Consider a finite group G. The product of conjugacy classes can be defined in natural way just by multiplying the representatives and counting multiplicities (see e.g. [MO 62088](https://mathoverflow.net/questions/62088/products-of-conjugacy-classes-in-s-n)). So we get ring with a basis and structure constants are natural numbers. Similar to what one has for product of irreps.
There are many analogies between conjugacy classes and irreps in particular see [this article](http://journals.cambridge.org/action/displayFulltext?type=1&fid=3077776&jid=PEM&volumeId=30&issueId=01&aid=3077768&bodyId=&membershipNumber=&societyETOCSession=).


Tanaka-Krein duality states that group can be reconstructed from the tensor category of its representations which is semisimple for finite groups, and hence carries the same information as ring + basis of irreps.


**Question:** Can one reconstruct a group having (ring + basis) made of conjugacy classes ?


If not - what partial information (e.g. character table) one can get ? 




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**Question:** Is there any relation between this ring and ring of irreps of the same group ? or may be some other group ?


(Remark. For abelian group they are isomorphic.)


**Question:** Are there any further analogies between ring of irreps and conjugacy classes except mentioned in the paper cited above ?



Product of conjugacy classes - is there an analog of Tanaka-Krein reconstruction ?
The answer to your first question is negative. For a concrete example, you can show that the conjugacy class rings of the nonisomorphic groups $Q\_8$ and $D\_8$ are isomorphic, via an isomorphism that pairs off the bases as follows: $[1] \leftrightarrow [1]$, $[-1] \leftrightarrow [r^2]$, $[i] \leftrightarrow [r]$, $[j] \leftrightarrow [s]$ and $[k] \leftrightarrow [rs]$.


As to your question about the relationship between the conjugacy class ring and the character ring, there are lots of partial results that can be stated. Nonetheless, the answer to the question of when these two rings are isomorphic is completely known. This turns out to be the case if and only if the group is $p$-nilpotent with abelian Sylow $p$-subgroup. More generally, for arbitrary finite groups $G$ and $G'$, the following two conditions are equivalent.


1. The character ring of $G$ is isomorphic to the conjugacy class ring of $G'$.
2. $G$ and $G'$ are $p$-nilpotent groups with abelian Sylow $p$-subgroups. Moreover, if $g\_1, \dots, g\_l$ and $g\_1',\ldots, g\_{l'}'$ are complete sets of representatives for the conjugacy classes of $p'$-elements of $G$ and $G'$, resp., and if $D\_i$ and $D\_i'$ are Sylow $p$-subgroups of $C\_G(g\_i)$ and $C\_{G'}(g\_i')$, resp., then $l=l'$ and $D\_i \cong D\_i'$.


This is due to Saksonov, *The ring of classes and the ring of characters of a finite group*. Mat. Zametki **26** (1979), no. 1, 3–14, 156.


