Isomorphism

neoscient · P versus NP Offline Joined: 01 May 2005 Posts: 26

Show that $(\mathbb{R}, +)$ is isomorphic to $(\mathbb{C},+)$ .

**Posted:** Sun May 08, 2005 3:41 pm

grobber

All we need to do is choose two bases of $\mathb R$ and $\mathbb C$ as vector spades over $\mathbb Q$ . The two bases have the same cardinality (the continuum cardinality), so the two vector spaces are both isomorphic to $\mathbb Q^c$ , where $c$ is the continuum cardinality.

**Posted:** Sun May 08, 2005 4:35 pm

grobber

Here's a nicer one:

Are $(\mathbb C^*,\cdot)$ and $(\mathbb S^1,\cdot)$ isomorphic? (here, $(\mathbb S^1,\cdot)$ is the multiplicative group of complex number with absolute value $1$ )

Enjoy Smile

**Posted:** Sun May 08, 2005 5:05 pm

neoscient · P versus NP Offline Joined: 01 May 2005 Posts: 26

hmm no. However, one can construct an isomorphism between

$(\mathbb{C}^{*}/ \mathbb{R}^{*}_{+},\cdot)$ and $(S^{1},\cdot)$

by the isomorphism theorem and using the map $f(z) = \frac{z}{|z|}$ .

**Posted:** Mon May 09, 2005 1:55 pm

grobber

How did you show that the two I asked about are not isomorphic?

**Posted:** Mon May 09, 2005 5:00 pm

neoscient · P versus NP Offline Joined: 01 May 2005 Posts: 26

well the problem is that you cant find a nice smooth function. the two dimensionalities are different, one is on a 2-manifold, the other's on a 1-manifold, so one cannot find a diffeomorphism between them. you can map the circle into $\mathbb{C}^{*}$ but it will be really jagged, and wont preserve any structure.

**Posted:** Mon May 09, 2005 9:16 pm

grobber

I don't want any smoothness properties for the isomorphism Wink

. The question is a purely algebraic one: is there a group isomorphism between the two or not? Smile

**Posted:** Mon May 09, 2005 9:19 pm

Leva1980 · Poincare Conjecture Offline Joined: 08 Jan 2005 Posts: 175

It seems that I found the solution, but it is rather long and boring.
May be I am missing something easy, but nevertheless here is it:

Suppose that in general we have an abelian groups $G', G''$ with a subgroups
$G' \supset N'$ , $G'' \supset N''$ , such that the following conditions
are satisfied :

* $N',N''$ are isomorphic

* For every $x' \in G'$ there exists $n \in \mathbb{N}$ such that $nx' \in N'$ .
For every $x'' \in G''$ there exists $n \in \mathbb{N}$ such that $nx'' \in N''$ .

* For every $y' \in G'$ and $n \in \mathbb{N}$ there exist exactly $n$ elements
$x' \in G'$ , such that $nx' = y'$ .
For every $y'' \in G''$ and $n \in \mathbb{N}$ there exist exactly $n$ elements
$x'' \in G''$ , such that $nx'' = y''$ .

Then $G',G''$ are isomorphic.
Let us prove this.
Suppose we have an isomorphism $f: N' \rightarrow N''$ , then we will show that
it can be extended to an isomorphism $G' \rightarrow G''$ .
Denote for a prime $p$ , by $N'_{p}$ the set of all $x' \in G'$ such that
$p^{k}x' = 0$ for large $k$ . Also denote by $N''_{p}$ the set of all
$x'' \in G'$ such that $p^{k}x'' = 0$ for large $k$ .
Then we have that every element $x \in G'$ can be written as a sum
$x = x_{1} + \ldots + x_{m}$ , where $x_{i} \in N'_{p_{i}}$ for some
collection of primes $p_{1},p_{2}, \ldots , p_{m}$ .
Indeed, take some $n \in \mathbb{N}$ , such that $nx \in N'$ ,
then if we take a prime decomposition $n = p_{1}^{k_{1}}p_{2}^{k_{2}} \ldots p_{m}^{k_{m}}$
and denote $n_{j} = \frac{n}{p_{j}^{k_{j}}}$ , then $gcd(n_{1},n_{2}, \ldots ,n_{k}) = 1$ ,
so we can find $a_{1},a_{2}, \ldots , a_{m} \in \mathbb{Z}$ , such that
$n_{1}a_{1} + n_{2}a_{2} + \ldots + n_{k} a_{k} = 1$ , then if we define
$x_{j} = a_{j}n_{j}x$ then of course $x = x_{1} + \ldots + x_{m}$ and
$x_{j} \in N_{p_{j}}'$ . Also note, that such representation is not unique, but if we have
$x = x_{1} + x_{2} + \ldots + x_{m} = y_{1} + y_{2} + \ldots + y_{m}$ , when
$x_{i},y_{i} \in N'_{p_{i}}$ , then $x_{i} - y_{i} \in N'$ for every $i$ .
All this is true also about $G''$ .
Take now some prime $p$ , and let's try to extend the isomorphism
$f : N' \rightarrow N''$ to homomorphism $N_{p}' \rightarrow N_{p}''$ .

Denote by $L_{p}'$ the group of all
$x' \in G'$ such that $px' = 0$ and by $L_{p}''$ the group of all
$x'' \in G''$ such that $px'' = 0$ . Then if we have $L_{p}' \subset N'$ then
also $L_{p}'' \subset N''$ , and $f(L_{p}') = L_{p}''$ .
Now, if $L_{p}' \nsubseteq N'$ , then $L_{p}' \cap N' = \emptyset$ and
$L_{p}'' \cap N'' = \emptyset$ , so $N' + L_{p}' \cong N' \oplus L_{p}'$
and $N'' + L_{p}'' \cong N'' \oplus L_{p}''$ , so because $L_{p}'$ , $L_{p}''$
are isomorphic, we can extend $f$ to an isomorphism
$f: N' + L_{p}' \rightarrow N'' + L_{p}''$ .

Suppose we have already extended $f$ to a monomorphism $f: K' \rightarrow N_{p}''$
for some $N' + L_{p}' \subset K' \subset N_{p}'$ . Denote $K'' := f(K') \subset N_{p}''$ .
Now suppose that $K' \subsetneq N_{p}'$ . Then there exists $x' \in N_{p}'$ , $x' \notin K'$ , such that $y':= px' \in K'$ . Denote $y'' = f(y') \in K''$ .
Take some $x'' \in N_{p}''$ , such that $px'' = y''$ . Then $x'' \notin K''$ , because
otherwise there exists $z \in K'$ such that $f(z) = x''$ , then
$f(pz) = y'' = f(y') = f(px')$ , so $f(pz-px') = 0$ so $pz-px' = 0$ so
$z-x' \in L_{p}'$ and $z \in K'$ , so $x' \in K' + L_{p}' = K'$ and we get a
contradiction. Therefore $x' \notin K'$ , $px' \in K'$
and $x'' \notin K''$ , $px'' \in K''$ . Therefore for every $k \in \mathbb{Z}$ ,
not divisible by $p$ , we have $kx' \notin K'$ , $kx'' \notin K''$ .
Now if we define $\overline{K}'$ to be the subgroup of $N_{p}'$ which is all
sums $kx' + z'$ , where $k \in \mathbb{Z}$ and $z' \in K'$ and
$\overline{K}''$ to be the subgroup of $N_{p}''$ which is all
sums $kx'' + z''$ , where $k \in \mathbb{Z}$ and $z'' \in K''$ .
Now for $y \in \overline{K}'$ if we represent it by $y = kx' + z'$ ,
where $k \in \mathbb{Z}$ and $z' \in K'$ , then define
$\widetilde{f} (y) := kx'' + f(z')$ . It is easy to see that $\widetilde{f}$
is well defined, and it is injective, because if we have
$f(y) = 0$ then $kx'' + f(z) = 0$ , so $kx'' \in K'$ , so $k$ is divisible by $p$ ,
therefore $kx' \in K'$ , so $y \in K'$ , and because $f$ is injective, we get $y = 0$ .

So we see that if we have a monomorphism $f : K' \rightarrow N_{p}''$ , and
$K' \subsetneq N_{p}'$ , then it can be always extended to a monomorphism, defined
on a bigger subgroup of $N_{p}'$ . Now applying the Zorn lemma, we get that
$f : N' + L_{p}' \rightarrow N'' + L_{p}''$ can be extended to a monomorphism
$f_{p} : N_{p}' \rightarrow N_{p}''$ . Then it must be also onto, because otherwise
we have an isomorphism $f_{p}^{-1} : Im(f_{p}) \subsetneq N_{p}'' \rightarrow N_{p}'$ , which cannot be extended to a monomorphism, defined on a
bigger subgroup.

Finally we see that for every prime $p$ , $f : N' \rightarrow N''$ can be extended
to an isomorphism $f_{p} : N_{p}' \rightarrow N_{p}''$ .
Now, define $F : G' \rightarrow G''$ in the following way:
if we take $x \in G'$ and a representation $x = x_{1} + \ldots + x_{m}$ ,
where $x_{i} \in N'_{p_{i}}$ for some collection of primes
$p_{1},p_{2}, \ldots , p_{m}$ , then define $F(x) := f_{p_{1}}(x_{1}) + f_{p_{2}}(x_{2}) + \ldots + f_{p_{m}}(x_{m})$ . Then $F$ is well defined,
because if we have two representations
$x = x_{1} + x_{2} + \ldots + x_{m} = y_{1} + y_{2} + \ldots + y_{m}$ , when
$x_{i},y_{i} \in N'_{p_{i}}$ , then $x_{i} - y_{i} \in N'$ for every $i$ ,
so $f_{p_{i}}(x_{i}-y_{i}) = f(x_{i} - y_{i})$ , so
$((f_{p_{1}}(x_{1}) + f_{p_{2}}(x_{2}) + \ldots + f_{p_{m}}(x_{m})) - (f_{p_{1}}(y_{1}) + f_{p_{2}}(y_{2}) + \ldots + f_{p_{m}...$ . Also it is clear that
$F$ is a homomorphism, and it is easy to see that it is one-to-one and onto.
Therefore $F$ is an isomorphism.

Let us now prove that $G' = ( \mathbb{S}^{1} , \cdot )$ and
$G'' = ( \mathbb{C}^{*} , \cdot )$
are isomorphic. Take a Hamel basis for the $\mathbb{R}$ over $\mathbb{Q}$ ,
which contains $1$ and denote by $\Lambda$ the set of all element of this basis,
except $1$ . Then $\Lambda$ has continuum cardinality and any non-trivial combination
with integer coefficients of elements of $\Lambda$ cannot be an integer number.
Now elements in $( \mathbb{S}^{1} , \cdot )$ , of the form $e^{ 2\pi i \lambda }$ ,
when $\lambda \in \Lambda$ . They are independent. Take now maximal set in
$( \mathbb{S}^{1} , \cdot )$ , which contains all those elements, and all elements
of it are independent. Denote by $N'$ the subgroup of $G' = ( \mathbb{S}^{1} , \cdot )$ ,
generated by elements of this set. Then $N'$ is a free abelian group with
continuum number of generators, and it is easy to see that the pair $(G',N')$
satisfies all the conditions above.
Now for $G'' = ( \mathbb{C}^{*} , \cdot )$ do the same thing :
take all elements of the form $e^{ 2\pi i \lambda }$ , when $\lambda \in \Lambda$ ,
then complete this set to a maximal set which contains independent elements,
and denote by $N''$ the subgroup of $G'' = ( \mathbb{C}^{*} , \cdot )$ ,
generated by this set. Then again, $N''$ is a free abelian group with
continuum number of generators, and the pair $(G'',N'')$ satisfies all conditions
above. Also, $N'$ , $N''$ are free abelian groups with continuum number of generators,
so they are isomorphic. Therefore applying what we have shown, we get that
$G' , G''$ are isomorphic.

**Posted:** Thu May 12, 2005 6:28 am

_________________
Gromov is the best!

grobber

I did not (and will not Wink

) read it, but I do believe we did about the same thing (when it comes to this problem; I see you have proven a very nice generalization Smile

).

**Posted:** Thu May 12, 2005 6:55 am

neoscient · P versus NP Offline Joined: 01 May 2005 Posts: 26

haha hmm.. i'll try to digest that one after exams.

**Posted:** Thu May 12, 2005 12:34 pm