Properties of first degree homogeneous functions

This is a math question, even if it may seem an economics one. I'll try to explain all the economics in this question.

I've got the following production function, where $Y$ is the product, $L$ is the labour used in production, while $K$ is the capital. The $M$ subscript just stands for "manifacturing sector".

$Y_M=f_M (L_M, K_M)$ (1)

which is assumed to be a first degree homogeneous function. With this assumption, we may write

$Y_M=L_M m (k_M)$ (2)

Where $k_M=\dfrac{K_M}{L_M}$ and $m$ is the average productivity of labour.

Which transformation has been performed in order to obtain (2) and what's the influence of the function's first-degree homogeneity in it?

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A homogeneous function $f(x,y)$ of degree one is any function $f(x,y)$ which satisfies the rule $f(kx,ky)=kf(x,y)$, for any choice of $k$. In your case, you want to pull out the $x$ to the front as a multiplier.
Now note that $$f(x,y)=f(x,x(y/x))$$ and this last has the common factor $x$ in both variable positions, so that $x$ is ready to be "pulled out" like the $k$ was in the above definition. This means we have $$f(x,y)=f(x,x(y/x))=xf(1,y/x).$$ We can now give a new name to the function $f(1,y/x)$ and say it is $g(u)$ where $u=y/x$. That is, to make it more clear, we can just define the function $g(u)$ in terms of the original $f(x,y)$ by the definition $g(u)=f(1,u)$. Then you will have $f(x,y)=xg(y/x)$.
To correspond things to your notation, drop the subscript $M$ on the $f$, let the numbers $L,K$ correspond to $x,y$, so that also $K/L$ corresponds to $y/x$. I hope that is something of the explanation you were looking for...
By the way, the reason this depends on homogeneous of specifically degree one is that, at the stage where the constant $k$ is moved in front, a homogeneous function of degree say 2 would satisfy $f(kx,ky)=k^2f(x,y)$, which doesn't match what you want.
Thanks. Almost clear, the only thing I don't understand is: where does that $m$ come from? It's the average product, so it should be $\frac{f(k)}{L}$ – Luigi Oct 29 '12 at 19:16
Dropping all the subscripts, since I have $Y=f(K,L)$, I pull out the L and I get $Y=Lf(\frac{K}{L}, 1)$ $Y=Lf(k)$ – Luigi Oct 29 '12 at 19:31
I think in your original description the letter $m$ is not a number but represents the name given to the function $f(x,1)$ or $f(1,x)$ [depending on which variable you choose to "pull out"]. So $m(x)$ is like the $g(x)$ in my "generic" description. – coffeemath Oct 29 '12 at 23:29