Expert: Paul Klarreich - 3/14/2007

Question
Thanks alot for your help.

i need your help in one part of thr following question:
Prove that isomorphism is an equivalence relation on any family of groups.
i only need a proof for the refexive property. Thanks alot again
-------------------------------------------
The text above is a follow-up to ...

-----Question-----
Hi,

i'm studying abstract algebra:

can you please help me with the following questions

1)prove if G and H are isomorphic groups and G is abelian, then H is abelian.

2)prove if G and H are isomorphic groups and G has a subgroup of order n, then H
has a subgroup of order n.

-----Answer-----
Questioner:   Paul
Category:  Calculus
 
Subject:  Isomorphism
Question:  Hi,

i'm studying abstract algebra:

can you please help me with the following questions

1)prove if G and H are isomorphic groups and G is abelian, then H is abelian.

2)prove if G and H are isomorphic groups and G has a subgroup of order n, then H has a subgroup of order n.

..........................................
Hi, Paul,

It has been some time, but I decided to go with this definition from the Wikipedia:
--------------- COPIED ----------------
In abstract algebra, a group isomorphism is a function between two groups that sets up a one-to-one correspondence between the elements of the groups in a way that respects the given group operations. If there exists an isomorphism between two groups, then the groups are called isomorphic. From the standpoint of group theory, isomorphic groups have the same properties and need not be distinguished.
---------------------------------------
which looks good to me.  

I use this notation to save typing:

1. I just use the word 'in' for 'is a member of'.  When you see  a in G, read 'a is a member of G'.
2. I use the prime, or apostrophe, for inverse, both in the group and for the function.  a' means the inverse of element a,  f' means the inverse function of f.  Easier than writing ^-1 all the time.

So given your two groups G and H, there exists a function f:G->H, having these properties:

A. f is a 1-1 correspondence. (So f' exists and is also a 1-1 corr and an isomorphism.)

B. If a,b in G,  then f(a)f(b) = f(ab).  [Respecting group operations, it says.]

C. If e is the identity in G, f(e) is the identity in H. [likewise]

D. If a in G, then f(a') = (f(a))' -- the images of inverses are preserved.

OK, on to your proofs:

1) prove if G and H are isomorphic groups and G is Abelian, then H is Abelian.  [Abel was a real person, so we capitalize Abelian.]

Let  p,q in H

Let  a = f'(p),  b = f'(q), and a,b in G.

pq = f(a)f(b) = f(ab) = f(ba) = f(b)f(a) = qp

H is Abelian.  

...................................
2)prove if G and H are isomorphic groups and G has a subgroup of order n, then H has a subgroup of order n.

Let Gn be the subgroup of G, and let Hn be the set of images of  elements of Gn.  That is:

Hn = { x in H such that there is  a in Gn with f(a) = x }

Clearly, since Gn has n elements, and f is 1-1, Hn has n elements.  Is Hn a group?

A. Closure:  

If  x,y in Hn, then  a = f'(x), b = f'(y) in Gn.
If a, b in Gn, then  ab in Gn.
If ab in Gn, then  f(ab) in Hn.
But f(ab) = f(a)f(b) = xy, so xy in Hn, and we have closure.

B. Identity.

e in Gn -->  f(e) in Hn.
But f(e) is the identity in H, so it is the identity in Hn, too.

C. Inverses.

If  a in Gn, then a' in Gn.  Then  f(a), f(a') in Hn.  But f(a') = (f(a))' from property D of isomorphism, so inverses are preserved.

I think that does it.


Answer
Questioner:   Paul
Category:  Calculus
Private:  No
 
Subject:  Isomorphism
Question:  Thanks alot for your help.

I need your help in one part of thr following question:
Prove that isomorphism is an equivalence relation on any family of groups.
I only need a proof for the refexive property. Thanks alot again
..........................................
Hi, Paul,

Thank you for the kind comments on the site.

About equivalence relations, as I recall, you have to prove:

If something, denoted <=>, is an equivalence relation, then for your family of groups G,

A. Reflexive:  If   g in G   g <=> g
B. Symmetric:  If  g <=> h, then  h <=> g
C. Transitive: If  g1 <=> g2 and g2 <=> g3, then  g1 <=> g3.

You have a family of groups G, such that  g, h in G means  g and h are isomorphic, meaning you can find  f:g -> h that satisfies all those properties we discussed last time.

For the reflexive property, all you have to do is prove that any group is isomprphic to itself.  Are we overlooking something?  It seems that all we have to do is define:

f: g -> g  as:

f(a) = a  for all a in g.  I.e. the identity map, maps every element onto itself.  This would have to be an isomorphism.  Shouldn't be too hard to prove.