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Given a finite Abelian group: $G=\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \mathbb{Z}_{n_3}$, where ${n_1},{n_2},{n_3}$ are arbitrary positive integers. ${n_1},{n_2},{n_3}$ may have or may not have common divisors.

Question: What is the smallest number of $N$ such that the following Lie group contains $G=\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \mathbb{Z}_{n_3}$ as a finite discrete subgroup? i.e. $N=$?

(1) SU($N$)

(2) SO($N$)

(3) SU($N$) $\times$ SU($N$)

(4) SO($N$) $\times$ SO($N$)

eg. for $G=\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$, I suppose the answer for the smallest number of $N$ is:

(1) SU($4$)

(2) SO($4$)

(3) SU($3$) $\times$ SU($3$)

(4) SO($3$) $\times$ SO($3$)

What will be the cases of $N$ for a generic $G=\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \mathbb{Z}_{n_3}$, then $N=$?

Thank you.

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    $\begingroup$ What do you mnean by "generic $G$"? As Alain points out, the 1-torus $S^1$ has a (unique) cyclic subgroup of each possible order. So rank $\geq 3$ is always safe; no need to enumerate special cases. But to refine $N$ you need to see how close $G$ is to being cyclic. For instance, if all $n_i$ are relatively prime, $G$ itself is cyclic and contained in any rank $\geq 1$ compact Lie group. But if all $n_i$ are equal, you need total rank at least 3. (Of course, rank of a product is sum of the rqnks in your list.) $\endgroup$
    – Jim Humphreys
    Commented Dec 16, 2013 at 0:48

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As soon as a compact connected Lie group has rank at least 3, i.e. it contains a 3-dimensional torus, it will contain your group $G$ (for all values of the parameters). So for example $SU(4), SO(6), SU(3)\times SU(3),SO(4)\times SO(4)$ do the job. Of course you must see whether these values are optimal... Where does the question come from?

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  • $\begingroup$ Dear Alain. Vielen Danke. Thanks for the answer. May I ask: So is it true that for any N larger than that value of your N, then SU(N>4) and SO(N>6), SU(N>3)×SU(N>3),SO(N>4)×SO(N>4) would contain G=Zn1×Zn2×Zn3? Thank you again. $\endgroup$
    – wonderich
    Commented Dec 15, 2013 at 22:47
  • $\begingroup$ @Idear: I think the answer to your question is in Jim's comment above. $\endgroup$
    – Alain Valette
    Commented Dec 16, 2013 at 10:24

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