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Let p be a prime and let $\mathbb Z_p$ denote the p-adic integers.

If n<m, then what are the embeddings $SL_n(\mathbb Z_p)\rightarrow SL_m(\mathbb Z_p)$? I am particularly interested in those which carry $SL_n(\mathbb Z)$ into $SL_m(\mathbb Z)$.

There are obvious "block" embeddings, e.g., carrying a matrix to the upper-left hand corner of a larger matrix. There are also certain conjugates of these. In general, the embeddings should come from representations of $SL_n(\mathbb Z_p)$, but I do not know where they are catalogued or what exactly to do with the catalog.

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Any such is induced from a map of p-algebraic groups $SL_n \rightarrow SL_m$ . You can see this by passage to the Lie algebra. Given such a map, the image of $SL_n(Z_p)$ is conjugate by $GL_m(Z_p)$ to a subgroup of $SL_m(Z_p)$ . However, there can be multiple conjugacy classes of maps inducing the same algebraic morphism. –  moonface Nov 30 '09 at 17:09
    
That should be $Q_p$-algebraic groups in the first line, and $GL_m(Q_p)$ in the third, sorry. Can't edit comments. –  moonface Nov 30 '09 at 17:19
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@moonface: regarding editing comments---you can delete them, so you can edit them by cut-and-pasting, editing, and then deleting the original. Regarding mathematics: SL_2(Z_p) maps onto SL_2(Z/pZ) which will I guess have a faithful representation into some SL_N(Z). Now tensor up with the identity map SL_2(Z_p)-->SL_2(Z_p) and don't I have a counterexample to your claim that all embeddings are algebraic (or did I miss something)? My guess is that the Lie algebra argument only proves they're locally algebraic. –  Kevin Buzzard Nov 30 '09 at 19:47
    
You're completely right, of course -- what I said above is FALSE and only true "virtually". Thank you for the correction! –  moonface Nov 30 '09 at 20:26
    
Thanks much for your responses! So the conjugator is in $GL_m(Q_p)$. But does my added hypothesis about carrying $SL_n(Z)$ into $Sl_m(Z)$ help us say anything about the conjugator? Thanks again. –  Samuel Coskey Dec 17 '09 at 19:00

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