Expert: Paul Klarreich - 10/6/2008

Question
QUESTION: (1) If I have a function phi mapped from G to G' which is an isomorphism of groups how do I explain why phi^{-1} from G to G exists and (2) how do I show phi^{-1} is an isomorphism of groups? I have lots of ideas but am having trouble putting the ideas into words. For instance, to be an isomorphism you need a homomorphism that is both onto and 1-1,(a bijection) which means that phi from G' to G is an isomorphism as well. I also know that phi(e) where e is the identity in G maps to e' which is the identity in G' and vice versa and that phi(a^{-1}) = [phi(a)]^{-1}. Somewhere in there, there is an explanation why my questions are true but how does one say it? Any help will be largely appreciated. Sombra

ANSWER: Questioner:   Sombra
Category:  Calculus
Private:  No
 
Subject:  group isomorphisms
Question:  
(1) If I have a function phi mapped from G to G' which is an isomorphism of groups how do I explain why phi^{-1} from G to G exists and

(2) how do I show phi^{-1} is an isomorphism of groups?

I have lots of ideas but am having trouble putting the ideas into words. For instance, to be an isomorphism you need a homomorphism that is both onto and 1-1,(a bijection) which means that phi from G' to G is an isomorphism as well.

I also know that phi(e) where e is the identity in G maps to e' which is the identity in G' and vice versa and that phi(a^{-1}) = [phi(a)]^{-1}.

Somewhere in there, there is an explanation why my questions are true but how does one say it? Any help will be largely appreciated. Sombra
...............................................
Hi, Sombra,

[I never liked algebra but it is good for me to go back and work on it.]  The following arguments might take a little time (2-3 weeks) to refine, but if you are in a hurry....

I will '*' for the inverse, to save typing.  Naturally, if I write  a*, it means the group inverse, not the 'inverse image', which is  phi*(a).

If phi : G -> G' is an isomorphism of groups.

Ah, that means:(I am repeating some of your stuff to review it for myself.)

0) phi is a mapping, so it takes every a in G to a unique phi(a) in G'.  So if it is 1-1 and onto, there is automatically an inverse map that is 1-1 and onto.

1) phi is 1-1.  i.e.  a /= b in G  --> phi(a) /= phi(b) in G'. [different elements in G --> different elements in G']
2) phi is onto.  i.e.  x in G' --> there is  a in G such that  phi(a) = x.
3) phi takes products into products:  phi(ab) = phi(a) phi(b)

From 1,2,3, we can show that:

4) phi(e in g) = e' in G' (as you noted.)

This comes from: phi(a) = phi(ea) = phi(e) phi(a), therefore  phi(e) = e'.

5) phi(a* in G) = phi(a)* in G'

Given a, phi(e) = phi(aa*)  phi(a) phi(a*) = e', therefore phi(a*) = phi(a)*

Parts 1 and 2) how do I explain why phi* from G to G exists AND IS AN ISOMORPHISM.

You prove it by showing that it satisfies the requirements.  Define phi*(x) as (naturally) the inverse image.

We have to show that:

1) phi* is 1-1.  This is so; a mapping, by definition, is unique.  i.e.  x/=y in G' --> phi*(x) /= phi*(y) and,
2) since phi is a mapping from G, phi* is onto G.

3) We know phi(ab) = phi(a) phi(b), to prove that  phi*(xy) = phi*(x) phi*(y)

Now x = phi(a), for some a, and  y = phi(b), for some b.
THIS IS THE KEY PROOF.

phi*(xy) = (phi*( phi(a) phi(b) )
        = phi*( phi(ab) )
        = ab
        = phi*(phi(a)) phi*(phi(b))
        = phi*(x) phi*(y)

so phi* takes products into products.

[whew!]

-- you do not have to formally prove 4 and 5; they are consequences of 1,2,3.



---------- FOLLOW-UP ----------

QUESTION: This may not seem like a follow up question but it is to the complete problem I am doing. What I asked above is just a small piece of what I am trying to solve. If I have two subgroups, say H and K to a group G and I know H intersect K is a subgroup and I also know that H and K are both finite but the gcd of the order of H and K is 1, how can I show their intersection is just {e}? In order for H and K, and H intersect K to be subgroups of a group G, the identity must be in all three so it is trivial that H intersect K is {e}. I must be missing a finer detail because I need to establish that H and K have nothing else in common. If I knew either or both were normal I could do it or if I knew that G was finite I could do it or if I knew that H and K  were in different partitions of G I could do it but I have none of these things to go on. What am I missing. Thanks! Sombra

ANSWER: Questioner:   Sombra
Category:  Calculus
Private:  No

---------- FOLLOW-UP ----------

This may not seem like a follow up question but it is to the complete problem I am doing. What I asked above is just a small piece of what I am trying to solve.

If I have two subgroups, say H and K to a group G and I know H intersect K is a subgroup and I also know that H and K are both finite but the gcd of the order (ORDERS?) of H and K is 1, how can I show their intersection is just {e}? In order for H and K, and H intersect K to be subgroups of a group G, the identity must be in all three so it is trivial that H intersect K is (CONTAINS?) {e}.

I must be missing a finer detail because I need to establish that H and K have nothing else in common. If I knew either or both were normal I could do it or if I knew that G was finite I could do it or if I knew that H and K  were in different partitions of G I could do it but I have none of these things to go on. What am I missing. Thanks! Sombra
 
..............................
Ok, let's see:

H  is a subg of G
K  is a subg of G
H int K is a subg of G; call it L.

gcd( n(H), n(K)) = 1

I seem to recall something about the order of a subgroup being a divisor of the order of the group, assuming that the group is finite.  (Some Theorem of Lagrange or something?  I can't recall.)  

Now H and K are finite.  Therefore:

order(L) divides order(H)
order(L) divides order(K)

Therefore,

order(L) divides gcd(order(H),order(K))

If this is correct, you can finish up.



---------- FOLLOW-UP ----------

QUESTION: Below is the solution I came up with, I thought you might be interested in my perspective. But while I am at it, I want to thank you for your insight as well, as it has helped me organize my thoughts tremendously.

Suppose that H and K are finite and gcd(|H|,|K|) = 1. Show  H intersect K ={e}.

H is finite implies there is an h in H such that h^m = e   for some integer m, equally as well K being finite implies there is a k in K such that k^n = e for some integer n. We already know e is in H, K, and H intersect K
since they are all subgroups of G. All e’s being equal we have e =h^m=k^n   which means either that n = m = 1 and h = k = e and H intersect K = {e} or that gcd(|H|,|K|) >1 which contradicts that the gcd(|H|,|K|) = 1. Therefore, if H and K are finite and gcd(|H|,|K|) = 1, H intersect K  = {e}.

Thanks again, Sombra


Answer
Hi, again, Sombra,

You're really making me work here. [It's good for me -- don't stop.]
..........
QUESTION: Below is the solution I came up with, I thought you might be interested in my perspective. But while I am at it, I want to thank you for your insight as well, as it has helped me organize my thoughts tremendously.

Suppose that H and K are finite and gcd(|H|,|K|) = 1. Show  H intersect K ={e}.

H is finite implies there is an h in H such that h^m = e   for some integer m,

>> Yes, that seems true.  If  h /= e then eventually h^m = e, otherwise H would be infinite.  Of course, that does not imply that Ord(H) = m, because finite groups are not necessarily cyclic.
.............

equally as well K being finite implies there is a k in K such that k^n = e for some integer n.

>> OK so far.
..............
We already know e is in H, K, and H ^ K since they are all subgroups of G.

All e’s being equal we have e = h^m = k^n   which means either n = m = 1 and h = k = e and H ^ K = {e} or that gcd(|H|,|K|) > 1

>> Wait a minute.  I don't see how that follows.  You have not proved that m,n are the orders of H,K.
...............
which contradicts that the gcd(|H|,|K|) = 1. Therefore, if H and K are finite and gcd(|H|,|K|) = 1, H intersect K  = {e}.

Thanks again, Sombra
........................

Sorry, I could be wrong, but I think your proof is flawed.  The direction I suggested would go this way:

1. (Thm) The order of a subgroup of a finite group is divisor of the order of the group.

To prove this, you just form the set of right or left cosets.  Prove that every coset has the same cardinality as the subgroup.  [Does not have to be normal.]

2. (Thm) If x | a  and  x | b, then x | gcd(a,b)  [See any basic number theory text.]

(As before)

H  is a finite subg of G
K  is a finite subg of G
L = H int K is a finite subg of H and of K.  [Any subset of a group that is also a group is a subgroup.]

gcd( order(H), order(K)) = 1

order(L) divides order(H)
order(L) divides order(K)

order(L) divides gcd(order(H),order(K)) = 1, therefore
order(L) = 1.