> > Subgroups of Z x Z
ballade1
not of the form nZ x mZ ('diagonal' subgroups spring to mind, namely those generated by elements of the form
(a, a)).
I read somewhere that *every* subgroup is generated
by at most 2 elements.
Why is this true?
Also, what name do we give to finitely generated groups all of whose subgroups are finitely generated?
Thanks in advance,
Arturo Magidin
> Clearly, there are subgroups of Z x Z (Z the integers)
> not of the form nZ x mZ ('diagonal' subgroups spring to mind, namely those generated by elements of the form
> (a, a)).
>
> I read somewhere that *every* subgroup is generated
> by at most 2 elements.
>
> Why is this true?
Because it is a free module over a PID of rank 2.
Basically: let H be a subgroup of Z x Z. If every element of H has
first coordinate equal to 0, then H is contained in {0} x Z, hence is
cyclic and generated by at most 1 element. Otherwise, let h1=(a,b) be
an element of H whose first coordinate is positive and as small as
possible; if there is more than one, pick the one with smallest |b|;
if there are two of those, pick the one with b>0.
Note that if x is in H, and x = (r,s), then a|r. This follows because
if you write r = ka+t, with 0<=t <a, then taking x - k(h1) we get an
element of H with first entry equal to t, and this implies that we
must have t=0.
If <h1> = H, then H is cyclic and you are done. Otherwise, there
exists an element in H, not in <h1>. By subtracting multiples of h1,
we obtain elements of the form (0,r) in H but not in <h1>. Let h2 be
the element of the form (0,r) not in H with smallest positive value of
r.
I claim that H = <h1, h2>. For let x = (u,v) in H. Then u = ka for
some integer k; thus, u-k*h1 = (0,v-kb). Write v-kb = zr + w, with 0<=
w < r. Then u - k*h1 - z*h2 = (0,w); This is in H, and the minimality
of r means that w=0. Thus, y = k*h1 + z*h2, and so H = <h1, h2>, as
claimed.
> Also, what name do we give to finitely generated groups all of whose subgroups are finitely generated?
"Finitely generated groups all of whose subgroups are finitely
generated".
--
Arturo Magidin
Arturo Magidin
Typo:
> Basically: let H be a subgroup of Z x Z. If every element of H has
> first coordinate equal to 0, then H is contained in {0} x Z, hence is
> cyclic and generated by at most 1 element. Otherwise, let h1=(a,b) be
> an element of H whose first coordinate is positive and as small as
> possible; if there is more than one, pick the one with smallest |b|;
> if there are two of those, pick the one with b>0.
>
> Note that if x is in H, and x = (r,s), then a|r. This follows because
> if you write r = ka+t, with 0<=t <a, then taking x - k(h1) we get an
> element of H with first entry equal to t, and this implies that we
> must have t=0.
>
> If <h1> = H, then H is cyclic and you are done. Otherwise, there
> exists an element in H, not in <h1>. By subtracting multiples of h1,
> we obtain elements of the form (0,r) in H but not in <h1>. Let h2 be
> the element of the form (0,r) not in H with smallest positive value of
> r.
That should be "in H, not in <h1>, with smallest ositive value of r".
--
Arturo Magidin
ballade1
u -> x
Then u - k*h1 - z*h2 = (0,w);
u -> x
This is in H,
> and the minimality> of r means that w=0. Thus, y = k*h1 + z*h2, and so H
> = <h1, h2>, as
> claimed.
Thanks!!
Mariano Suárez-Alvarez
> Clearly, there are subgroups of Z x Z (Z the integers)
> not of the form nZ x mZ ('diagonal' subgroups spring to mind, namely those generated by elements of the form
> (a, a)).
>
> I read somewhere that *every* subgroup is generated
> by at most 2 elements.
>
> Why is this true?
Since Z is noetherian and Z x Z is finitely generated,
every subgroup of Z x Z is itself finitely generated.
Let G be a subgroup of Z x Z. Since G has no torsion
and is finitely generated, it is free, so G is isomorphic
to a group of the form Z x ... x Z for some number of factors.
So we have an injective morphism of groups
Z x ... x Z --> Z x Z
This is impossible unless the domain has at most two factors,
and in that case G is generated by at most two elements.
-- m
ballade1
This seems intuitively clear; but, just to drive home the point: why can't there be an injective homomorphism from 3 factors, say?
W. Dale Hall
Tensor with Q. Look at the rank.
Dale
ballade1
I think this follows from:
If F is a free abelian group of finite rank n, and H is a nontrivial subgroup of F, then H is free abelian
with rank at most n.
Arturo Magidin
Well, yes; but in fact that's precisely what you are trying to show
when you say that there can be no injection from, say, 3 factors into
2. So invoking this is a circular argument.
Tensoring to Q and looking at the dimension as vector spaces, as Dale
Hall suggests, does work, however.
--
Arturo Magidin