| |
 |
|
| Science Forum Index » Mathematics Forum » groups |
|
Page 1 of 1 |
|
| Author |
Message |
| texas mathgal |
 Posted: Mon Feb 21, 2005 6:10 pm |
|
|
|
Guest
|
please help the following:
Let H be an additive group having an abelian normal subgroup B.
(a) Prove that if
alpha belta
{0} ----------> B --------> H -------->G -------> {0} is a
split extension, where v: G -->H satisfies (belta) o v = I_G, then
alpha(B) intercept v(G) = {0} and alpha(B) + v(G) = H.
(b) In this case, prove that each h belong to H has a unique expression
h = alpha(b) + v(g), where b belong to B and g belong to G.
(c) Let B and G be subgroups of H with B normal to H. Prove that if H =
B x G, then B intercept G = {0}, B +G = H, and each h belong to H has a
unique expression h = b+g, where
b belong to B and g belong to G.
Thanks |
|
|
| Back to top |
|
|
|
| Arturo Magidin |
| Posted: Mon Feb 21, 2005 6:10 pm |
|
|
|
Guest
|
In article <1109027457.774446.169440@z14g2000cwz.googlegroups.com>,
texas mathgal <greatsmileforever@yahoo.com> wrote:
[quote:ddcd54d79f]please help the following:
[/quote:ddcd54d79f]
What have you managed to prove? What are you having trouble with?
[quote:ddcd54d79f]Let H be an additive group having an abelian normal subgroup B.
[/quote:ddcd54d79f]
Does that mean that we do not assume H commutative, though we write it
additively, or that H is an abelian group (in which case, why did we
specify that B was abelian an normal?)
Probably the former...
[quote:ddcd54d79f](a) Prove that if
alpha belta
{0} ----------> B --------> H -------->G -------> {0} is a
split extension, where v: G -->H satisfies (belta) o v = I_G, then
alpha(B) intercept v(G) = {0} and alpha(B) + v(G) = H.
[/quote:ddcd54d79f]
The diagram you have is a "short exact sequence". That means that:
(i) ker(alpha) = {0};
(ii) im(alpha) = ker(beta);
(iii) im(beta) = G.
(I'm using "beta", the usual name for the second greek letter)
Pick an x element of alpha(B) intersect v(G). That means that x =
alpha(b) for some b in B, and x=v(g) for some g in G.
Since beta(alpha(b))=0, that means that beta(x)=0. Since beta(v(g))=g
for every g in G, we also know that beta(v(g))=g.
But v(g)=alpha(b). What can you conclude about x now?
To show that H = alpha(B) + v(G), the only thing you need (after
showing the two subgroups intersect trivially) is to show that every h
in H can be written as an element of alpha(B) plus an element of v(G).
Well, you know that beta(h) is in G, so v(beta(h)) is in H.
Since beta(h - v(beta(h))) = beta(h) -beta(v(beta(h)))
= beta(h) - I_G(beta(h))
= beta(h) - beta(h) = 0
that means that h - v(beta(h)) is in ker(beta). So it is in im(alpha).
Can you now write h as the sum of something in im(alpha) plus
something in v(G)?
[quote:ddcd54d79f](b) In this case, prove that each h belong to H has a unique expression
h = alpha(b) + v(g), where b belong to B and g belong to G.
[/quote:ddcd54d79f]
Assume that alpha(b)+v(g) = alpha(b') + v(g').
Then
-alpha(b') + alpha(b) = v(g) - v(g')
or
alpha(-b'+b) = v(g-g').
What did you know about the intersection of the images of beta and v?
[quote:ddcd54d79f](c) Let B and G be subgroups of H with B normal to H. Prove that if H =
B x G, then B intercept G = {0}, B +G = H, and each h belong to H has a
unique expression h = b+g, where
b belong to B and g belong to G.
[/quote:ddcd54d79f]
Surely you can do this one on your own?
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu |
|
|
| Back to top |
|
|
|
| texas mathgal |
| Posted: Fri Feb 25, 2005 3:34 am |
|
|
|
Guest
|
[quote:9688f7f781]Let H be an additive group having an abelian normal subgroup B.
(a) Prove that if
alpha delta
{0} ----------> B --------> H -------->G -------> {0} is a
split extension, where v: G -->H satisfies (belta) o v = I_G, then
alpha(B) intercept v(G) = {0} and alpha(B) + v(G) = H.
The diagram you have is a "short exact sequence". That means that:
(i) ker(alpha) = {0};
(ii) im(alpha) = ker(beta);
(iii) im(beta) = G.
(I'm using "beta", the usual name for the second greek letter)
Pick an x element of alpha(B) intersect v(G). That means that x =
alpha(b) for some b in B, and x=v(g) for some g in G.
Since beta(alpha(b))=0, that means that beta(x)=0. Since beta(v(g))=g
for every g in G, we also know that beta(v(g))=g.
But v(g)=alpha(b). What can you conclude about x now?
To show that H = alpha(B) + v(G), the only thing you need (after
showing the two subgroups intersect trivially) is to show that every
h
in H can be written as an element of alpha(B) plus an element of
v(G).
Well, you know that beta(h) is in G, so v(beta(h)) is in H.
Since beta(h - v(beta(h))) = beta(h) -beta(v(beta(h)))
= beta(h) - I_G(beta(h))
= beta(h) - beta(h) = 0
that means that h - v(beta(h)) is in ker(beta). So it is in
im(alpha).
Can you now write h as the sum of something in im(alpha) plus
something in v(G)?
(b) In this case, prove that each h belong to H has a unique
expression
h = alpha(b) + v(g), where b belong to B and g belong to G.
Assume that alpha(b)+v(g) = alpha(b') + v(g').
Then
-alpha(b') + alpha(b) = v(g) - v(g')
[/quote:9688f7f781]
are you sure this is right? i get: -alpha(b') + alpha(b) = - v(g) +
v(g')
[quote:9688f7f781]
or
alpha(-b'+b) = v(g-g').
[/quote:9688f7f781]
show it's homo?
[quote:9688f7f781]
What did you know about the intersection of the images of beta and v?
[/quote:9688f7f781]
it is equal to {0}, so?!
[quote:9688f7f781]
(c) Let B and G be subgroups of H with B normal to H. Prove that if
H =
semi-direct of B & G, then B intercept G = {0}, B +G = H, and each h
belong to H has a unique expression h = b+g, where b belong to B and g[/quote:9688f7f781]
belong to G.
[quote:9688f7f781]
Surely you can do this one on your own?
[/quote:9688f7f781]
well i can show then B intercept G = {0} easily, but not sure how to
show B + G = H. and to prove the second part of c) i am still not
clear enough...
thanks |
|
|
| Back to top |
|
|
|
| Arturo Magidin |
| Posted: Fri Feb 25, 2005 4:46 am |
|
|
|
Guest
|
In article <1109320460.469267.285510@o13g2000cwo.googlegroups.com>,
texas mathgal <greatsmileforever@yahoo.com> wrote:
[.snip.]
[quote:ce6babdc8c](b) In this case, prove that each h belong to H has a unique
expression
h = alpha(b) + v(g), where b belong to B and g belong to G.
Assume that alpha(b)+v(g) = alpha(b') + v(g').
Then
-alpha(b') + alpha(b) = v(g) - v(g')
are you sure this is right? i get: -alpha(b') + alpha(b) = - v(g) +
v(g')
[/quote:ce6babdc8c]
Although I was not quite right, neither are you.
If your group H is not commutative, despite being written
additively, the from
alpha(b) + v(g) = alpha(b') + v(g')
you must subtract alpha(b') on the left and subtract v(g) on the
right:
-alpha(b') + alpha(b) + v(g) - v(g) = -alpha(b') + alpha(b') +v(g') - v(g)
to get
-alpha(b') + alpha(b) = v(g') - v(g).
[quote:ce6babdc8c]or
alpha(-b'+b) = v(g-g').
show it's homo?
[/quote:ce6babdc8c]
Show what? You ->already know<- that alpha and v are
homomorphisms. That was part of the given information.
[quote:ce6babdc8c]What did you know about the intersection of the images of beta and v?
it is equal to {0}, so?!
[/quote:ce6babdc8c]
Think!
alpha(-b'+b) is in the image of beta. So is v(g'-g). If the
intersection is trivial, then both alpha(-b'+b) and v(g'-g) must be
equal to 0. That means that alpha(b) = alpha(b') and v(g) = v(g').
Which in turn means that the "two" ways you had of writing the
element, namely
alpha(b) + v(g)
and
alpha(b') + v(g')
are really the same. Which means the expression is UNIQUE, which, oh
wonder of wonders, is what you were supposed to be proving.
[quote:ce6babdc8c](c) Let B and G be subgroups of H with B normal to H. Prove that if
H =
semi-direct of B & G, then B intercept G = {0}, B +G = H, and each h
belong to H has a unique expression h = b+g, where b belong to B and g
belong to G.
Surely you can do this one on your own?
well i can show then B intercept G = {0} easily, but not sure how to
show B + G = H. and to prove the second part of c) i am still not
clear enough...
[/quote:ce6babdc8c]
Perhaps you should either spend some time thinking about it, or else
tell us how far you've gotten and where you are stuck, so we don't
think (perish the thought) that you are hoping someone will do the
entirety of your homework for you.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Sun Nov 08, 2009 3:55 am
|
|