Ring structure on $R = \oplus_N R^n$ where $R^n:=R(S_n)$ is the vector space of class functions defined on $S_n$. Define $R = \oplus_N R^n$ where $R^n:=R(S_n)$ is the vector space of class functions defined on $S_n$. According to Sagan, we can define a ring structure on $R$ that makes $R$ into a graded algebra, isomorphic to the algebra of symmetric functions.
For each $n\in \mathbb{N}$, the irreducible characters of $S_n$ form a basis for $R^n$ so it is enough to define a product of characters and then extend bilinearly.
For characters $\chi, \psi$ of $S_n$ and $S_m$, respectively, we define their product as
$$
\chi \cdot \psi := (\chi \otimes \psi)\uparrow_{S_n\times S_m}^{S_{n+m}}.
$$
My question is: why this product is associative? I have tried computing, when $\theta$ is an additional character for $S_k$,
\begin{align*}(\chi \cdot \psi)\cdot \theta &= \left((\chi \otimes \psi)\uparrow_{S_n\times S_m}^{S_{n+m}} \otimes \theta\right)\uparrow_{S_{n+m}\times S_k}^{S_{n+m+k}}
\end{align*}
but I can't see how the "induced part" interacts with the tensor product.
 A: Write $V_n, V_m, V_k$ for the representations corresponding to the characters. I will leave the names of the groups we're inducing between implicit to shorten notation and just write $\uparrow$, and also write them on the left. We want to show
$$\uparrow (\uparrow (V_n \boxtimes V_m) \boxtimes V_k) \cong \, \uparrow(V_n \boxtimes \uparrow (V_m \boxtimes V_k))$$
where I use $\boxtimes$ for the external tensor product to avoid confusion with the tensor product of $G$-representations for a fixed $G$. Let me also write $\downarrow$ for the restriction of representations. By definition the induced representation satisfies
$$\text{Hom}(\uparrow V, W) \cong \text{Hom}(V, \downarrow W)$$
(Frobenius reciprocity) and so the LHS satisfies
$$\text{Hom}(\uparrow (\uparrow (V_n \boxtimes V_m) \boxtimes V_k), -) \cong \text{Hom}(\uparrow (V_n \boxtimes V_m) \boxtimes V_k, \downarrow(-)).$$
At this point we need to say something about the universal property of the external tensor product $\boxtimes$. In general, if $R$ and $S$ are rings and $M$ and $N$ are modules, the external tensor product $M \boxtimes N$ is the tensor product equipped with the structure of an $R \otimes S$-module. As an $R \otimes S$-module it satisfies the following universal property: $R \otimes S$-linear maps $M \boxtimes N \to V$ correspond to bilinear maps $M \times N \to V$ which are $R$-linear in the first variable and $S$-linear in the second variable. Equivalently,
$$\text{Hom}_{R \otimes S}(M \boxtimes N, V) \cong \text{Hom}_R(M, \text{Hom}_S(N, V)).$$
This gives (again, the groups are implicit)
$$\text{Hom}(\uparrow (V_n \boxtimes V_m) \boxtimes V_k, \downarrow(-)) \cong \text{Hom}(\uparrow (V_n \boxtimes V_m), \text{Hom}(V_k, \downarrow(-))$$
and now by a second application of Frobenius reciprocity
$$\text{Hom}(\uparrow (V_n \boxtimes V_m), \text{Hom}(V_k, \downarrow(-)) \cong \text{Hom}(V_n \boxtimes V_m, \downarrow \text{Hom}(V_k, \downarrow(-))$$
and now by a second application of the universal property of the external tensor product
$$\text{Hom}(V_n \boxtimes V_m, \downarrow \text{Hom}(V_k, \downarrow(-)) \cong \text{Hom}(V_n, \downarrow \text{Hom}(V_m, \downarrow \text{Hom}(V_k, \downarrow(-)))).$$
This may look a bit intimidating but what it says concretely is that a map out of $\uparrow (\uparrow (V_n \boxtimes V_m) \boxtimes V_k)$ to another representation $W$ of $S_{n+m+k}$ is the same thing as a trilinear map $V_n \times V_m \times V_k \to W$ which is $S_n$-linear in the first variable, $S_m$-linear in the second variable, and $S_k$-linear in the third variable. A third application of Frobenius reciprocity and the universal property of the external tensor product gives that this is the same thing as the triple product $\uparrow (V_n \boxtimes V_m \boxtimes V_k)$, and then basically the same argument shows that $\uparrow(V_n \boxtimes \uparrow (V_m \boxtimes V_k))$ is also the triple product (by the Yoneda lemma).
This is a generalization of what is, to my mind, the cleanest way to show that the ordinary tensor product is associative (up to isomorphism): you show that $(U \otimes V) \otimes W \cong U \otimes (V \otimes W)$ by showing that both sides represent the functor given by trilinear maps out of $U \times V \times W$ and then using Yoneda.
A: Here is a rather concrete approach:
Suppose $\chi_m,\chi_n,\chi_k$ are characters of $S_m,S_n,S_k$, respectively. To show that the induction product is associative, it suffices to show that
$$ \uparrow_{S_{m+n}\ \ \times S_k}^{S_{m+n+k}}((\uparrow_{S_m\times S_n}^{S_{m+n}}\chi_m\times\chi_n)\times\chi_k)=\uparrow_{S_{m}\times S_n \times S_k}^{S_{m+n+k}}(\chi_m\times\chi_n\times \chi_k).\tag{1} $$
Notice that the right side can be written as: $$ \uparrow_{S_{m}\times S_n \times S_k}^{S_{m+n+k}}(\chi_m\times\chi_n\times \chi_k)=\uparrow_{S_{m+n}\ \ \times S_k}^{S_{m+n+k}}(\uparrow_{S_m\times S_n\times S_k}^{S_{m+n}\ \ \times S_k}\chi_m\times\chi_n\times\chi_k). \tag{2}$$
Compare this with the left side of (1), it now suffices to show that
$$ (\uparrow_{S_m\times S_n}^{S_{m+n}}\chi_m\times\chi_n)\times\chi_k=\uparrow_{S_{m}\times S_n \times S_k}^{S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n\times \chi_k)\tag{3}. $$
But this follows exactly from the definition of the induced character. Indeed, for any $h\times h'\in S_{m+n}\times S_k$, we want to show that (3) holds by showing that both sides of (3) give the same value evaluated at $h\times h'$.
We compute that:
\begin{align}
((\uparrow_{S_m\times S_n}^{S_{m+n}}\chi_m\times\chi_n)\times\chi_k)(h\times h')&=\left(\frac{1}{m!n!}\sum_{g\in S_{m+n}}(\chi_m\times\chi_n)^\circ\underset{\in S_m\times S_n}{(g^{-1}hg)}\right)\times\chi_k(h'),
\end{align}
whereas,
\begin{align}
(\uparrow_{S_{m}\times S_n \times S_k}^{S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n\times \chi_k))(h\times h')&=\frac{1}{m!n!k!}\sum_{g\times g'\in S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n\times\chi_k)^\circ\underset{\in S_m\times S_n\times S_k}{((gg')^{-1}hh'gg')}\\
&=\frac{1}{m!n!k!}\sum_{g\times g'\in S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n\times\chi_k)^\circ\underset{\in S_m\times S_n\times S_k}{((gg')^{-1}hgh'g')}\\
&=\frac{1}{m!n!k!}\sum_{g\times g'\in S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n\times\chi_k)^\circ\underset{\in S_m\times S_n\times S_k}{((gg')^{-1}h(gg')(g'^{-1}h'g'))}\\
&=\frac{1}{m!n!k!}\sum_{g\times g'\in S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n)^\circ\underset{\in S_m\times S_n}{(g'^{-1}(g^{-1}hg)g')}\times\chi_k(h')\\
&=\frac{1}{m!n!k!}\sum_{g\times g'\in S_{m+n}\ \ \times S_k}(\chi_m\times\chi_n)^\circ\underset{\in S_m\times S_n}{(g^{-1}hg)}\times\chi_k(h')\\
&=\frac{k!}{m!n!k!}\sum_{g\times g'\in S_{m+n}}(\chi_m\times\chi_n)^\circ\underset{\in S_m\times S_n}{(g^{-1}hg)}\times\chi_k(h')\\
&=\left(\frac{1}{m!n!}\sum_{g\in S_{m+n}}(\chi_m\times\chi_n)^\circ\underset{\in S_m\times S_n}{(g^{-1}hg)}\right)\times\chi_k(h')\\
&=((\uparrow_{S_m\times S_n}^{S_{m+n}}\chi_m\times\chi_n)\times\chi_k)(h\times h')
\end{align}
where we have used the commutativity of elements from two different components in the internal direct product of groups for several times, and finally we are done.
