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Here is another variation on the other answers (particularly, the Cartier approach and DeBruyn's current approach). Let $G$ be a group acting on the left of a finite set $X$, let $A$ an abelian group and let $c\colon G\times X\to A$ be a "$1$-cocycle", meaning that $c(gh,x) = c(g,hx)c(h,x)$ for all $x\in X$ and $g,h\in H$. You can then define a homomorphism $\theta\colon G\to A$ by the rule $$\theta(g) = \prod_{x\in X}c(g,x).$$ Although we are supposed to rule out computation, this is easy: $$\theta(gh)=\prod_{x\in X}c(gh,x)=\prod_{x\in X}c(g,hx)c(h,x) =\prod_{y\in X}c(g,y)\prod_{x\in X}c(h,x) =\theta(g)\theta(h)$$ by putting $y=hx$.

Now apply this to $S_n$ acting on the set $X$ of $2$-element subsets of $\{1,\ldots, n\}$ with $A=\{\pm 1\}$. Put $$c(\sigma,\{i,j\})= \frac{\sigma(i)-\sigma(j)}{i-j}$$ (this depends only on the pair $\{i,j\}$ and not the ordering). So $c$ checks if $\sigma$ inverts $\{i,j\}$) and observe that the $1$-cocycle condition just says that $\sigma\tau$ inverts $i,j$ if and only if exactly one of $\tau$ inverting $i,j$ and $\sigma$ inverting $\tau(i),\tau(j)$ occurs. Formally, it boils down to $$c(\sigma\tau,\{i,j\}) = \frac{\sigma\tau(i)-\sigma\tau(j)}{i-j} = \frac{\sigma\tau(i)-\sigma\tau(j)}{\tau(i)-\tau(j)}\cdot \frac{\tau(i)-\tau(j)}{i-j} = c(\sigma,\{\tau(i),\tau(j)\})c(\tau,\{i,j\}).$$ Then $c((1,2),\{i,j\})=-1$ iff $\{i,j\}=\{1,2\}$ and so $\theta((1,2))=-1$.

Categorical remark. In categorical terms, you can build the transformation groupoid (aka Grothendieck construction, aka category of elements) $G\ltimes X$ and a $1$-cocycle is precisely a functor, that is, an element of $H^1(G\ltimes X,A)\cong H^1(G,A^X)$ (by Shapiro's lemma), the latter of which maps to $H^1(G,A)$ via the augmentation $A^X\to A$ sending $f$ to $\prod_{x\in X}f(x)$ like in my other answer. In fancier terms, $B(G\ltimes X)$ is the finite sheet covering space of $BG$ corresponding to the $G$-set $X$, and so you can use the transfer map to get $\mathrm{Hom}(G\ltimes X,A)\cong H^1(B(G\ltimes X),A)\to H^1(BG,A)\cong \mathrm{Hom}(G,A)$.

Topological remark. As essentially mentioned in DeBruyn's latest version of his answer (in categorical language), we can view this all topologically (but then I get switched to right actions). Let $BS_n$ be the classifying space of $S_n$ built via the bar construction. Let $X$ the right $S_n$-set of all two-element subsets of $\{1,\ldots, n\}$. Then this is a transitive $S_n$-set so corresponds to a connected finite sheet covering $E\to BS_n$ whose fundamental group is isomorphic to a stabilizer, which in turn is isomorphic to $S_2\times S_{n-2}$. Moreover, $E$ is a classifying space for its fundamental group since its universal covering space, that of $BS_n$, is contractible. Since $H^1(S_2\times S_{n-2},\{\pm 1\})$ is obviously nontrivial via the projection to $S_2$, this gives a nontrivial $1$-cocycle $c$ on $E$. Edges of $E$ look like $(e,\sigma)\colon e\to e\sigma$ where $e\in X$, $\sigma \in S_n$ and the corresponding $2$-cocycle gives $-1$ if $\sigma$ inverts $e$ and $1$, else. Now the transfer map sends this cocycle to an element of $H^1(BS_n,\{\pm 1\})=\mathrm{Hom}(S_n,\{\pm 1\})$ which is easily verified by the construction of transfer to send $\sigma$ to the product of the $c(e,\sigma)$ over all $e$, which is $-1$ raised to the number of inversions.

Here is another variation on the other answers (particularly, the Cartier approach). Let $G$ be a group acting on the left of a finite set $X$, let $A$ an abelian group and let $c\colon G\times X\to A$ be a "$1$-cocycle", meaning that $c(gh,x) = c(g,hx)c(h,x)$ for all $x\in X$ and $g,h\in H$. You can then define a homomorphism $\theta\colon G\to A$ by the rule $$\theta(g) = \prod_{x\in X}c(g,x).$$ Although we are supposed to rule out computation, this is easy: $$\theta(gh)=\prod_{x\in X}c(gh,x)=\prod_{x\in X}c(g,hx)c(h,x) =\prod_{y\in X}c(g,y)\prod_{x\in X}c(h,x) =\theta(g)\theta(h)$$ by putting $y=hx$.

Now apply this to $S_n$ acting on the set $X$ of $2$-element subsets of $\{1,\ldots, n\}$ with $A=\{\pm 1\}$. For $i<j$, put $$c(\sigma,\{i,j\})= \begin{cases} 1, & \text{if}\ \sigma(i)<\sigma(j)\\ -1, &\text{if}\ \sigma(i)>\sigma(j)\end{cases}$$ (so $c$ checks if $\sigma$ inverts $\{i,j\}$) and observe that the $1$-cocycle condition just says that $\sigma\tau$ inverts $i,j$ if and only if exactly one of $\tau$ inverting $i,j$ and $\sigma$ inverting $\tau(i),\tau(j)$ occurs. Then $c((1,2),\{i,j\})=-1$ iff $\{i,j\}=\{1,2\}$ and so $\theta((1,2))=-1$.

Categorical remark. In categorical terms, you can build the transformation groupoid (aka Grothendieck construction, aka category of elements) $G\ltimes X$ and a $1$-cocycle is precisely a functor, that is, an element of $H^1(G\ltimes X,A)\cong H^1(G,A^X)$ (by Shapiro's lemma), the latter of which maps to $H^1(G,A)$ via the augmentation $A^X\to A$ sending $f$ to $\prod_{x\in X}f(x)$ like in my other answer. In fancier terms, $B(G\ltimes X)$ is the finite sheet covering space of $BG$ corresponding to the $G$-set $X$, and so you can use the transfer map to get $\mathrm{Hom}(G\ltimes X,A)\cong H^1(B(G\ltimes X),A)\to H^1(BG,A)\cong \mathrm{Hom}(G,A)$.

Topological remark. As essentially mentioned in DeBruyn's latest version of his answer (in categorical language), we can view this all topologically (but then I get switched to right actions). Let $BS_n$ be the classifying space of $S_n$ built via the bar construction. Let $X$ the right $S_n$-set of all two-element subsets of $\{1,\ldots, n\}$. Then this is a transitive $S_n$-set so corresponds to a connected finite sheet covering $E\to BS_n$ whose fundamental group is isomorphic to a stabilizer, which in turn is isomorphic to $S_2\times S_{n-2}$. Moreover, $E$ is a classifying space for its fundamental group since its universal covering space, that of $BS_n$, is contractible. Since $H^1(S_2\times S_{n-2},\{\pm 1\})$ is obviously nontrivial via the projection to $S_2$, this gives a nontrivial $1$-cocycle $c$ on $E$. Edges of $E$ look like $(e,\sigma)\colon e\to e\sigma$ where $e\in X$, $\sigma \in S_n$ and the corresponding $2$-cocycle gives $-1$ if $\sigma$ inverts $e$ and $1$, else. Now the transfer map sends this cocycle to an element of $H^1(BS_n,\{\pm 1\})=\mathrm{Hom}(S_n,\{\pm 1\})$ which is easily verified by the construction of transfer to send $\sigma$ to the product of the $c(e,\sigma)$ over all $e$, which is $-1$ raised to the number of inversions.

Here is another variation on the other answers (particularly, the Cartier approach). Let $G$ be a group acting on the left of a finite set $X$, let $A$ an abelian group and let $c\colon G\times X\to A$ be a "$1$-cocycle", meaning that $c(gh,x) = c(g,hx)c(h,x)$ for all $x\in X$ and $g,h\in H$. You can then define a homomorphism $\theta\colon G\to A$ by the rule $$\theta(g) = \prod_{x\in X}c(g,x).$$ Although we are supposed to rule out computation, this is easy: $$\theta(gh)=\prod_{x\in X}c(gh,x)=\prod_{x\in X}c(g,hx)c(h,x) =\prod_{y\in X}c(g,y)\prod_{x\in X}c(h,x) =\theta(g)\theta(h)$$ by putting $y=hx$.

Now apply this to $S_n$ acting on the set $X$ of $2$-element subsets of $\{1,\ldots, n\}$ with $A=\{\pm 1\}$. For $i<j$, put $$c(\sigma,\{i,j\})= \begin{cases} 1, & \text{if}\ \sigma(i)<\sigma(j)\\ -1, &\text{if}\ \sigma(i)>\sigma(j)\end{cases}$$ (so $c$ checks if $\sigma$ inverts $\{i,j\}$) and observe that the $1$-cocycle condition just says that $\sigma\tau$ inverts $i,j$ if and only if exactly one of $\tau$ inverting $i,j$ and $\sigma$ inverting $\tau(i),\tau(j)$ occurs. Then $c((1,2),\{i,j\})=-1$ iff $\{i,j\}=\{1,2\}$ and so $\theta((1,2))=-1$.

Categorical remark. In categorical terms, you can build the transformation groupoid (aka Grothendieck construction, aka category of elements) $G\ltimes X$ and a $1$-cocycle is precisely a functor, that is, an element of $H^1(G\ltimes X,A)\cong H^1(G,A^X)$ (by Shapiro's lemma), the latter of which maps to $H^1(G,A)$ via the augmentation $A^X\to A$ sending $f$ to $\prod_{x\in X}f(x)$ like in my other answer. In fancier terms, $B(G\ltimes X)$ is the finite sheet covering space of $BG$ corresponding to the $G$-set $X$, and so you can use the transfer map to get $\mathrm{Hom}(G\ltimes X,A)\cong H^1(B(G\ltimes X),A)\to H^1(BG,A)\cong \mathrm{Hom}(G,A)$.

Topological remark. As essentially mentioned in DeBruyn's latest version of his answer (in categorical language), we can view this all topologically (but then I get switched to right actions). Let $BS_n$ be the classifying space of $S_n$ built via the bar construction. Let $X$ the right $S_n$-set of all two-element subsets of $\{1,\ldots, n\}$. Then this is a transitive $S_n$-set so corresponds to a connected finite sheet covering $E\to BS_n$ whose fundamental group is isomorphic to a stabilizer, which in turn is isomorphic to $S_2\times S_{n-2}$. Moreover, $E$ is a classifying space for its fundamental group since its universal covering space, that of $BS_n$, is contractible. Since $H^1(S_2\times S_{n-2},\{\pm 1\})$ is obviously nontrivial via the projection to $S_2$, this gives a nontrivial $1$-cocycle $c$ on $E$. Edges of $E$ look like $(e,\sigma)\colon e\to e\sigma$ where $e\in X$, $\sigma \in S_n$ and the corresponding $2$-cocycle gives $-1$ if $\sigma$ inverts $e$ and $1$, else. Now the transfer map sends this cocycle to an element of $H^1(BS_n,\{\pm 1\})=\mathrm{Hom}(S_n,\{\pm 1\})$ which is easily verified by the construction of transfer to send $\sigma$ to the product of the $c(e,\sigma)$ over all $e$, which is $-1$ raised to the number of inversions.

Here is another variation on the other answers (particularly, the Cartier approach and DeBruyn's current approach). Let $G$ be a group acting on the left of a finite set $X$, let $A$ an abelian group and let $c\colon G\times X\to A$ be a "$1$-cocycle", meaning that $c(gh,x) = c(g,hx)c(h,x)$ for all $x\in X$ and $g,h\in H$. You can then define a homomorphism $\theta\colon G\to A$ by the rule $$\theta(g) = \prod_{x\in X}c(g,x).$$ Although we are supposed to rule out computation, this is easy: $$\theta(gh)=\prod_{x\in X}c(gh,x)=\prod_{x\in X}c(g,hx)c(h,x) =\prod_{y\in X}c(g,y)\prod_{x\in X}c(h,x) =\theta(g)\theta(h)$$ by putting $y=hx$.

Now apply this to $S_n$ acting on the set $X$ of $2$-element subsets of $\{1,\ldots, n\}$ with $A=\{\pm 1\}$. Put $$c(\sigma,\{i,j\})= \frac{\sigma(i)-\sigma(j)}{i-j}$$ (this depends only on the pair $\{i,j\}$ and not the ordering). So $c$ checks if $\sigma$ inverts $\{i,j\}$) and observe that the $1$-cocycle condition just says that $\sigma\tau$ inverts $i,j$ if and only if exactly one of $\tau$ inverting $i,j$ and $\sigma$ inverting $\tau(i),\tau(j)$ occurs. Formally, it boils down to $$c(\sigma\tau,\{i,j\}) = \frac{\sigma\tau(i)-\sigma\tau(j)}{i-j} = \frac{\sigma\tau(i)-\sigma\tau(j)}{\tau(i)-\tau(j)}\cdot \frac{\tau(i)-\tau(j)}{i-j} = c(\sigma,\{\tau(i),\tau(j)\})c(\tau,\{i,j\}).$$ Then $c((1,2),\{i,j\})=-1$ iff $\{i,j\}=\{1,2\}$ and so $\theta((1,2))=-1$.

Categorical remark. In categorical terms, you can build the transformation groupoid (aka Grothendieck construction, aka category of elements) $G\ltimes X$ and a $1$-cocycle is precisely a functor, that is, an element of $H^1(G\ltimes X,A)\cong H^1(G,A^X)$ (by Shapiro's lemma), the latter of which maps to $H^1(G,A)$ via the augmentation $A^X\to A$ sending $f$ to $\prod_{x\in X}f(x)$ like in my other answer. In fancier terms, $B(G\ltimes X)$ is the finite sheet covering space of $BG$ corresponding to the $G$-set $X$, and so you can use the transfer map to get $\mathrm{Hom}(G\ltimes X,A)\cong H^1(B(G\ltimes X),A)\to H^1(BG,A)\cong \mathrm{Hom}(G,A)$.

Topological remark. As essentially mentioned in DeBruyn's latest version of his answer (in categorical language), we can view this all topologically (but then I get switched to right actions). Let $BS_n$ be the classifying space of $S_n$ built via the bar construction. Let $X$ the right $S_n$-set of all two-element subsets of $\{1,\ldots, n\}$. Then this is a transitive $S_n$-set so corresponds to a connected finite sheet covering $E\to BS_n$ whose fundamental group is isomorphic to a stabilizer, which in turn is isomorphic to $S_2\times S_{n-2}$. Moreover, $E$ is a classifying space for its fundamental group since its universal covering space, that of $BS_n$, is contractible. Since $H^1(S_2\times S_{n-2},\{\pm 1\})$ is obviously nontrivial via the projection to $S_2$, this gives a nontrivial $1$-cocycle $c$ on $E$. Edges of $E$ look like $(e,\sigma)\colon e\to e\sigma$ where $e\in X$, $\sigma \in S_n$ and the corresponding $2$-cocycle gives $-1$ if $\sigma$ inverts $e$ and $1$, else. Now the transfer map sends this cocycle to an element of $H^1(BS_n,\{\pm 1\})=\mathrm{Hom}(S_n,\{\pm 1\})$ which is easily verified by the construction of transfer to send $\sigma$ to the product of the $c(e,\sigma)$ over all $e$, which is $-1$ raised to the number of inversions.

Here is another variation on the other answers (particularly, the Cartier approach). Let $G$ be a group acting on the left of a finite set $X$, let $A$ an abelian group and let $c\colon G\times X\to A$ be a "$1$-cocycle", meaning that $c(gh,x) = c(g,hx)c(h,x)$ for all $x\in X$ and $g,h\in H$. You can then define a homomorphism $\theta\colon G\to A$ by the rule $$\theta(g) = \prod_{x\in X}c(g,x).$$ Although we are supposed to rule out computation, this is easy: $$\theta(gh)=\prod_{x\in X}c(gh,x)=\prod_{x\in X}c(g,hx)c(h,x) =\prod_{y\in X}c(g,y)\prod_{x\in X}c(h,x) =\theta(g)\theta(h)$$ by putting $y=hx$.

Now apply this to $S_n$ acting on the set $X$ of $2$-element subsets of $\{1,\ldots, n\}$ with $A=\{\pm 1\}$. For $i<j$, put $$c(\sigma,\{i,j\})= \begin{cases} 1, & \text{if}\ \sigma(i)<\sigma(j)\\ -1, &\text{if}\ \sigma(i)>\sigma(j)\end{cases}$$ (so $c$ checks if $\sigma$ inverts $\{i,j\}$) and observe that the $1$-cocycle condition just says that $\sigma\tau$ inverts $i,j$ if and only if exactly one of $\tau$ inverting $i,j$ and $\sigma$ inverting $\tau(i),\tau(j)$ occurs. Then $c((1,2),\{i,j\})=-1$ iff $\{i,j\}=\{1,2\}$ and so $\theta((1,2))=-1$.

Categorical remark. In categorical terms, you can build the transformation groupoid (aka Grothendieck construction, aka category of elements) $G\ltimes X$ and a $1$-cocycle is precisely a functor, that is, an element of $H^1(G\ltimes X,A)\cong H^1(G,A^X)$ (by Shapiro's lemma), the latter of which maps to $H^1(G,A)$ via the augmentation $A^X\to A$ sending $f$ to $\prod_{x\in X}f(x)$ like in my other answer. In fancier terms, $B(G\ltimes X)$ is the finite sheet covering space of $BG$ corresponding to the $G$-set $X$, and so you can use the transfer map to get $\mathrm{Hom}(G\ltimes X,A)\cong H^1(B(G\ltimes X),A)\to H^1(BG,A)\cong \mathrm{Hom}(G,A)$.

Topological remark. As essentially mentioned in DeBruyn's latest version of his answer (in categorical language), we can view this all topologically (but then I get switched to right actions). Let $BS_n$ be the classifying space of $S_n$ built via the bar construction. Let $X$ the right $S_n$-set of all two-element subsets of $\{1,\ldots, n\}$. Then this is a transitive $S_n$-set so corresponds to a connected finite sheet covering $E\to BS_n$ whose fundamental group is isomorphic to a stabilizer, which in turn is isomorphic to $S_2\times S_{n-1}$. Moreover, $E$ is a classifying space for its fundamental group since its universal covering space, that of $BS_n$, is contractible. Since $H^1(S_2\times S_{n-1},\{\pm 1\})$ is obviously nontrivial via the projection to $S_2$, this gives a nontrivial $1$-cocycle $c$ on $E$. Edges of $E$ look like $(e,\sigma)\colon e\to e\sigma$ where $e\in X$, $\sigma \in S_n$ and the corresponding $2$-cocycle gives $-1$ if $\sigma$ inverts $e$ and $1$, else. Now the transfer map sends this cocycle to an element of $H^1(BS_n,\{\pm 1\})=\mathrm{Hom}(S_n,\{\pm 1\})$ which is easily verified by the construction of transfer to send $\sigma$ to the product of the $c(e,\sigma)$ over all $e$, which is $-1$ raised to the number of inversions.

Here is another variation on the other answers (particularly, the Cartier approach). Let $G$ be a group acting on the left of a finite set $X$, let $A$ an abelian group and let $c\colon G\times X\to A$ be a "$1$-cocycle", meaning that $c(gh,x) = c(g,hx)c(h,x)$ for all $x\in X$ and $g,h\in H$. You can then define a homomorphism $\theta\colon G\to A$ by the rule $$\theta(g) = \prod_{x\in X}c(g,x).$$ Although we are supposed to rule out computation, this is easy: $$\theta(gh)=\prod_{x\in X}c(gh,x)=\prod_{x\in X}c(g,hx)c(h,x) =\prod_{y\in X}c(g,y)\prod_{x\in X}c(h,x) =\theta(g)\theta(h)$$ by putting $y=hx$.

Now apply this to $S_n$ acting on the set $X$ of $2$-element subsets of $\{1,\ldots, n\}$ with $A=\{\pm 1\}$. For $i<j$, put $$c(\sigma,\{i,j\})= \begin{cases} 1, & \text{if}\ \sigma(i)<\sigma(j)\\ -1, &\text{if}\ \sigma(i)>\sigma(j)\end{cases}$$ (so $c$ checks if $\sigma$ inverts $\{i,j\}$) and observe that the $1$-cocycle condition just says that $\sigma\tau$ inverts $i,j$ if and only if exactly one of $\tau$ inverting $i,j$ and $\sigma$ inverting $\tau(i),\tau(j)$ occurs. Then $c((1,2),\{i,j\})=-1$ iff $\{i,j\}=\{1,2\}$ and so $\theta((1,2))=-1$.

Categorical remark. In categorical terms, you can build the transformation groupoid (aka Grothendieck construction, aka category of elements) $G\ltimes X$ and a $1$-cocycle is precisely a functor, that is, an element of $H^1(G\ltimes X,A)\cong H^1(G,A^X)$ (by Shapiro's lemma), the latter of which maps to $H^1(G,A)$ via the augmentation $A^X\to A$ sending $f$ to $\prod_{x\in X}f(x)$ like in my other answer. In fancier terms, $B(G\ltimes X)$ is the finite sheet covering space of $BG$ corresponding to the $G$-set $X$, and so you can use the transfer map to get $\mathrm{Hom}(G\ltimes X,A)\cong H^1(B(G\ltimes X),A)\to H^1(BG,A)\cong \mathrm{Hom}(G,A)$.

Topological remark. As essentially mentioned in DeBruyn's latest version of his answer (in categorical language), we can view this all topologically (but then I get switched to right actions). Let $BS_n$ be the classifying space of $S_n$ built via the bar construction. Let $X$ the right $S_n$-set of all two-element subsets of $\{1,\ldots, n\}$. Then this is a transitive $S_n$-set so corresponds to a connected finite sheet covering $E\to BS_n$ whose fundamental group is isomorphic to a stabilizer, which in turn is isomorphic to $S_2\times S_{n-2}$. Moreover, $E$ is a classifying space for its fundamental group since its universal covering space, that of $BS_n$, is contractible. Since $H^1(S_2\times S_{n-2},\{\pm 1\})$ is obviously nontrivial via the projection to $S_2$, this gives a nontrivial $1$-cocycle $c$ on $E$. Edges of $E$ look like $(e,\sigma)\colon e\to e\sigma$ where $e\in X$, $\sigma \in S_n$ and the corresponding $2$-cocycle gives $-1$ if $\sigma$ inverts $e$ and $1$, else. Now the transfer map sends this cocycle to an element of $H^1(BS_n,\{\pm 1\})=\mathrm{Hom}(S_n,\{\pm 1\})$ which is easily verified by the construction of transfer to send $\sigma$ to the product of the $c(e,\sigma)$ over all $e$, which is $-1$ raised to the number of inversions.