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Do there exist families of isomorphism classes of groups $G_{s}$ such that $\forall s\in[0,1], |G_{s}|=\mathfrak{c}$, where $\mathfrak{c}$ is the cardinality of the continuum, and for any $a,b\in[0,1], a\neq b$, $G_{a}$≇$G_{b}$?

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Given your tags: presumably you are asking for these to be Lie groups? and if so, by (non)isomorphism, do you mean in the sense of bicontinuous group isomorphism? –  user16299 Sep 22 '11 at 7:34
    
I don't think I actually require that they be Lie groups -- that would be a nice bonus, though, and yes, if they were Lie groups, then I do mean in the sense of a bicontinuous group isomorphism. –  deoxygerbe Sep 22 '11 at 7:40
    
Well, if you didn't need them to be Lie groups, my guess would be that one could cook up discrete examples, although I don't claim to see exactly how to do it –  user16299 Sep 22 '11 at 8:27
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2 Answers

up vote 3 down vote accepted

I think you are asking if there are families $\{G_t:t\in[0,1]\}$ of Lie groups of positive dimension whose elements are non-isomorphic in pairs.

There are examples of curves of non-isomorphic Lie groups in the class of nilpotent groups. In fact, as soon as the dimension is greater or equal to $7$, there are continuously-many non-isomorphic filiform Lie algebras, so the same thing happens for groups (filiform Lie algebras are "maximally non-abelian" nilpotent Lie algebras) This is discussed in detail, for example, in the book by Goze on Nilpotent Lie algebras.

As the dimension grows, we find families of unbounded dimension. In general, the nilpotent part of the world is filled with many, many examples.

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(Notice that a bicontinuous isomorphism between Lie groups is in fact a bidifferentiable, so looking at Lie algebras is enough to tell groups apart) –  Mariano Suárez-Alvarez Sep 22 '11 at 8:00
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Apparently the question was about bicontinuous isomorphism classes of Lie groups, and that question was answered by Mariano Suarez-Alvarez. Now, I suspect that many of these Lie-groups are in fact isomorphic algebraically, just not by a bicontinuous isomorphism.

Let me answer the question literally: There is a family of continuum many, pairwise non-isomorphic groups of size continuum. Actually, there is such a family of countable groups.

Namely, let $\mathbb P$ denote the set of prime numbers. For each infinite set $S\subseteq\mathbb P$ let $G_S$ be the direct sum of the abelian groups $\mathbb Z_p$, $p\in S$. The groups $G_S$ are pairwise non-isomorphic since if $S,T\subseteq\mathbb P$ and $p\in S\setminus T$, then $G_T$ has no element of order $p$ but $G_S$ does.

Now, the groups $G_S$ are countable. If you want size continuum, you can either take the cartesian product instead of direct sum, or simply multiply everything with $\mathbb R$. In either case you get a family of pairwise nonisomorphic groups of size $2^{\aleph_0}$.

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