## In search of an isomorphism

I’m working on some group-theoretic stuff related to *n* copies of a unital quantum channel and am in search of some isomorphism between the dihedral group and the unitary group U(*n*).

Now, what’s interesting is that the unitary group U(1) is isomorphic to the circle group, i.e. the geometric circle, while can be thought of as representing the discrete rotations of an *n*-gon. So one would think that as , would become isomorphic to U(1) since the *n*-gon geometrically approaches a circle. But is isomorphic to which geometrically is a line infinite in both directions.

Technically, for what I’m interested in, I suspect U(1) would be too restrictive anyway. I’m more interested in finding a direct isomorphism between and U(*n*) or, at the very least, something broader than U(1) and, preferably, broader than SU(2) as well.

So if you’re a regular reader of this blog and you know of any such isomorphism, post it here!

February 5, 2010 at 7:16 pm

In general, one cannot take a “limit” of a sequence of finite groups, especially without choosing a family of homomorphisms between them.

One way to do this would be to use the family of surjective homomorphisms D_n -> D_m existing when m divides n. In terms of generators and relations

D_n =

so that if m | n, simply sending a to a and b to b induces a homomorphism. If you took the inverse limit of this system, you would get the profinite dihedral group, which is an uncountable group which is compact and totally disconnected with respect to a natural topology.

It is more common to define

D_{\infty} =

and then each D_n is a quotient of D_{\infty}. Note that D_{\infty} is the isometry group of the integers viewed as a subset of R. But as a group, D_{\infty} is not isomorphic to Z, as you claim: it couldn’t be, because the relation shows that it is nonabelian. It is in fact, a semidirect product of Z with the cyclic group of order 2.

February 5, 2010 at 7:18 pm

Unfortunately the symbols with generators and relations disappeared.

In plainer text, D_n is generated by elements a and b and subject to the relation that a^n = b^2 = 1, b a b^{-1} = a^{-1}, and D{\infty} has the same generating set but omits the relation a^n = 1.

February 5, 2010 at 10:54 pm

Pete,

Huh. Wonder where I read that (about being isomorphic to ). Thanks for the clarification.