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For $d < n$, define an injective function $\mathbb Z_d \rightarrow S_n$ preserving the operation, that is, such that the sum of equivalence classes in $\mathbb Z_d$ corresponds to the product of the corresponding permutations.

Is the question asking me define a function $\phi: \mathbb Z_d \rightarrow S_n$ such that $\phi(i)=\sigma_i$, where $\phi $ is injective and $\phi(i+j)=\phi(i)\phi(j)$?

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Yes to the very last part. The part about $\phi(i) = \sigma_i$ does not really make sense unless you define $\sigma_i$. –  Tobias Kildetoft Feb 14 '13 at 14:58

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Since $\mathbb Z_d$ is a cyclic group and $\phi:\mathbb Z_d\to S_n$ is an injective group homomorphism, then $\phi(\hat 1)$ is an element of $S_n$ of order $d$. Such elements there exist only if $d\mid n!$. In this case one can take $\sigma\in S_n$ an element of order $d$ and define $\phi:\mathbb Z_d\to S_n$ by $\phi(\hat k)=\sigma^k$.

On the other side, if $d\mid n!$ we don't need to have an element of order $d$ in $S_n$: take $d=6$ and $n=4$. Thus, in this case, one can't find an injective group homomorphism $\phi:\mathbb Z_6\to S_4$.

Observation. A natural question comes out: for which $d\mid n!$ are there elements of order $d$ in $S_n$?

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Fix $d$ points $x_1,..,x_d$ in the $n$ point set where the permutations of $S_n$ act, and consider $\phi(1)$ as the cycle $x_1\mapsto x_2,\ x_2\mapsto x_3,\ \dots,\ x_d\mapsto x_1$.

Then, what the rest $\phi(2),...,\phi(d-1),\phi(0)$ has to be?

Note: In general, any element of order $d$ will be a good choice for $\phi(1)$.

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This does not answer the actual question asked. –  Tobias Kildetoft Feb 14 '13 at 15:03
    
I take it is already answered in the comment (by you). –  Berci Feb 14 '13 at 15:05

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