Adem relations in one of J.F. Adams's proof I am reading now J.F. Adams's paper "The Sphere, considered as an H-space mod p", and I have the following problem with the proof of it's first lemma.
Assume that we have an $X$ - associative H-space, which cohomology ring is $H^*(X;\mathbb{Z}_p)$ is a $\Lambda(x)$ - external algebra over $\mathbb{Z}_p$, with one generator of degree $n$. Now, by the work of Dold and Lashof, we can construct a "classifying space", which cohomology ring is $\mathbb{Z}_p[y]$ with $y$ of degree $n+1$.
We have $y^p\neq 0$. In other words, $P^f_p(y)\neq 0$, where $P_p^f$ denotes reduced element of the Steenrod algebra of reduced $p$-th powers and $f=\frac{n+1}{2}$.
Now we claim, that 

by the Adem relations  $P^f_p$ can befactorized in terms of operations $P^g$, where $g=p^h$.

Why?
 A: The general statement is that every $P^n$ decomposes into a sum of compositions of the elements $P^{p^i}$ (where if $n$ is a power of $p$ this decomposition is trivial).   Note that these expressions will use inadmissible sequences of operations.
For simplicity, I will give  examples of relevant Adem relations for the case $p=2$.  This case is usually mentioned in discussions of the Hopf invariant one problem, because it implies that if $x\in H^n(X;\Bbb Z/2\Bbb Z)$ for a space $X$ with $H^i(x;\Bbb Z/2\Bbb Z)= 0$ for $i = n+1, \ldots, 2n-1$, then $x^2$ can only be nonzero if $n$ is a power of 2 (since $\textrm{Sq}^{2^j}x = 0$ for $2^j < n$).


*

*$\textrm{Sq}^1\textrm{Sq}^2 = \textrm{Sq}^3$

*$\textrm{Sq}^1\textrm{Sq}^4 = \textrm{Sq}^5$

*$\textrm{Sq}^2\textrm{Sq}^4 = \textrm{Sq}^5\textrm{Sq}^1+ \textrm{Sq}^6 $

*$\textrm{Sq}^1\textrm{Sq}^6 = \textrm{Sq}^7$


We can get expressions in the desired form from these relations. 
For instance, for $\textrm{Sq}^7$ we have 
$$\textrm{Sq}^7 = \textrm{Sq}^1\textrm{Sq}^6 = \textrm{Sq}^1(\textrm{Sq}^2\textrm{Sq}^4+\textrm{Sq}^5\textrm{Sq}^1) = \textrm{Sq}^1\textrm{Sq}^2\textrm{Sq}^4+\textrm{Sq}^1(\textrm{Sq}^1\textrm{Sq}^4)\textrm{Sq}^1 = \textrm{Sq}^1\textrm{Sq}^2\textrm{Sq}^4 .$$
From this kind of iterated substitution we see that it inductively suffices to find any relation with a nonzero coefficient for $\textrm{Sq}^n$, since any other term will be of the form $\textrm{Sq}^i\textrm{Sq}^j$ with $i,j<n$.
At this point, we will shift gears and consider the case of $p$ arbitrary.  Consider the Adem relation
$$
P^aP^b = \sum_{i}(-1)^{a+i}\binom{(p-1)(b-i)-1}{a-pi}P^{a+b-i}P^i
$$
for $pb>a$.
For a given $n$ that is not a power of $p$, we wish to find values of $a,b$ such that $a+b = n$ and with nonzero binomial coefficient mod $p$ for the $i=0$ term corresponding to $P^nP^0 = P^n$.  In other words, we need to find $a$ such that $p(n-a)>a$ and the binomial coefficient $$ \binom{(p-1)(n-a)-1}{a}$$ is nonzero.
Write $n = mp^k$ where $(m,p) =1$.  Note that $n$ not being a power of $p$ implies that $m\ge 2$.  Now set $a = p^k$, and note that $p(n-a) = (m-1)p^{k+1} > p^k = a$.  Our binomial coefficient is now $$\binom{(p-1)(mp^k-p^k)-1}{p^k} = \binom{(m-1)p^{k+1} - (m-1)p^k - 1}{p^k} .$$  At this point this can directly shown to not be divisible by $p$ since the value $y = (m-1)p^{k+1} - (m-1)p^k - 1$ satisfies $p^k \mid y+1, p^{k+1}\not\mid y-(p^k-1)$.  One can also appeal to Lucas's theorem on binomial coefficients mod a prime.
