# An exercise about groups

Posted by Mathoman, Sunday 20 December 2009 at 12:45 - Riddles And Exercises

A *topological group* is a set *G* with a group structure and a topology such that the binary law

**Question**

Are there topological groups which are isomorphic as groups and homeomorphic as topological spaces but not isomorphic as topological groups?

**Answer**

Yes. The proof goes in three stages:

**Let***G*and*H*be dense subsets of and a monotonic bijection. Then*f*is a homeomorphism.

Without restriction of generality we can suppose that*f*is increasing. We prove that it is continuous. Let and . Since*H*is dense in we have . Thus there exists . Similarly there exists . Since*f*is onto we can write with . Let . Then for all*x*in*G*which proves that*f*is continuous in . The proof of the continuity of the inverse function is the same.

**Let***G*and*H*be countable and dense subsets of . Then they are homeomorphic.

First we write We will now list*G*and*H*in a new way, and The aim is to make a monotonic bijection (then it is automatically a homeomorphism). For this we proceed as follows.

We take*k*=0.

We take . To choose we look at the order of and .*k*=1.

If then we take as any element in*H*which is smaller than .

If then we take as any element in*H*which is greater than .

We take as the first element of listed in (**). To choose we look at the order of .*k*=2.

If is smaller than and we take as any element in*G*which is smaller than and .

If is greater than and we take as any element in*G*which is greater than and .

If is between and we take as any element in*G*which is between and .

We take as the first element of listed in (*). To choose we look at the order of . There are 24 possible orders for these four numbers.*k*=3.

If we take as any element in*H*which is smaller than .

If we take as any element in*H*which is bewteen and .

And so forth.

**The topological groups and fulfill our wishes.**

After what has been said before,*G*and*H*are homeomorphic as topological spaces. Clearly they are isomorphic groups. But they are not isomorphic topological groups. In fact, suppose that there exists an isomorphism of topological groups . One proves easily by induction that*f*(*n*)=*nf*(1) for every integer*n*, and then that*f*(*r*)=*rf*(1) for every rational*r*. Therefore by continuity a contradiction in .

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