Apparent inconsistency in geometric product associativity With a little bit of work, I have proven to myself that the geometric product between three vectors is associative ($a$, $b$, and $c$ are 1-vectors):
$$\begin{aligned}(ab)c &= a(bc)  \\ &= (b \cdot c) a - (a \cdot c) b + (a \cdot b) c + a \wedge b \wedge c.\end{aligned}$$
And, by extension, that it is consistent for four vectors ($a$, $b$, $c$, $d$) as:
$$\begin{aligned}   (abc)d    &= a(bcd) \\    &= (b \cdot c)(a \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot b)(c \cdot d) \\    &\quad+ (b \cdot c)a \wedge d - (a \cdot c)b \wedge d + (a \cdot b)c \wedge d \\    &\quad+ b \wedge c(a \cdot d) - a \wedge c(b \cdot d) + a \wedge b(c \cdot d) \\    &\quad+ a \wedge b \wedge c \wedge d\end{aligned}$$
However, the following, which I would expect to match the rules for associativity, is quite different:
$$\begin{aligned}   (ab)(cd) &= (a \cdot b + a \wedge b)(c \cdot d + c \wedge d) \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + (a \wedge b)(c \wedge d) \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + (a \wedge b).(c \wedge d) + a \wedge b \wedge c \wedge d \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + a \cdot (b \cdot (c \wedge d)) + a \wedge b \wedge c \wedge d \\    &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) \\       &\quad + (a \cdot d)(b \cdot c) - (a \cdot c)(b \cdot d) \\       &\quad + a \wedge b \wedge c \wedge d \\ &= (b \cdot c)(a \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot b)(c \cdot d) \\ &\quad + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) \\ &\quad + a \wedge b \wedge c \wedge d.\end{aligned}$$
Which, when compared with $a(bcd)$ and $(abc)d$ has all pure scalars and the quad-vector, but is missing four of the six scaled bi-vectors.
Is it that my understanding of associativity is incorrect in that $(ab)(cd)$ is not the same as $a(bcd)$ or $(abc)d$, did I make a mistake in my algebra, or is there something else going on?
My digging into this comes from the desire to understand $e_{12} e_{23}$ which is short-hand for $(e_1 \wedge e_2)(e_2 \wedge e_3)$ (note, geometric product for $()()$) where $e_1$, $e_2$, $e_3$ are orthonormal basis vectors and I have been unable to determine whether the result should be $0$ or $e_{13}$. Also, that $e_1 e_2 e_2 e_3$ comes up in my working out of geometric products when $e_1 e_2$ is used as shorthand for $e_1 \wedge e_2$.
Follow-up:
Many thanks to Somos for encouraging me to keep looking and to Peeter Joot for his hint about ${\langle ( \wedge )( \wedge ) \rangle}_2$ as the apparent circular definition made me dig into just what $(a \wedge b)(c \wedge d)$ looks like.
Extrapolating from appendix C.1 of Geometric Algebra for Computer Science (Dorst, Fontijne, Mann).
$$
(a \wedge b)(c \wedge d) = \frac{1}{4} (ab - ba)(cd - dc)
$$
This expands out into four permutations of $abcd$:
$$
abcd - abdc - bacd + badc
$$
I won't go into the full expansion, but it results in 12 grade-0 terms (4 of which cancel), 24 grade-2 terms (8 of which cancel), and 4 grade-4 terms. The terms that don't cancel add together, thus requiring the $\frac{1}{4}$. The terms that do cancel are exactly those terms in my $(ab)(cd)$ expansion remaining after removing $(a \wedge b)(c \wedge d)$.
Thus, while the geometric product is defined to be associative, this exercise shows that associativity in action, which is what I needed.
 A: It looks like you are missing a factor in your expansion of $ \left( {a \wedge b} \right) \left( {c \wedge d} \right) $.  In general, a product of bivectors $ A B $, should have scalar, bivector, and quadvector components:
$$A B = A \cdot B + {\left\langle{{ A B }}\right\rangle}_{2} + A \wedge B,$$
so for $ A = a \wedge b, B = c \wedge d $, we have
$$   \left( { a \wedge b } \right)   \left( { c \wedge d } \right)   =   \left( { a \wedge b } \right)   \cdot   \left( { c \wedge d } \right)   +   {\left\langle{{   \left( { a \wedge b } \right)   \left( { c \wedge d } \right)}}\right\rangle}_{2}   +   \left( { a \wedge b } \right)   \wedge   \left( { c \wedge d } \right).$$
You can drop the braces in the final wedge, but that grade-two term should provide the missing bivector terms.
On your rationale.
If you only want to understand $(e_1 \wedge e_2)(e_2 \wedge e_3)$, it's helpful to revert to the GA axioms, two of which are:

*

*Associativity

*Contraction: $x^2 = x \cdot x$, for any vectors x.

With those, plus $x \wedge y = x y$, for vectors $x, y$ where $x \cdot y = 0$, you have
$$\begin{aligned}\left( {e_1 \wedge e_2} \right) \left( { e_2 \wedge e_3 } \right)&=\left( {e_1 e_2} \right) \left( { e_2 e_3 } \right) \\ &=e_1 e_2 e_2 e_3  \\ &=e_1 \left( { e_2 e_2 } \right) e_3  \\ &=e_1 e_3.\end{aligned}$$
As the dot and wedges of a pair of bivectors are the grade-0 and grade-4 terms respectively (by definition), you can immediately conclude that
$$\left( {e_1 \wedge e_2} \right) \cdot \left( { e_2 \wedge e_3 } \right) = 0,$$
$$\left( {e_1 \wedge e_2} \right) \wedge \left( { e_2 \wedge e_3 } \right) = 0,$$
and
$$   {\left\langle{{\left( {e_1 \wedge e_2} \right) \left( { e_2 \wedge e_3 } \right)}}\right\rangle}_{2} = e_1 e_3.$$
One expansion.
Here's one way to expand the product of four vectors, grouping the first and last pairs.
$$\begin{aligned}a b c d &=\left( { a \cdot b + a \wedge b } \right) \left( { c \cdot d + c \wedge d } \right) \\ &=\left( { a \cdot b } \right) \left( { c \cdot d } \right) \\ &+\quad\left( { a \cdot b } \right) \left( { c \wedge d } \right)+\left( { c \cdot d } \right) \left( { a \wedge b } \right)  \\ &+\quad \left( { a \wedge b } \right)\left( { c \wedge d } \right).\end{aligned}$$
The first term of this bivector-bivector product can be expanded with application of $ a b = a \cdot b + a \wedge b $ in reverse
$$\begin{aligned}\left( { a \wedge b } \right)\left( { c \wedge d } \right)&=\left( { a b - a \cdot b } \right)\left( { c \wedge d } \right) \\ &=a b \left( { c \wedge d } \right) - \left( { a \cdot b } \right) \left( { c \wedge d } \right) \end{aligned}$$
The product of $ b $ with $ c \wedge d $ is
$$\begin{aligned}b \left( { c \wedge d } \right) &=b \cdot\left( { c \wedge d } \right) +b \wedge\left( { c \wedge d } \right)  \\ &=\left( { b \cdot c } \right) d- \left( { b \cdot d } \right) c+ b \wedge c \wedge d.\end{aligned}$$
Multiplying this by $ a $ on the left, we have a scalar grade
$$   \left\langle{{   a b \left( { c \wedge d } \right) }}\right\rangle= \left( { b \cdot c } \right) \left( { a \cdot d } \right) - \left( { b \cdot d } \right) \left( { a \cdot c } \right),$$
a bivector grade
$$\begin{aligned}{\left\langle{{   a b \left( { c \wedge d } \right) }}\right\rangle}_{2}&=\left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right)+ a \cdot \left( { b \wedge c \wedge d } \right) \\ &=\left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right) \\ &\quad + \left( { a \cdot b } \right) \left( { c \wedge d } \right) - \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right),\end{aligned}$$
and a grade-four component
$${\left\langle{{   a b \left( { c \wedge d } \right) }}\right\rangle}_{4}=   a \wedge b \wedge c \wedge d.$$
Putting all the pieces together, here are all the grades of the $ a b c d $ product
$$\begin{aligned}   \left\langle{{( a b)( c d) }}\right\rangle    &=\left( { a \cdot b } \right) \left( { c \cdot d } \right) +\left( { b \cdot c } \right) \left( { a \cdot d } \right) - \left( { b \cdot d } \right) \left( { a \cdot c } \right) \\    {\left\langle{{ (a b)( c d) }}\right\rangle}_{2}    &=  \left( { a \cdot b } \right) \left( { c \wedge d } \right)+ \left( { c \cdot d } \right) \left( { a \wedge b } \right)  - \left( { a \cdot b } \right) \left( { c \wedge d } \right) \\ &\quad+ \left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right)+ \left( { a \cdot b } \right) \left( { c \wedge d } \right) \\ &\quad- \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right) \\    &=  \left( { a \cdot b } \right) \left( { c \wedge d } \right)+ \left( { c \cdot d } \right) \left( { a \wedge b } \right)  + \left( { b \cdot c } \right) \left( { a \wedge d } \right) \\ &\quad - \left( { b \cdot d } \right) \left( { a \wedge c } \right)- \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right) \\    {\left\langle{{ (a b)( c d) }}\right\rangle}_{4}    &=   a \wedge b \wedge c \wedge d.\end{aligned},$$
which matches your result, finishing your experimental proof of associative multiplication for a product of four vectors.
