# Product of $\geq 2^r$ elements can be written as a simple product of $r$ elements?

This comes from Jacobson's Basic Algebra I, Exercise 4 of Section 1.4, found on page 42. I don't understand the following problem.

For a given binary composition define a simple product of the sequence of elements $a_1,a_2,\dots,a_n$ inductively as either $a_1u$ where $u$ is a simple product of $a_2,\dots,a_n$ or as $va_n$ where $v$ is a simple product of $a_1,\dots,a_{n-1}$. Show that any product of $\geq 2^r$ elements can be written as a simple product of $r$ elements (which are themselves products).

I thought about approaching this by induction on $r$. But then for $r=1$, it seems to ask that any product of $\geq 2$ elements can be written as a simple product of $1$ element, which sounds unusual to me. The inductive definition given doesn't make sense to me, is there a better way to approach this problem?

-
This requires knowing what he means by the unqualified term product; without that we don’t know what is meant by a ‘product of $\ge 2^r$ elements’. –  Brian M. Scott May 21 '12 at 5:36
This looked so weird, I was sure you copied the problem wrong, so I had a look at the book. Nope, you copied it quite correctly. I hope someone figures out what's up here, as now you've got me wondering. –  Gerry Myerson May 21 '12 at 5:38
@GerryMyerson The thing which comes to my mind is that this could be a question about inserting brackets (associativity not assumed)? –  Mark Bennet May 21 '12 at 5:44
@Mark: That was my immediate thought as well. Even so I suspect that the definition is defective and should specify that all $ab$ with $a$ and $b$ not products are simple products, and that the induction is for $n>2$. –  Brian M. Scott May 21 '12 at 5:50
@Mark, certainly associativity is not assumed. For anyone else who wants to check, this is Exercise 4 in section 1.4, page 42 in the first, 1974, edition. The section concerns "generalized associativity, and commutativity". –  Gerry Myerson May 21 '12 at 6:19

The definition is confusing because Jacobson gives here two different definitions at the same time. First one is this:

Simple product of $a_1, a_2, \ldots a_n$ is defined as $a_1u$, where $u$ is a simple product of $a_2, \ldots, a_n$.

The other one is this:

Simple product of $a_1, a_2, \ldots a_n$ is defined as $va_n$, where $v$ is a simple product of $a_1, \ldots, a_{n-1}$.

In the first case the product is of the form $a_1(a_2(a_3(\ldots(a_{n-1}a_n)\ldots)))$ and in the second case it is of the form $(((\ldots(a_1a_2)a_3)\ldots)a_{n-1})a_n$. Both are "simple products" as Jacobson defines them.

Suppose $b_1, b_2, \ldots, b_k$ are elements defined on some binary relation such that $k \geq 2^r$. Let $P$ be some product of these elements, ie. $b_1b_2\ldots b_k$ bracketed in some way. What the question is asking you to show is that there exist elements $a_1, a_2, \ldots a_r$ such that the simple product of $a_1, a_2, \ldots, a_r$ equals $P$.

Now notice that $P = XY$, where $X$ is some product of $b_1, \ldots, b_i$ and $Y$ is some product of $b_{i+1}, \ldots, b_k$. Then either $X$ or $Y$ has $\geq 2^{r-1}$ elements in the product, so by induction one of them is a simple product of $r-1$ elements. In both cases $P$ is a simple product of $r$ elements of one of the two forms.

-
I understood it in the way that for example $a_1((a_2a_3)a_4)$ is also a simple product of four items, but $(a_1a_2)(a_3a_4)$ of only two. –  Jyrki Lahtonen May 21 '12 at 10:05
@JyrkiLahtonen I think you can express $(a_1a_2)(a_3a_4)$ as a simple product of three items - $(a_1a_2)$ and [$a_3$ and $a_4$] –  Mark Bennet May 21 '12 at 13:26
@Mark, you're right! –  Jyrki Lahtonen May 21 '12 at 13:51
Thanks m.k., clearing up what the definitions mean was very helpful. –  Adelaide Dokras May 21 '12 at 17:53
@JyrkiLahtonen: I think you might be right, that makes a bit more sense. Jacobson might have just written it as $a_1(a_2(a_3(\ldots(a_{n-1}a_n)\ldots)))$ if the way I defined it here is right. When I was writing this answer, for some reason I thought that with the way you understood it every product of elements would also be a simple product of those elements, but as you said this is not the case with $(a_1a_2)(a_3a_4)$.. the same proof works with both definitions though. –  Mikko Korhonen May 22 '12 at 6:41

This is done by induction (if I understood it correctly). Let us assume that $z$ is a product of at least $2^r$ elements. Because $z$ is a product, it is of the form $z=xy$. Obviously either $x$ or $y$ has at least $2^{r-1}$ factors. By induction hypothesis that factor is a simple product of at least $(r-1)$ elements (most of those products themselves). There $z$ is a simple product of one of the two forms: the shorter of $x$ and $y$ is the $r^{th}$ factor.

The starting point $r=1$ being obvious.

-
Thank you Jyrki. –  Adelaide Dokras May 21 '12 at 17:54