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. 



Embedding group algebra $F[S_m X S_n]$ into a  group algebra $F[S_{m+n}]$
**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 $n$.


