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Let $k$ and $k'$ and $n_{1},\ldots,n_{k}$ and $m_{1},\ldots,m_{k'}$ be natural numbers. Let $f_{1}\leq \ldots \leq f_{k}$ and $e_{1} \leq \ldots \leq e_{k'}$ be power primes, such that the following equations hold:

$\prod_{j=1}^{j=k} \prod_{i=0}^{n_{j}-1}(f_{j}^{n_{j}}-f_{j}^{i})=\prod_{j=1}^{j=k'} \prod_{i=0}^{m_{j}-1}(e_{j}^{m_{j}}-e_{j}^{i})$

and

$\prod_{j=1}^{j=k}f_{j}^{n_{j}^{2}}=\prod_{j=1}^{j=k'}e_{j}^{m_{j}^{2}}.$

Is it achievable that $k=k'$ and for each $i=1,\ldots,k$ we have $f_{i}=e_{i}$ and $n_{i}=m_{i}$?

Actually, this question arises from the question that, if two finite semisimple ring $R$ and $R'$ have the same number of members and the same number of unit members, then can we say that these rings are isomorphic together?

If there is some counterexample what additional conditions should be employed to obtain that $R\simeq R'$?

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1 Answer 1

up vote 3 down vote accepted

Dear Mohsen, the answer to the ring question seems to be "no". Indeed, take $\mathbb Z_{18}\times\mathbb Z_{24}$ and $\mathbb Z_{12}\times\mathbb Z_{36}$: both have 432 elements and 48 units, while the first one has elements of order 8, and the second one hasn't. -- Mike

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