Does anyone know the formula for a Hall polynomial $g_{u,v}^{\lambda}(p)$ when $v$ is the type of cyclic subgroup (ie. $v=(v_{1})$ ) . http://en.wikipedia.org/wiki/Hall_algebra
I was hoping this particular case would be simple enough to describe .
Does anyone know the formula for a Hall polynomial $g_{u,v}^{\lambda}(p)$ when $v$ is the type of cyclic subgroup (ie. $v=(v_{1})$ ) . http://en.wikipedia.org/wiki/Hall_algebra I was hoping this particular case would be simple enough to describe . 


Let's say you want to compute the Hall polynomial $g^\lambda_{(r),\mu}(p)$. According to [Dutta and Prasad, Degenerations and orbits in finite abelian groups], the orbits under the automorphism group of a finite abelian group are given by $\{O_II\subset J(P_\lambda)\}$, where $J(P_\lambda)$ denotes the lattice of order ideals in a certain poset $P_\lambda$. The same paper also gives a formula for $O_I$. Clearly, the Hall polynomial that you are looking for is $(p^rp^{r1})^{1}\sum_{I} O_I$, the sum being over all $I$ for which the order of an element in $O_I$ is $p^r$ and for which the quotient of the group of type $\lambda$ by any element of $O_I$ is of type $\mu$. As pointed out by David Speyer in the comments to an earlier version of this answer, $I$ is uniquely determined by these conditions. So a final form of the answer is obtained by explaining how to obtain $I$ from $\lambda$ and $\mu$. Given an element $(p^{v_1},p^{v_2},\dotsc)$, the type of the quotient is found by computing the Smith canonical form of the matrix $\begin{pmatrix} p^{\lambda_1} & & &\\ & p^{\lambda_2} & &\\ & & \ddots & \\& & &p^{\lambda_l}\\ p^{v_1} & p^{v_2} & \cdots & p^{v_l} \end{pmatrix}$. By the characterization of order ideals in $P_\lambda$, we have that $v_i\leq v_{i1}\leq v_i+(\lambda_{i1}\lambda_i)$. Proposition. Let $I\subset P_\lambda$ be an order ideal. Let $\mu$ be the type of the group obtained by going modulo an element of $O_I$. Then $\mu_l=v_l$ $\mu_{l1}=\lambda_l+v_{l1}v_l$ $\mu_{l2}=\lambda_{l1}+v_{l2}v_{n1}$ $\vdots$ $\mu_{i}=\lambda_{i+1}+v_iv_{i+1}$ $\vdots$ $\mu_1=\lambda_2+v_1v_2$. Proof. The gcd of $i\times i$ minors of the above matrix can be seen to be $v_{li+1}+\lambda_{li+2}+\dotsb+\lambda_l$ (using the inequalities on $v_i$). Therefore, we get $\mu_{li+1}+\dotsb+\mu_l=v_{li+1}+\lambda_{li+2}+\dotsb+\lambda_l$, from which the above identities follow.QED. This allows us to recover $(v_1,v_2,\dotsc)$ once we know $\lambda$ and $\mu$, where (this also follows from the above proposition), $\mu$ is a partition obtained from $\lambda$ by removing a horizontal strip of length $r$. In particular, $v$ is uniquely determined by $\lambda$ and $\mu$. We get $v_l=\mu_l$, and $v_i=\mu_i[(\lambda_{i+1}+\dotsb+\lambda_l)(\mu_{i+1}+\dotsb+\mu_l)]$ for $i<l$. Let $I\subset P_\lambda$ be the order ideal defined by $(v_1,v_2,\dotsc)$. We get
It easily follows (from the formula for $O_I$ in our paper, which says that $O_I$ is a monic polynomial in $p$ of degree $\sum_i (\lambda_i\nu_i)$) that $g^\lambda_{(r)\mu}(p)$ is monic in $p$ of degree $n(\lambda)n(\mu)$. This could perhaps give another approach to Hall's theorem (in analogy with the proof that MacDonald gave). 


I'll put down what I figured out. Example 2.4 of Schiffman's Lectures on Hall Algebras tells how to multiply by the partition $1^r$. It is a generalization of the Pieri rule. If we set $p=1$, there is a symmetry of the ring of symmetric polynomials which sends a partition to its transpose, so we can just transpose this formula. But I don't know if there is an analogous symmetry for Hall polynomials. (The ring of Hall polynomials has an antipode, which morally should do this, but I couldn't find a statement of whether it does in a quick search.) The formula for multiplying by $1^r$ also appears as Lemma 2.4 in this paper. According to this, the formula occurs on page 341 of Macdonald's book. I don't have the book available, but if I did I would look there to see whether the transpose formula also occurs. Finally, I share your intuition that this computation should be doable by hand. If I were you, I'd look at Schifman's proof and see if I could adapt it. 

