# Why and how is G* isomorphic to G?

• Feb 14th 2013, 05:53 AM
x3bnm
Why and how is G* isomorphic to G?
There is an example on pg-155-156 of Pinter's abstract algebra book which is:

The Problem:

Let $G$ and $H$ be any groups and consider their product $G \times H$.

Remember that $G \times H$ consists of all the ordered pairs $(x,y)$ as $x$ ranges over $G$ and $y$ ranges over $H$. You multiply ordered pairs by multiplying corresponding
components; that is, the operation on $G \times H$ is given by

$(a,b) \cdot (c,d) = (ac, bd)$

Now, let $f$ be the function from $G \times H$ onto $H$ given by

$f(x, y) = y$

It is easy to check that $f$ is a homomorphism. Furthermore, $(x, y)$ is in the kernel of $f$ if and only if $f(x,y) = y = e$. This means that the kernel of $f$ consists of all the ordered pairs whose
second component is $e$. Call this kernel $G^{*}$; then

$G^{*} = \{(x,e) : x \in G \}$

We symbolize all this by writing

$f: (G \times H) \xrightarrow[ G^{*}\,\,\,\,\,\,\, ]{\,\,} H$

By the fundamental homomorphism theorem, we deduce that $H \cong (G \times H)/G^{*}$.

[It is easy to see that $G^{*}$ is an isomorphic copy of $G$; thus identifying $G^{*}$ with $G$, we have shown that, roughly speaking, $(G \times H)/G \cong H$.]

My question is: why and how is that $G^{*}$ isomorphic to $G$?

Can anyone kindly tell me: why's that $G^{*}$ here isomorphic to $G$?
• Feb 14th 2013, 06:59 AM
emakarov
Re: Why and how is G* isomorphic to G?
You are honestly claiming that you don't see an isomorphism between $\{(x,e_H)\mid x\in G\}$ and $\{x\mid x\in G\}$ (here $e_H$ is the unit of $H$)?
• Feb 14th 2013, 07:33 AM
x3bnm
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by emakarov
You are honestly claiming that you don't see an isomorphism between $\{(x,e_H)\mid x\in G\}$ and $\{x\mid x\in G\}$ (here $e_H$ is the unit of $H$)?

Sorry that my analyzing capability is little slow. Here are my problems.

According to the definition of isomorphism(Pinter's AA book on pg-89):

Let $G_1$ and $G_2$ be groups. A bijective function $f : G_1 \to G_2$ with the property that for any two elements $a$ and $b$ in $G_1$,

$f(ab) = f(a)f(b)$

is called an isomorphism from $G_1$ to $G_2$.

If there exists an isomorphism from $G_1$ to $G_2$, we say that $G_1$ is isomorphic to $G_2$.

I can't find a way to show that $f$ is both one-to-one and onto. I know that $f(ab) = f(a)f(b)$ because of homomorphism.

In other words how do I prove that:

1) If $f(a, e_1) = f(b, e_2)$ then $(a, e_1) = (b, e_2)$ and
2) There is a function $y = f(x)$ for every $y$?
• Feb 14th 2013, 07:46 AM
emakarov
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by x3bnm
I can't find a way to show that $f$ is both one-to-one and onto.

This depends on what f is. How do you define it?

Quote:

Originally Posted by x3bnm
I know that $f(ab) = f(a)f(b)$ because of homomorphism.

Are you given that f is a homomorphism, or are you trying to prove it? In the second case, you obviously don't know that f is a homomorphism yet.

I feel almost silly offering to consider the map $G^*\ni(x,e)\mapsto x\in G$ as a potential isomorphism.
• Feb 14th 2013, 08:56 AM
x3bnm
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by emakarov
This depends on what f is. How do you define it?

Are you given that f is a homomorphism, or are you trying to prove it? In the second case, you obviously don't know that f is a homomorphism yet.

I feel almost silly offering to consider the map $G^*\ni(x,e)\mapsto x\in G$ as a potential isomorphism.

I think I'm causing confusion. Suppose there is an example like below:

-----------------------------------------------------------------------------------------------------------------------------------
Let $G$ and $H$ be any groups and consider their product $G \times H$.

Remember that $G \times H$ consists of all the ordered pairs $(x,y)$ as $x$ ranges over $G$ and $y$ ranges over $H$. You multiply ordered pairs by multiplying corresponding
components; that is, the operation on $G \times H$ is given by

$(a,b) \cdot (c,d) = (ac, bd)$

Now, let $f$ be the function from $G \times H$ onto $H$ given by

$f(x, y) = y$

It is easy to check that $f$ is a homomorphism. Furthermore, $(x, y)$ is in the kernel of $f$ if and only if $f(x,y) = y = e$. This means that the kernel of $f$ consists of all the ordered pairs whose
second component is $e$. Call this kernel $G^{*}$; then

$G^{*} = \{(x,e) : x \in G \}$

...
-----------------------------------------------------------------------------------------------------------------------------

Now how do I mathematically prove that $G^{*}$ is isomorphic to $G$? Is it possible that $G^{*}$ is isomorphic to $G$?

But the author(Pinter) said that on pg-156:

"...... $G^{*}$ is a isomorphic copy of $G$....." . Then where am I wrong?
• Feb 14th 2013, 09:23 AM
emakarov
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by x3bnm
Now how do I mathematically prove that $G^{*}$ is isomorphic to $G$?

As I said,
Quote:

Originally Posted by emakarov
I feel almost silly offering to consider the map $G^*\ni(x,e)\mapsto x\in G$ as a potential isomorphism.

• Feb 14th 2013, 09:53 AM
x3bnm
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by emakarov
As I said,

But the author Pinter on his book(pg-156) "A Book of Abstract Algebra" second edition said that " .......[tex]G^{*} is a isomorphic copy of $G$....".

The word "isomorphic" and "isomorphic copy" are they same thing or two different things?
• Feb 14th 2013, 09:55 AM
emakarov
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by x3bnm
The word "isomorphic" and "isomorphic copy" are they same thing or two different things?

The same. The word "isomorphic" is a technical term, but "copy" is just a part of the natural language description.
• Feb 14th 2013, 10:25 AM
Deveno
Re: Why and how is G* isomorphic to G?
consider the mapping:

h:G*--->G given by: h(g,eH) = g.

although it is very simple, we can easily prove h is an isomorphsm of G* with G.

1) h is a homomorphism:

h((g,eH)*(g',eH)) = h((gg',eH)) = gg' = h((g,eH))h(g',eH))

2) h is injective:

suppose that h((g,eH)) = h((g',eH)). this means that g = g', so (g,eH) = (g',eH) (both coordinates are equal).

3) h is surjective:

suppose g is ANY element of G. then (g,eH) is an element of G* that h maps to g.

****************

it may make more sense to consider the inverse isomorphism h-1:G-->G* given by: h-1(g) = (g,eH). if we extend the co-domain to GxH this is an example of an embedding of G in GxH

(an embedding is a monomorphism from one group into another). even though G* isn't strictly EQUAL to G, as groups they are isomorphic (the second coordinate in G* is always eH, which pretty much sits around and does nothing all day).

we use embeddings all the time, and don't even think about it, here is another example:

the integers and rational numbers are two "different kinds of things". an integer is a series of 1's or -1's:

5 = 1+1+1+1+1
-4 = (-1)+(-1)+(-1)+(-1)

a rational numbers is an equivalence class of pairs of integers:

a/b = {(c,d) in ZxZ*: ad = bc}.

nevertheless, it is common to embed the integers in the rationals by the map n--> [(n,1)] (we think of the integer 5 as 5/1 = 10/2 = 15/3 =....etc.)

*********************

with the direct product group GxH the two homomorphisms:

p1(g,h) = g
p2(g,h) = h

are very important: these are called PROJECTIONS.

it turns out that these projections actually DEFINE GxH:

suppose we have a third group K, with two group homomphisms φ1: K-->G, and φ2:K-->H.

i claim there is a UNIQUE homomorphism ψ:K-->GxH with:

p1(ψ(k)) = φ1(k) and
p2(ψ(k)) = φ2(k), for all k in K.

how should we define ψ? it would seem the obvious way would be to set ψ(k) = (φ1(k),φ2(k)) (convince yourself that this works, and IS a homomorphism).

in fact the properties:

p1∘ψ = φ1
p2∘ψ = φ2

force this definition on us. just as GxH is called the (direct) PRODUCT of G and H, we can call ψ the product of the homomorphisms φ1 and φ2: ψ = φ1 x φ2.

this is very similar to (in some ways "just the same") the projection of a point in the plane (x,y) onto either the x-axis, or the y-axis. in fact, it turns out that as (additive/abelian) groups, we have:

R2 ≅ R x R. the analogue of G* is in this case, the x-axis, and perhaps it should be clearer to you that the x-axis (the points in the plane (x,0)) is indeed isomorphic to the real line.
• Feb 14th 2013, 10:47 AM
x3bnm
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by Deveno
consider the mapping:

h:G*--->G given by: h(g,eH) = g.

although it is very simple, we can easily prove h is an isomorphsm of G* with G.

1) h is a homomorphism:

h((g,eH)*(g',eH)) = h((gg',eH)) = gg' = h((g,eH))h(g',eH))

2) h is injective:

suppose that h((g,eH)) = h((g',eH)). this means that g = g', so (g,eH) = (g',eH) (both coordinates are equal).

3) h is surjective:

suppose g is ANY element of G. then (g,eH) is an element of G* that h maps to g.

****************

it may make more sense to consider the inverse isomorphism h-1:G-->G* given by: h-1(g) = (g,eH). if we extend the co-domain to GxH this is an example of an embedding of G in GxH

(an embedding is a monomorphism from one group into another). even though G* isn't strictly EQUAL to G, as groups they are isomorphic (the second coordinate in G* is always eH, which pretty much sits around and does nothing all day).

we use embeddings all the time, and don't even think about it, here is another example:

the integers and rational numbers are two "different kinds of things". an integer is a series of 1's or -1's:

5 = 1+1+1+1+1
-4 = (-1)+(-1)+(-1)+(-1)

a rational numbers is an equivalence class of pairs of integers:

a/b = {(c,d) in ZxZ*: ad = bc}.

nevertheless, it is common to embed the integers in the rationals by the map n--> [(n,1)] (we think of the integer 5 as 5/1 = 10/2 = 15/3 =....etc.)

*********************

with the direct product group GxH the two homomorphisms:

p1(g,h) = g
p2(g,h) = h

are very important: these are called PROJECTIONS.

it turns out that these projections actually DEFINE GxH:

suppose we have a third group K, with two group homomphisms φ1: K-->G, and φ2:K-->H.

i claim there is a UNIQUE homomorphism ψ:K-->GxH with:

p1(ψ(k)) = φ1(k) and
p2(ψ(k)) = φ2(k), for all k in K.

how should we define ψ? it would seem the obvious way would be to set ψ(k) = (φ1(k),φ2(k)) (convince yourself that this works, and IS a homomorphism).

in fact the properties:

p1∘ψ = φ1
p2∘ψ = φ2

force this definition on us. just as GxH is called the (direct) PRODUCT of G and H, we can call ψ the product of the homomorphisms φ1 and φ2: ψ = φ1 x φ2.

this is very similar to (in some ways "just the same") the projection of a point in the plane (x,y) onto either the x-axis, or the y-axis. in fact, it turns out that as (additive/abelian) groups, we have:

R2 ≅ R x R. the analogue of G* is in this case, the x-axis, and perhaps it should be clearer to you that the x-axis (the points in the plane (x,0)) is indeed isomorphic to the real line.

Deveno, you must be a good teacher. How do you know I've problem with embeddings and projections? Thanks for taking the time explaining this and thanks for the proof. It is such a great help.
• Feb 20th 2013, 09:27 AM
x3bnm
Re: Why and how is G* isomorphic to G?
Sorry Deveno I've a question.

You said:

Quote:

Originally Posted by Deveno

consider the mapping:

h:G*--->G given by: h(g,eH) = g.
......

Is the choice of the function $h$ here arbitrary?

How do one choose a function $h$ if they suspect that two groups may be isomorphic? Is there any rule for that?

I've this question because I for this instance had difficulty figuring out the isomorphic function $h$.

So my question: is there any rule that I need to follow for assigning one group's element to the other?

or is it always straightforward like you demonstrated in your answer to my question like this:

$h(g,e_H) = g$?

Are all the isomorphic functions this simple?
• Feb 25th 2013, 12:04 PM
x3bnm
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by x3bnm
Sorry Deveno I've a question.

You said:

Is the choice of the function $h$ here arbitrary?

How do one choose a function $h$ if they suspect that two groups may be isomorphic? Is there any rule for that?

I've this question because I for this instance had difficulty figuring out the isomorphic function $h$.

So my question: is there any rule that I need to follow for assigning one group's element to the other?

or is it always straightforward like you demonstrated in your answer to my question like this:

$h(g,e_H) = g$?

Are all the isomorphic functions this simple?

I found the answer. Not all isomorphic functions are this simple.

For example, the group of positive real numbers $\mathbf{R}^{+}$ is isomorphic to the group of real numbers $\mathbf{R}$

The isomorphic function $f$ is $\log_{b}:\mathbf{R}^{+} \to \mathbf{R}$

So all isomorphic functions are not simple and that's the answer to my question. You've to use trial and error to find the isomorphic function.
• Feb 27th 2013, 01:11 AM
Deveno
Re: Why and how is G* isomorphic to G?
yes, and no.

proving two groups are isomorphic is not always as easy as it sounds. often, given two groups G and G', it's easier to show they AREN'T isomorphic, then it is to find an isomorphism.

some practical things to check first:

are the two groups of the same size (if one is infinite and one is finite, that settles it)?

are both groups cyclic, or non-cyclic?

are both groups abelian, or non-abelian? (if they are both abelian and finitely-generated, there is a structure theorem that allows us to tell if they are isomorphic or not).

the "bad case" for many practical purposes is two finite non-abelian groups of the same order: we may have two "presentations" of the same group (that is the group may be defined on a generating set with relations), and it can be darned difficult to TELL.

in THIS particular case, the isomorphism of G* = Gx{eH} with G is pretty easy to figure out. such is usually NOT the case, and is one of the reasons a direct product is a "nice" way to make a bigger group from two smaller ones (we can recover the "factors" fairly easily). so if we know a group K = GxH, where G and H are groups we understand well, we've got a good grip on what K is like.
• Feb 28th 2013, 07:20 AM
x3bnm
Re: Why and how is G* isomorphic to G?
Quote:

Originally Posted by Deveno
yes, and no.

proving two groups are isomorphic is not always as easy as it sounds. often, given two groups G and G', it's easier to show they AREN'T isomorphic, then it is to find an isomorphism.

some practical things to check first:

are the two groups of the same size (if one is infinite and one is finite, that settles it)?

are both groups cyclic, or non-cyclic?

are both groups abelian, or non-abelian? (if they are both abelian and finitely-generated, there is a structure theorem that allows us to tell if they are isomorphic or not).

the "bad case" for many practical purposes is two finite non-abelian groups of the same order: we may have two "presentations" of the same group (that is the group may be defined on a generating set with relations), and it can be darned difficult to TELL.

in THIS particular case, the isomorphism of G* = Gx{eH} with G is pretty easy to figure out. such is usually NOT the case, and is one of the reasons a direct product is a "nice" way to make a bigger group from two smaller ones (we can recover the "factors" fairly easily). so if we know a group K = GxH, where G and H are groups we understand well, we've got a good grip on what K is like.

Thank you Deveno for your informational post. I understand what you mean. Sorry for late reply. I edit my last post here. I told to use "trial and error" to find isomorphism. After reading your(Deveno) post I think my "trial and error" method is wrong. Thanks for correcting my concept about this.