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Let $q$ denote a prime power and $\text{GL}_n(q)$ and $\text{U}_n(q^2)$ the general linear and unitary group, respectively. Then $\text{U}_n(q^2)$ is naturally a subgroup of $\text{GL}_{n}(q^2)$, so one kind of groups can be embedded into the other. Let $C(g)$ be the conjugacy class of an element $g$ in its respective group. Then we can define the length of $g$ to be $\ell(g):=\frac{\log|C(g)|}{\log|G|}$. The above embedding only changes the length by a constant factor.

Note that the length function induces a biinvariant metric by $d(g,h):=\ell(gh^{-1}$.

My question is if there are any functions $f,g:\mathbb{N}\rightarrow \mathbb{N}$ such that we can always find an embedding of $\text{GL}_n(q)$ into $\text{U}_{f(n)}(q^{g(n)})$, which doesn't change the above length function to much. It would be nicest if the distortion of length would only be a constant factor.

My feeling is that this is not possible, but I somehow fail to find an explanation.

A further question is what happens if we exchange the unitary groups for orthogonal or symplectic groups.

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Well, of course; doesn't $S_n$ embed into $U_n(q^2)$? Use Cayley's theorem. –  Qiaochu Yuan Oct 21 '11 at 13:39
@Qiaochu Yuan: Right, this is of course true. I forgot an important fact in my question and will update it. Sorry for that! –  Abel Stolz Oct 21 '11 at 14:49

1 Answer 1

up vote 4 down vote accepted

I guess by $U_n(q^2)$ you mean the general unitary group in which the field of representation has order $q^2$? That is often denoted by ${\rm GU}_n(q)$, but I will use your notation.

In general ${\rm GL}_n(q^2)$ embeds into $U_{2n}(q^2)$, by acting on a totally isotropic space of dimension $n$, and it does not embed in $U_m(q^k)$ for any $k$ with $m < 2n$.

Similarly ${\rm GL}_n(q)$ embeds into ${\rm Sp}_{2n}(q)$ and into ${\rm GO}^+_{2n}(q)$. For ${\rm GO}^-_{2n}(q)$, the largest totally isotropic spaces have dimension $n-1$, so to embed ${\rm GL}_n(q)$, we need to go up to ${\rm GO}^-_{2n+2}(q)$.

I am not exactly sure what you looking for with your length function. For a large proportion of elements $g \in {\rm GL}_n(q)$ the centralizer of $g$ in the larger group will be only twice as large as the centralizer in the smaller group, so $\ell(g)$ will not change much. But elements of ${\rm GL}_n(q)$ with a large fixed point space will have a much larger centralizer in the large group, so there will be distortion of $\ell(g)$.

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Thank you for your answer, which was quite useful so far. Can you give me any references for the contents of your answer? –  Abel Stolz Nov 7 '11 at 15:46

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