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I'm trying to generate to monomial symmetric polynomial $$ m_{(1,2,3,4)}(X_1,X_2,\dots,X_{10})=X_1^1X_2^2X_3^3X_4^4 + \text{all permutations,} $$ starting from the first element by applying all necessary permutations.

Since $m_{(1,2,3,4)}(X_1,X_2,\dots,X_{10})$ has $\frac{10!}{6!}=5040$ elements, I think that not all elements of $S_{10}$ are necessary and I wonder if there is a subgroup of it.

I found $S_7$ to have the right number of elements, but first this doesn't make sense to me and second my next example would be $m_{(1,2,3,4,5)}(X_1,X_2,\dots,X_{15})$, where $S_9$ doesn't match (see here for a related question).

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draks ...
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  • What you're looking at is a subset of $S_{10}$, but it's not a sub_group_ in any natural sense. If you just want all of the relevant ordered combinations of that length, there are better ways of getting there than by walking through the elements of $S_{10}$; searching on 'generating all combinations' should provide useful information. (Note that once you're iterating over all combinations, getting all ordered combinations is easy.) – Steven Stadnicki Jul 18 '13 at 22:15
  • @StevenStadnicki ah ok, thanks, I expected something like this. So the answer is: No...? – draks ... Jul 18 '13 at 22:17
  • well, I freely admit I don't actually know the answer to the posted question as it's stated (are there any other subgroups of $S_{10}$ of that size?)! I'm just noting that your particular structure is not, in any natural sense that I can see, a group. But it seems likely that whether it's a group or not is wholly irrelevant to the question of how best to iterate through them. – Steven Stadnicki Jul 18 '13 at 22:21
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    The answer to the question that was asked is yes: $A_7 \times C_2$. – Derek Holt Jul 18 '13 at 22:31
  • @StevenStadnicki do the subsets create a kind of coset structure? – draks ... Jul 20 '13 at 20:34
  • @StevenStadnicki maybe along the lines shown here? – draks ... Aug 04 '13 at 21:49
  • You seem to have stumbled across the equation $6!7!=10!$, which is nice but rather exceptional. – Marc van Leeuwen Aug 28 '13 at 13:24

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