Hey everybody,


I think this question might be just a simple oversight on my part, but this has been bugging me a few days. 


I am reading Hatcher's [Spectral Sequences book](http://www.math.cornell.edu/~hatcher/SSAT/SSch2.pdf), and trying to understand his example where he computes $\pi\_\*^s$ for $p=2$ (page 21-23), and I'm a bit confused about a certain step. He claims that the element corresponding to $h\_3^2$ must have order 2 in $\pi\_{14}^s$, because of "the commutativity property of the composition product, since $h\_3$ has odd degree". Now, I see why $h\_3^2$ can have order at most 4, because $h\_3^2h\_0^2=0 \in E\_2$, but why must it have order 2 exactly? What does the odd degree have to do with it? If I am not mistaken, the Yoneda product on $Ext\_A(Z/2,Z/2)$ induces the composition product on $\pi\_\*^s$, which, mod 2, is commutative, but the Yoneda product has $h\_3h\_0=h\_0h\_3$ in the $E\_2$ page, so I can't from that derive the induced composition product is 0. Do I need to use a fact about $\pi\_s^\*$ that doesn't come from this spectral sequence? 


Thanks for the help everybody!
-Joseph Victor



How do you know when something must die in the Adams Spectral Sequence for $\pi_*^s$
The point is that the element $h\_3$ in odd degree lifts to an element (typically called $\sigma$) in $\pi\_7^s$ whose square must be 2-torsion. This is true for all elements in odd degree because the stable homotopy groups of spheres are graded-commutative. It does not necessarily have to be exactly 2-torsion a priori - it could be 1-torsion (i.e. trivial) - but it must be annihilated by 2.


