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Here's a question I've been thinking about, it's a curiosity that I don't know how to answer. There could be a simple counterexample, or it could be true and I don't know how difficult it would be to prove.

If we fix $m$, is it always possible to find a sufficiently large $n$ satisfying the conditions of the following question: [Note: My original question was to determine whether it is true for arbitrary $m,n$ which was answered below negatively; I have edited it to make the question more interesting].

Define $ \phi: S_m \rightarrow S_{m+n}$ is a canonical embedding, and $\phi^{*} : F[S_m] \rightarrow F[S_{m+n}]$ and similarly embeddings $\theta: S_{n} \rightarrow S_{m+n}$, and the induced map, such that $\phi(S_{m}) \times \theta(S_{n})$ is a direct product.
Given an element $x \in \phi^{*}(F[S_m]), x \neq 0$, is it necessary that there exist an element $x' \in F[S_{m+n}]$ so that the product $xx' \in \theta^{*}(F[S_n]), xx' \neq 0$. It seemed true in the cases that I have tried, but they are quite small so I'm not certain if this is true.

Making the assumption $ \text{char} F = 0$ would make it easier I'm sure, but even in this case I can't prove it.

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    $\begingroup$ What conditions are you wanting to place on x'? (I assume that it is at least nonzero.) $\endgroup$
    – Tyler Lawson
    Commented Nov 29, 2009 at 12:56
  • $\begingroup$ Just how many cases have you tried, by the way? and what kinds of examples? $\endgroup$
    – Yemon Choi
    Commented Nov 30, 2009 at 2:23
  • $\begingroup$ By the way, if $n\geq m$ then it is easy to find elements $x', x''$ in $F[S_{m+n}]$ such that $x''xx'$ is a nonzero element in the range of $\theta^*$. But I guess you know that already and this is why you asked about the one-sided case. $\endgroup$
    – Yemon Choi
    Commented Nov 30, 2009 at 3:07
  • $\begingroup$ yes that seems pretty easy, just take $x''=x'^{-1}$ that conjugates $S_{m}$ into $S_{n}$ right? $\endgroup$
    – Puraṭci Vinnani
    Commented Nov 30, 2009 at 6:31
  • $\begingroup$ @vinoth Exactly. I still don't know about your one-sided case. $\endgroup$
    – Yemon Choi
    Commented Nov 30, 2009 at 8:20

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NOTE: This answer was given in response to an earlier version of the question above, and so probably seems completely irrelevant if you can't see the old version. (More strikethrough and less deletion in editing, please?)

Firstly, I'm going to assume you want $x'$ to be nonzero. Secondly, do you want $x$ and $xx'$ to also both be non-zero?

If so, then there is a silly counterexample -- take $n=1$, and take $x$ to be $id - t$ where $t$ is the permutation $1\to 2\to \dots \to m\to 1$. Then $x$ lies in the augmentation ideal of $F[S_{m+n}]$ and so if $xx'$ lies in $\theta^*(F[S_1]) = F$ then it would have to be zero.

I strongly suspect that one could play similar tricks with higher values of $n$.

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  • $\begingroup$ Oh right, i see. Yeah that would certainly work given n=1, but for n arbitrarily large I can't see a way of generalizing that counter-example to go through. All we get that it has to lie inside the augmentation ideal of $F[S_n]$ which doesn't seem to be an immediate contradiction. I've changed the question slightly now: for $m$ fixed, does there necessarily exist a sufficiently large $n$ satisfying the given condition? $\endgroup$
    – Puraṭci Vinnani
    Commented Nov 29, 2009 at 22:12

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