasigroup $(Q,\circ)$, where \begin{align*} R_{\beta(b)}^\circ(z)=z\circ \beta(b) \quad \text{for every } z \in Q. \end{align*} Therefore $R_b^*$ is bijective. Hence $(Q,*)$ is a quasigroup. [guided] The operation $*$ is obtained by feeding re-labeled inputs into the original quasigroup operation. We must verify that this new operation is still a quasigroup, meaning that left and right division are uniquely solvable. Fix $a \in Q$. Define the left translation by $a$ in the new operation to be the map $L_a^*: Q \to Q$ given by \begin{align*} L_a^*(y)=a*y \quad \text{for every } y \in Q. \end{align*} Using the definition of $*$, this becomes \begin{align*} L_a^*(y)=\alpha(a)\circ \beta(y) \quad \text{for every } y \in Q. \end{align*} Now $\beta:Q\to Q$ is bijective by construction. Since $(Q,\circ)$ is a quasigroup, the left translation $L_{\alpha(a)}^\circ: Q \to Q$ defined by \begin{align*} L_{\alpha(a)}^\circ(z)=\alpha(a)\circ z \quad \text{for every } z \in Q \end{align*} is also bijective. Therefore \begin{align*} L_a^* = L_{\alpha(a)}^\circ \circ \beta \end{align*} is a composition of bijections, hence is bijective. This proves that for every $a,c \in Q$, the equation $a*y=c$ has a unique solution $y \in Q$. Fix $b \in Q$. Define the right translation by $b$ in the new operation to be the map $R_b^*: Q \to Q$ given by \begin{align*} R_b^*(x)=x*b \quad \text{for every } x \in Q. \end{align*} Again using the definition of $*$, we have \begin{align*} R_b^*(x)=\alpha(x)\circ \beta(b) \quad \text{for every } x \in Q. \end{align*} The map $\alpha:Q\to Q$ is bijective by construction. Since $(Q,\circ)$ is a quasigroup, the right translation $R_{\beta(b)}^\circ: Q \to Q$ defined by \begin{align*} R_{\beta(b)}^\circ(z)=z\circ \beta(b) \quad \text{for every } z \in Q \end{align*} is bijective. Therefore \begin{align*} R_b^* = R_{\beta(b)}^\circ \circ \alpha \end{align*} is a composition of bijections, hence is bijective. This proves that for every $b,c \in Q$, the equation $x*b=c$ has a unique solution $x \in Q$. Since both left and right translations for $*$ are bijections, $(Q,*)$ is a quasigroup. [/guided] [/step] [step:Verify that $e=p\circ q$ is the identity for the transported operation] For every $x \in Q$, using $\beta(e)=q$, we obtain \begin{align*} x*e=\alpha(x)\circ \beta(e)=\alpha(x)\circ q=x. \end{align*} For every $y \in Q$, using $\alpha(e)=p$, we obtain \begin{align*} e*y=\alpha(e)\circ \beta(y)=p\circ \beta(y)=y. \end{align*} Thus $e$ is a two-sided identity element for $(Q,*)$. Since $(Q,*)$ is a quasigroup with a two-sided identity element, $(Q,*)$ is a [loop](/page/Loop). [/step] [step:Read the construction as an isotopy] Let $\gamma: Q \to Q$ be the identity map $\operatorname{id}_Q$, so \begin{align*} \gamma(x)=x \quad \text{for every } x \in Q. \end{align*} The maps $\alpha,\beta,\gamma:Q\to Q$ are bijections, and by the definition of $*$, for all $x,y \in Q$, \begin{align*} \alpha(x)\circ \beta(y)=x*y=\gamma(x*y). \end{align*} Hence $(Q,\circ)$ is [isotopic](/page/Isotopy) to the loop $(Q,*)$, with isotopy data $(\alpha,\beta,\gamma)$ in the displayed convention. This proves that every finite quasigroup is isotopic to a loop. [/step]