Homotopy Groups of $S^l \vee S^k$ 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?

 A: First, if $k,l=1$, then $S^1\vee S^1$ is the Eilenberg Mac Lane space $K(\mathbb{Z}\ast\mathbb{Z},1)$ amd $S^1\times S^1$ is the Eielenberg Mac Lane space $K(\mathbb{Z}\oplus\mathbb{Z},1)$. It follows that $\pi_1(S^1\times S^1)$ is the abelianisation of $\pi_1(S^1\vee S^1)$, and neither the wedge nor product have any other homotopy groups.
If one of $k,l$ is equal to $1$ (say, $k$), and the other is greater than one, then you can use cover space theory to calculate the homotopy groups of $S^1\vee S^l$. I believe its universal cover may be the infinite wedge $\bigvee^\infty_{i=1} S^l$, but i may not be remembering that correctly. In any case $\pi_*(S^1\times S^l)\cong\pi_*(S^1)\oplus\pi_*(S^l)$ is $\pi_*S^l$ with an extra copy of $\mathbb{Z}$ in degree $1$, whilst $\pi_*(S^1\vee S^l)$ is exceptionally complicated.
Therefore, in the following, we assume that $k,l$ are both $>1$.
Let $j:S^k\vee S^l\hookrightarrow S^k\times S^l$ be the inclusion, and let $F_j$ be its homotopy fiber. Then, after looping, there is a homotopy fibration
$$\Omega F_j\xrightarrow{\Omega i} \Omega(S^k\vee S^l)\xrightarrow{\Omega j} \Omega(S^k\times S^l),$$
where $i:F_j\rightarrow S^k\vee S^l$ is the fibre inclusion.
The projections onto each factor induce a homeomorphism 
$\Psi:\Omega(S^k\times S^l)=Map_*(S^1,S^k\times S^l)\xrightarrow{\cong}Map_*(S^1,S^k)\times Map_*(S^1,S^l)=\Omega S^k\times \Omega S^l,$
sending $\omega\mapsto (pr_1\circ\omega,pr_2\circ \omega)$. Composing with $\Omega j$, this map acts as
$\Psi\circ \Omega j(\omega)=\Theta(j\circ \omega)=(pr_1,circ j\circ \omega,pr_2\circ j\circ \omega=(q_1\circ \omega,q_2\circ \omega),$
where $q_1:S^k\vee S^l\rightarrow S^k$, $q_2:S^k\vee S^l\rightarrow S^l$ are the pinch maps. Thus we are left with the homotopy fibration sequence
$$\Omega F_j\rightarrow \Omega(S^k\vee S^l)\xrightarrow{(\Omega q_1,\Omega q_2)} \Omega S^k\times \Omega S^l.$$
Now the pinch maps $q_1,q_2$ individually have sections, namely the inclusions $i_1:S^k\hookrightarrow S^k\vee S^l$, $i_2:S^l\hookrightarrow S^k\vee S^l$, and with these we use the monoid structure on $\Omega (S^k\vee S^l)$ to define a map $\Omega i_1\circ pr_1+\Omega i_2\circ pr_2:\Omega S^k\times\Omega S^l\rightarrow \Omega(S^k\vee S^l)$, $(\omega_1,\omega_2)\mapsto\Omega i_1(\omega_1)+\Omega i_2(\omega_2)$. Observe that 
$(\Omega q_1,\Omega q_2)\circ(\Omega i_1\circ pr_1+\Omega i_2\circ pr_2)(\omega_1,\omega_2)=(\Omega q_1,\Omega q_2)(\Omega i_1(\omega_1)+\Omega i_2(\omega_2))=(\Omega(q_1)(\Omega i_1(\omega_1)+\Omega i_2(\omega_2),\Omega q_2(\Omega i_1(\omega_1)+\Omega i_2(\omega_2))=(\Omega(q_1\circ i_1)(\omega_1)+\Omega(q_1\circ i_2)(\omega_2),\Omega(q_2\circ i_1)(\omega_1)+\Omega(q_2\circ i_2)(\omega_2))=(\Omega(id_{S^1})(\omega_1)+\Omega(\ast)(\omega_2),\Omega(\ast)(\omega_1)+\Omega(id_{S^l})(\omega_2))=(\omega_1+\ast,\ast+\omega_2)\simeq (\omega_1,\omega_2).$
The point is that the map we have just produced is a homotopy section of $\Omega j$. Using again the monoid structure we define a map 
$\Theta:\Omega S^k\times\Omega S^l\times\Omega F_j\rightarrow \Omega(S^k\vee S^l)$
by $\Theta=\Omega i_1\circ pr_1+\Omega i_2\circ pr_2+\Omega i\circ pr_3$. 
Turning to study the long exact sequence of homotopy groups of the homotopy fibration $F_i\xrightarrow{i}S^k\vee S^l\xrightarrow{j} S^k\times S^l$ we find that the map $\Theta$ is a weak homotopy equivalence. Thus
$$\pi_r(S^k\vee S^l)\cong \pi_{r-1}(\Omega(S^k\vee S^l))\cong \pi_{r-1}(\Omega S^k\times\Omega S^l\times\Omega F_j)\cong \pi_r(S^k\times S^l)\oplus\pi_r(F_i)$$
so for $k,l>1$, the homotopy module $\pi_*(S^k\times S^l)$ is always a direct summand of $\pi_*(S^k\vee S^l)$. Sadly this doesn't really make the task of computing $\pi_*(S^k\vee S^l)$ any easier, since we still need to find $\pi_*(F_i)$. What is this space?
Well, we know from homology that 
$H_*(S^k\vee S^l)=\begin{cases}H_*(S^k\vee S^l)&*<k+l\\\mathbb{Z}& *=k+l\\0&\text{otherwise}\end{cases}$
so since both spaces are simply connected we find that the map $j$ is a $(k+l-1)$-equivalence. In fact the attaching map for the top cell of $S^k\vee S^l$ is the Whitehead product $[i_1,i_2]:S^{k+l-1}\rightarrow S^k\vee S^l$. In any case, the $F_i$ must be $(k+l-1)$-connected, and its not difficult to argue that $\pi_{k+l-1}F_i\cong\pi_{k+l-1}S^{k+l-1}\cong\mathbb{Z}$. 
Unfortunately there is no short answer to what $F_i$ is, so I'll just write down
$F_i\simeq (\Omega S^k)\ast(\Omega S^l)\simeq \Sigma(\Omega S^k\wedge \Omega S^l),$
and refer you to Jeff Strom's book "Modern Classical Homotopy Theory" for more details.
A: 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)
