$Theorem$ Let K be a commutative ring with identity and let V be a module over K. Then the exterior product is an associative operation on the alternating multilinear forms on V. In other words, if L, M, and N are alternating multilinear forms on V of degrees r, s, and t, respectively, then

$(Lwedge M)wedge N=Lwedge (Mwedge N)$

$proof$

Let $G(r,s,t)$ be the subgroup of $S_{r+s+t}$ that consists of the permutations which permute the sets

{1,…,r}, {r + 1,…,r + s}, {r+s+1,…,r+s+t}

within themselves. Then $(sgnmu){(Lotimes Motimes N)}_mu$ is the same multilinear function for all $mu $ in a given left coset of $G(r,s,t)$ Choose one element from each left coset of $G(r,s,t)$, and let E be the sum of the corresponding terms $(sgnmu){(Lotimes Motimes N)}_mu$. Then E is independent of the way in which the representatives ju are chosen, and

$r!s!t!E=pi _{r+s+t}(Lotimes Motimes N)$.

We shall show that $(Lwedge M)wedge N$ and $Lwedge (Mwedge N)$ are both equal to $E$.

Let $G(r+s,t)$ be the subgroup of $S_{r+s+t}$ that permutes the sets

{1,…,r + s}, {r+s+1,…,r+s+t}

within themselves.

Let $T$ be any set of permutations of {1,…,r+s+t} which contains exactly one element from each left coset of $G(r+s,t)$.

By (5-50)

$(Lwedge M)wedge N$=$sum (sgntau )$$((Lwedge M)wedge N)_tau $

extended over $tau$ in $T$. Now let $G(r,s)$ be the subgroup of $S_{r+s}$ that permutes the sets

{1,…,r},{r+1,…,r+s}

within themselves. Let S be any set of permutations of {1,…,r+s} which contains exactly one element from each left coset of $G(r,s)$. From (5-50) and what we have shown above, it follows that

$(Lwedge M)wedge N$=$sum (sgnsigma )(sgntau ){({(Lotimes M)}_{sigma }otimes N)}_{tau }$

extended over all pairs $(sigma ,tau )$ in $Stimes T$. If we agree to identify each $sigma $ in $S_{r+s}$, with the element of $S_{r+s+t}$ which agrees with $sigma $ on {1,. . . ,r+s} and is the identity on {r+s+1,…,r+s+t}, then we may write

$(Lwedge M)wedge N$=$sum sgn(sigma tau ){({(Lotimes Motimes N)}_{sigma })}_{tau }$

But,

${({(Lotimes Motimes N)}_{sigma })}_{tau }$=${{(Lotimes Motimes N)}}_{tau sigma }$

Therefore

$(Lwedge M)wedge N$=$sum sgn(sigma tau ){{(Lotimes Motimes N)}}_{tau sigma }$

Now suppose we have

$tau _1sigma _1=tau _2sigma _2gamma $

with $sigma _i$ in $S$, $tau _i$ in $T$, and $gamma $ in $G(r,s,t)$. Then ${tau _2}^{-1}tau _1=sigma _2gamma {sigma _1}^{-1}$, and since $sigma _2gamma {sigma _1}^{-1}$ lies in $G(r+s,t)$, it follows that $tau _1$ and $tau _2$ are in the same left coset of $G(r+s,t)$. Therefore, $tau _1=tau _2$, and $sigma _1=sigma _2gamma $. But this implies that $sigma _1$ and $sigma _2$ (regarded as elements of $S_{r+s}$) lie in the same coset of $G(r, s)$ ; hence $sigma _1=sigma _2$. Therefore, the products $tau sigma $ corresponding to the

$frac{(r+s+t)!}{(r+s)!t!}frac{(r+s)!}{r!s!}$

pairs ($tau ,sigma $)in $Ttimes S$ are all distinct and lie in distinct cosets of $G(r, s, t)$.

Since there are exactly

$frac{(r+s+t)!}{r!s!t!}$

left cosets of $G(r,s,t)$ in $S_{r+s+t}$, it follows that $(Lwedge M) wedge N)=E$ .By an analogous argument, $Lwedge (Mwedge N)=E$ as well.The end

$question$

I have a question about the following part of this proof:

**Now suppose we have

$tau _1sigma _1=tau _2sigma _2gamma $

with $sigma _i$ in $S$, $tau _i$ in $T$, and $gamma $ in $G(r,s,t)$.**

What is the reason for assuming that ” $tau _1sigma _1=tau _2sigma _2gamma $” ? And I don’t understand how this assumption suggests ” the products $tau sigma $ corresponding to the

$frac{(r+s+t)!}{(r+s)!t!}frac{(r+s)!}{r!s!}$

pairs ($tau ,sigma $)in $Ttimes S$ are all distinct and lie in distinct cosets of $G(r, s, t)$.”