Re: Adjoining square roots to a group

In article <1155131630.394812.145420@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Grothendieck-Hirzebruch <Grothendieck-Hirzebruch@xxxxxx> wrote:
If g is of finite order n, then consider the amalgamated free product
of G with a cyclic group C of order 2n, generated by x, amalgamated
over the subgroups <g> = <x^2>. If g is of infinite order, then take
the amalgamated free product of G with an infinite cyclic group
generated by x, and again amalgamate over the subgroups <g>=<x^2>.

This produces an infinite group in which g has a square root; similar
arguments show you can adjoing an n-th roots to any (finite) family of
elements of G, by doing this sequentially. If G is finite, then one
can also produce a ->finite<- group with the property, though the
result is a bit harder.

Thanks! It's not immediate to me, though, why G injects in this
amalgam.

Which one? I assume you mean the free product with amalgamation.

It is not trivial, but it is not hard. Probably the easiest way to do
it is simply to construct the amalgam (maybe using van der Waerden's
trick, or explicitly). I will give you the more general construction
below, though you can simplify the choices for C in this case.

Let {H_i} be the (right) cosets of <g> in G, and let {h_i} be a family
of coset representatives, picking the identity to represent the coset
of {g}. Likewise, let {D_i} be the (right) cosets of
<x^2> in C, and let {d_i} be a family of coset representatives picking
the identity to represent the coset of x^2. (In
this case, we have that the cosets are D_0 = {x^{2k}:k in Z} and
D_1 = {x^{2k+1} : k in Z}, so we could take d_0=1 and d_1=x).

Consider the set of all words (finite length strings) in elements of
{h_i: h_i<>1} \/ {d_i: d_i<>i} \/ <g>, which alternate nontrivial
elements of {h_i} with nontrivial elements of {d_i}, and then have an
element of <g> (possibly the identity) at the end.

Now let me show you how to multiply two of these strings. Say you have

h_{i_1}*d_{j_1}*h_{i_2}g^{c_1} and h_{i_3}*d_{j_2}*g^{c_2}

and you want to multiply them together. First, concatenate them:

h_{i_1}*d_{j_1}*h_{i_2}*g^{c_1}*h_{i_3}*d_{j_2}*g^{c_2}

Next we do a rewriting process. First, g^c_1*h_{i_3} is an element of
G, hence it can be written uniquely as h_{k_1}*g^{e_1} for some coset
representative h_{k_1} and some integer e_1. Replace then
g^{c_1}h_{i_3} with h_{k_1}*g^{e_1} to get

h_{i_1}*d_{j_1}*h_{i_2}*h_{k_1}*g^{e_1}*d_{j_2}*g^{c_2}

Now, replace g^{e_1} with x^{2e_1} (since we want g = x^2
anyway). Then x^{2e_1}*d_{j_2} is an element of C, hence can be
written uniquely as d_{m_2}x^{2e_2} for some coset representative
d_{m_2} and some integer e_2. Replacing
g^{e_1}*d_{j_2}=x^{2e_1}d_{j_2} with d_{m_2}x^{2e_2} we have

h_{i_1}*d_{j_1}*h_{i_2}*h_{k_1}*d_{m_2}*x^{2e_2}*g^{c_2}

Now replace x^{2e_2} with g^{e_2} and multiply by g^{c_2} to get

h_{i_1}*d_{j_1}*h_{i_2}*h_{k_1}*d_{m_2}*g^{e_2+c_2}

This is not quite in the form we want yet, since it has two h's
together. We can rewrite h_{i_2}*h_{k_1} with h_{k_2}*g^{e_3} for some
coset representative h_{k_2} and some integer e_3. We get

h_{i_1}*d_{j_1}*h_{k_2}*g^{e_3}*d_{m_2}*g^{e_2+c_2}

Next, replace g^{e_3} with x^{2e_3}, and then replace x^{2e_3}*d_{m_2}
with d_{m_3}x^{2e_4}, replace x^{2e_4} with g^{e_4}, and multiply by
g^{e_2+c_2} to get

h_{i_1}*d_{j_1}*h_{k_2}*d_{m_3}*g^{e_4+e_2+c_2}

Now, if h_{k_2} and d_{m_3} are nontrivial, you are done. If h_{k_2}
is nontrivial but d_{m_3} is trivial, drop the latter. If h_{k_2} is
trivial and d_{m_3} is nontrivial, drop h_{k_2} and then replace
d_{j_1}*d_{m_3} with d_{m_4}x^{2e_5} and do the obvious thing. Etc.

Using van der Waerden's trick and letting G*C/N (where G*C is the free
product, and N is the normal subgroup generated by gx^{-2}) act on
this set of words, it is easy to verify that the collection of words
with this product forms a group. This group contains a copy of G,
namely all words of the form h_i g^k, with k an integer. It also
contains a copy of C, namely, all words of the form d_i g^k, with k an
integer. Also, in this group, x^2=g. Thus, there is a map from G*C/N
to this group, induced by the universal properties of the free product
and the quotient. Since the map sends elements of g to the elements
h_ig^k and elements of C to the elements d_ig^k, it follows that
composing the canonical maps from G and C to G*C with the quotient
yields embeddings from G and from C into G*C/N, hence G*C/N contains
copies of G and of C.

In fact, it can be shown that the group of words described has the
corresponding universal property, and so that G*C/N is isomorphic to
the group of words under concatenation/simplification as described
above, which give you a normal form for elements of G*C/N. This is, I
believe, how Schreier established his proofs, via the normal form
described.


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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