# conjugate subgroups of the symmetric group

Let $S_n$ be the symmetric group. let $H=S_{d_1}\times\cdots \times S_{d_k}$ such that $d_1+\cdots + d_k=n$.

What does the following statement means:

every two embeddings of $H$ in $S_n$ are conjugate.

In my opinion this means that changing the order of $d_i$ gives conjugate subgroups. For example $S_{d_1}\times S_{d_2}\times S_{d_3}$ is conjugate to $S_{d_2}\times S_{d_1}\times S_{d_3}$ and to $S_{d_3}\times S_{d_1}\times S_{d_2}$ and so on ...

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In general, when faced with a term you don't know, it's best to ask the source (whether text or person) for a definition rather than assuming a meaning arbitrarily. – user92843 Jul 3 '11 at 9:20

I think that your interpretation is correct. It is certainly not true that any subgroups of $S_n$ that are abstractly isomorphic to $H$ are conjugate. The most famous example is that of two non-conjugate $S_5$ sitting inside $S_6$. So the statement must refer to the canonical embedding of $H$ in $S_n$, in which case the only way I can think of to obtain different embeddings is by permuting the $d_i$ (and then the statement is almost trivially true).

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I think it's more likely that the interpretation is incorrect, that the standard meaning of embedding is intended, and that the quoted statement is false (perhaps as part of a true/false question). This is entirely conjectural since neither of us has the initial source, unlike palio. – user92843 Jul 3 '11 at 9:23
@Alex B. : thanks!! please where can i read on the example of two non conjugate $s_5$ sitting in $S_6$? – palio Jul 3 '11 at 10:04
@user92843 : $S_{d_i}$ is embedded in $S_n$ as the subgroup permuting the $d_i$ first elements of $(x_1,...,x_{d_i},x_{d_{i+1}},...,x_n)$ – palio Jul 3 '11 at 10:24
@palio See e.g. JA Todd, The Odd Number 6, Math. Proc. Camb. Phil. Soc. 41 (1945) or en.wikipedia.org/wiki/… – Alex B. Jul 3 '11 at 10:32