# Amalgam of groups

A family of groups $ G _ \alpha , \alpha \in I $,
that satisfies the condition that the intersection $ G _ \alpha \cap G _ \beta $
is a subgroup in $ G _ \alpha $
and $ G _ \beta $
for any $ \alpha , \beta $
from $ I $.
An example of an amalgam of groups is an arbitrary family of subgroups of a given group. An imbedding of an amalgam of groups $ A = \{ {G _ \alpha } : {\alpha \in I } \} $
into a group $ G $
is a one-to-one mapping of the union $ \cup _ {\alpha \in I } G _ \alpha $
into $ G $
whose restriction to each $ G _ \alpha $
is an isomorphism. An amalgam of groups in which all intersections $ G _ \alpha \cap G _ \beta $
are identical (and equal to, say, a subgroup $ H $)
is imbeddable in the group that is the free product of the groups $ G _ \alpha $
with the amalgamated subgroup $ H $.
On the other hand, there exists an amalgam of four Abelian groups that is not imbeddable in a group. The principal problem concerning amalgams of groups is, generally speaking, as follows. Let $ \sigma , \tau $
be possible properties of groups. The question to be answered is the nature of the conditions under which an amalgam of groups with the property $ \sigma $
is imbeddable in a group with the property $ \tau $.
It was found that all amalgams of two finite groups are imbeddable in a finite group. An amalgam of three Abelian groups is imbeddable in an Abelian group. An amalgam of four Abelian groups imbedded in a group is contained in an Abelian group. There exists an amalgam of five Abelian groups which is imbeddable in a group, but not in an Abelian group. Another problem that has been studied is the imbeddability of an amalgam of groups if $ \sigma , \tau $
denote solvability, nilpotency, periodicity, local finiteness, etc. (in different combinations).

#### Comments

In the definition of amalgam above, think of the $ G _ \alpha $ as all being subsets of some large set $ X $. The amalgamated product of groups $ G _ {i} $" over a common subgroup U" is constructed as follows. Let $ G _ {i} $ be a set of groups indexed by the set $ I $. For each $ i $ let $ U _ {i} $ be a subgroup of $ G _ {i} $ and for each $ i $ let there be an isomorphism $ \phi _ {i} : U _ {i} \rightarrow U $ identifying $ U _ {i} $ and $ U $. Consider the set $ \widetilde{G} $ of all words

$$ a _ {1} \dots a _ {t} $$

with each $ a _ {i} $ from some $ G _ {j} $, and consider the following elementary equivalences

1) if $ a _ {i} = 1 $ then $ a _ {1} \dots a _ {i-1} a _ {i} a _ {i+1} {} \dots a _ {t} $ is equivalent to $ a _ {1} \dots a _ {i-1} a _ {i+1} \dots a _ {t} $;

2) if $ a _ {i} $ and $ a _ {i+1} $ belong to the same group $ G _ {j} $ and $ a _ {i} a _ {i+1} = a _ {i} ^ \prime $ in $ G _ {j} $ then $ a _ {1} \dots a _ {i-1} a _ {i} a _ {i+1} \dots a _ {t} $ is equivalent to $ a _ {1} \dots a _ {i-1} a _ {i} ^ \prime a _ {i+2} \dots a _ {t} $;

3) if $ a _ {i} = u _ {i} \in U _ {j} \subseteq G _ {j} $ and $ b _ {i} = u _ {k} \in U _ {k} \subseteq G _ {k} $ and $ \phi _ {k} ( u _ {k} ) = u = \phi _ {i} ( u _ {i} ) \in U $, then $ a _ {1} \dots a _ {i-1} a _ {i} a _ {i+1} \dots a _ {t} $ is equivalent to $ a _ {1} \dots a _ {i-1} b _ {i} a _ {i+1} \dots a _ {t} $.

Let $ \sim $ be the equivalence relation generated by these elementary equivalences, then $ \widetilde{G} / \sim $ is the amalgamated product of the $ G _ {i} $, more precisely of the $ ( G _ {i} , U _ {i} ) $( i.e. the free product with amalgamated subgroup of the $ G _ {i} $ $ U $); the group law is induced by concatenation.

Amalgamated products are non-trivial. This follows from the following canonical form theorem. For each $ i $ select a set $ R _ {i} $ of left coset representatives of $ U $ in $ G _ {i} $. Then each word is equivalent to precisely one of the form $ uz _ {1} \dots z _ {t} $ with each $ z _ {i} $ in some $ R _ {j} $, $ u \in U $ and $ z _ {1} $ and $ z _ {i+1} $ belonging to different $ G _ {j} $' s for $ i = 1 \dots t-1 $. If $ U = \{ e \} $ one obtains of course the free product of the $ G _ {i} $. A subgroup of a free product is itself a free product (Kurosh' theorem). Subgroups of a product with an amalgamated subgroup need not be themselves of this type. The reason is that if $ U $ is the amalgamated subgroup, then one can take subgroups $ H _ {i} $ of the $ G _ {i} $ with different intersections with $ U $ so that the $ H _ {i} $ will amalgamate in various different ways. This leads to generalized amalgamated products and the notion of amalgam as defined above. The theory of these is still incomplete.

#### References

[a1] | M. Hall jr., "The theory of groups" , Macmillan (1959) |

[a2] | H. Neumann, "Generalized free products with amalgamated subgroups I" Amer. J. Math. , 70 (1948) pp. 590–625 |

[a3] | H. Neumann, "Generalized free products with amalgamated subgroups II" Amer. J. Math. , 71 (1949) pp. 491–540 |

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Amalgam of groups.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Amalgam_of_groups&oldid=45095