TITLE: Homotopy Groups of $S^l \vee S^k$
QUESTION [5 upvotes]: Let $S^l \vee S^k$ wedge sum of two spheres $S^l, S^k$.
How can I simplify the calculation of homotopy groups $\pi_n(S^l \vee S^k)$?
I know that $\pi_n(S^l \times S^k) = \pi_n(S^l) \oplus \pi_n(S^k)$,
as well $H_n(S^l \vee S^k) = \pi_n(S^l) \oplus H_n(S^k)$ holds, therefore we have $\pi_n ^{ab}(S^l \times S^k) = \pi_n ^{ab}(S^l) \oplus \pi_n ^{ab}(S^k)$,
but do we have similar reduction rules for $\pi_n(S^l \vee S^k)$?
If, yes, why, but I suppose not generally.

What other strategy can I use to simplify the calculation?

REPLY [2 votes]: This solution Hilton's theorem, from his paper ON THE HOMOTOPY GROUPS OF THE UNION OF SPHERES. It makes the calculation of $\pi_n(S^{r_1+1}\vee \cdots \vee S^{r_m+1})$ for $r_i\geq 1$ easy to do by hand for small $n, m$.
In the Hilton's paper there is Theorem A
$$
\pi_n(T)=\sum_{i=1}^{\infty} \pi_n(S^{q_{i}+1})
$$
where $T=S^{r_1+1}\vee \cdots \vee S^{r_m+1}$, $r_i\geq 1$.
He inductively defines so-called basic generators $p_1,p_2,\cdots$. For each basic generator we get an integer $q_s$, called height.
The basic generators of weight $1$ are $\iota_1,\cdots,\iota_m$ where $\iota_i$ is the generatror of $\pi_{r_i+1}(S^{r_i+1})$ and can be taken to be the generator of a direct summand in $\pi_{r_i+1}(T)$. Order the generators arbitrarily $\iota_1<\cdots<\iota_m$.
Assume the basic generators of weight less then $w$ have been defined and ordered. The basic generators of weight $w$ are $[a,b]$ where $a,b$ are the basic generators of order $u$ and $v$ respectively, $u+v=w$ and $[-,-]$ is the Whitehead product. Furthermore we demand that $a<b$ and if $b=[c,d]$ then $c\leq a$.
Order the basic products of weight $w$ among themselves arbitrarily, and set them to be larger then any basic product of smaller weight.
Any basic product $p_q$ is then some string of basic generators of weight $1$ and some brackets. Define its height as  $q_s=\sum_{i=1}^k r_iw_i$ where $w_i$ is the number of occurences of $\iota_i$ in $p_s$.
Let us have a basic product $p_s$ with height $q_s$. There is then an embedding as a direct summand $\pi_n(S^{q_s+1})\hookrightarrow \pi_n(T)$ induced by the map $p_s\in\pi_{q_s+1}(T)$(an easy excercise that $p_s$ is from that group)