Identifying Two Groups Along a Common Group
Suppose two groups contain copies of the same group \(C\). To combine them while identifying those copies, we start with the free product, which keeps the two groups' internal multiplication rules but imposes no relations between them. We then add exactly the relations that equate the two images of each element of \(C\). This construction is the free product with amalgamation.
The maps from \(C\) need not be inclusions. In applications, including those arising from intersections of topological spaces, they are often homomorphisms induced by inclusions and may fail to be injective. The quotient construction handles both injective and noninjective maps.
A subgroup is normal if it is unchanged by conjugation by every element of the group. The subgroup \(N\) exists: it is the intersection of all normal subgroups containing the displayed elements. This intersection is itself normal and contains those elements. The definition identifies the two images of \(C\), but does not add relations unless they follow from these identifications and the multiplication already present in \(A\) and \(B\).
The Free Product and Its Universal Property
For clarity, the free product \(A*B\) can be specified by generators and relations. Use a symbol for each element of \(A\) and a separate symbol for each element of \(B\), with the two identity symbols identified with the identity. Impose the relations that encode multiplication within \(A\) and within \(B\), and impose no cross-relations. Thus, for example, the symbols for \(a,a'\in A\) multiply as the symbol for \(aa'\), and the corresponding rule holds for elements of \(B\).
Proof. Send the generator corresponding to \(a\in A\) to \(f(a)\), and the generator corresponding to \(b\in B\) to \(g(b)\). The relations for multiplication in \(A\) are respected because \(f\) is a homomorphism: \(f(a)f(a')=f(aa')\). The relations for multiplication in \(B\) are respected because \(g(b)g(b')=g(bb')\). Thus the assignments define a homomorphism from the group given by these generators and relations, namely \(A*B\), into \(H\). It has the required restrictions. Any homomorphism with these restrictions must send every generator to the specified value, and the generators determine the whole group, so it is unique. \(\square\)
The same principle gives the defining property of the amalgamated product: maps out of it are precisely pairs of maps out of \(A\) and \(B\) that agree on the images of \(C\).
Proof. First suppose such an \(h\) exists. For each \(c\in C\), $$ f(i(c))=h(\alpha(i(c)))=h(\beta(j(c)))=g(j(c)), $$ so \(f\circ i=g\circ j\).
Conversely, suppose \(f\circ i=g\circ j\). By the free product property, \(f\) and \(g\) determine a homomorphism \(\Phi:A*B\to H\). For every \(c\in C\), the defining relation maps to the identity: $$ \Phi\!\left(\iota_A(i(c))\iota_B(j(c))^{-1}\right) =f(i(c))\,g(j(c))^{-1}=1_H. $$ The kernel of \(\Phi\) is a normal subgroup containing every defining relation, so it contains \(N\). Therefore the rule \(h(wN)=\Phi(w)\) is well-defined: if \(wN=w'N\), then \(w'^{-1}w\in N\subseteq\ker\Phi\), which gives \(\Phi(w)=\Phi(w')\). This rule defines a homomorphism with the required restrictions. Finally, any such homomorphism must agree with \(\Phi\) on the images of \(A\) and \(B\), which generate the quotient, so it is unique. \(\square\)
This property explains the role of the relations: they are exactly what makes the two maps out of the factors agree on \(C\). For a topological application, the groups associated to two overlapping pieces play the roles of \(A\) and \(B\), while homomorphisms induced from the intersection play the roles of \(i\) and \(j\). The Seifert–van Kampen theorem gives conditions under which this algebraic construction computes the fundamental group of the union.
Worked Constructions
Worked Example: A Trivial Common Group
Let \(A=\mathbb{Z}/2\mathbb{Z}=\langle a\mid a^2=1\rangle\) and \(B=\mathbb{Z}/3\mathbb{Z}=\langle b\mid b^3=1\rangle\). Take \(C\) to be the trivial group, whose only element maps to the identity in each factor. The only amalgamation relation is \(1=1\), so it adds no new relation. Therefore $$ A *_C B = \langle a,b\mid a^2=1,\ b^3=1\rangle. $$
To check the construction by its universal property, choose any group \(H\). A homomorphism from \(A\) to \(H\) is determined by an element \(x\in H\) with \(x^2=1_H\); a homomorphism from \(B\) is determined by an element \(y\in H\) with \(y^3=1_H\). The maps from the trivial group agree automatically. Hence the maps from the factors extend uniquely to the displayed group, with no condition relating \(x\) and \(y\). This confirms that the trivial common group imposes no cross-relation.
Check: taking \(H=A *_C B\), \(x=a\), and \(y=b\) satisfies \(x^2=1\) and \(y^3=1\), exactly the two relations in the presentation.
Worked Example: Identifying Subgroups of the Integers
Let \(A=\langle a\rangle\cong\mathbb{Z}\) and \(B=\langle b\rangle\cong\mathbb{Z}\). Let \(C=\langle c\rangle\cong\mathbb{Z}\), and define \(i(c^n)=a^{2n}\) and \(j(c^n)=b^{3n}\) for every integer \(n\). These are homomorphisms; both are injective, since \(2n=0\) or \(3n=0\) in \(\mathbb{Z}\) implies \(n=0\).
The groups \(A\) and \(B\) have no internal relations other than the group laws for infinite cyclic groups. The amalgamation relation for \(c\) is \(a^2=b^3\). The relations for \(c^n\) follow by taking powers of this equality, so the presentation is $$ A *_C B=\langle a,b\mid a^2=b^3\rangle. $$
To verify the relation direction and its full force, let \(f:A\to H\) and \(g:B\to H\) be homomorphisms, and write \(x=f(a)\), \(y=g(b)\). Agreement on \(C\) means that for every integer \(n\), \(x^{2n}=y^{3n}\). This condition follows from \(x^2=y^3\) by taking integer powers. Conversely, agreement for every \(n\) includes \(n=1\), so it implies \(x^2=y^3\). Thus maps from this presented group into \(H\) are exactly the pairs satisfying the required agreement.
Check: if \(x^2=y^3\), then \(x^{2n}=(x^2)^n=(y^3)^n=y^{3n}\) for each integer \(n\), including negative \(n\), since group elements have inverses.
Worked Example: Noninjective Maps Can Force Extra Collapse
Let \(A=\langle a\mid a^4=1\rangle\cong\mathbb{Z}/4\mathbb{Z}\), \(B=\langle b\mid b^6=1\rangle\cong\mathbb{Z}/6\mathbb{Z}\), and \(C=\langle c\rangle\cong\mathbb{Z}\). Define \(i(c^n)=a^n\) and \(j(c^n)=b^n\) for every integer \(n\). These are homomorphisms, but neither is injective: \(i(c^n)=1\) exactly when \(4\) divides \(n\), and \(j(c^n)=1\) exactly when \(6\) divides \(n\).
The relation for \(c\) identifies \(a\) with \(b\). Replacing both by a single generator \(x\) gives $$ A *_C B=\langle x\mid x^4=1,\ x^6=1\rangle. $$ The two power relations imply \(x^2=1\), because \(x^6(x^4)^{-1}=x^2\). Conversely, \(x^2=1\) implies \(x^4=(x^2)^2=1\) and \(x^6=(x^2)^3=1\). Hence this presentation is equivalent to \(\langle x\mid x^2=1\rangle\).
This last group has exactly two elements: every word reduces to \(1\) or \(x\), and the homomorphism sending \(x\) to the nonidentity element of \(\mathbb{Z}/2\mathbb{Z}\) shows that \(x\ne1\). Thus the amalgamated product is \(\mathbb{Z}/2\mathbb{Z}\). The noninjective maps have forced the two factors' generators to coincide, and their original order relations then reduce the common generator's order.
Check: in \(\mathbb{Z}/2\mathbb{Z}\), the image of \(x\) has fourth and sixth powers equal to the identity, and the images of \(a\) and \(b\) agree. Thus the defining relations hold in the computed group.
A Common Relation Error
Common pitfall. The identification relation is \(\iota_A(i(c))\iota_B(j(c))^{-1}=1\), not \(\iota_A(i(c))\iota_B(j(c))=1\). The inverse matters: equating two elements means multiplying the first by the inverse of the second.
For the integer example, the correct relation is \(a^2=b^3\). An incorrect calculation might instead impose \(a^2b^3=1\). To see that these conditions differ, take the target group \((\mathbb{Z},+)\), and define homomorphisms by \(a\mapsto 3\) and \(b\mapsto -2\). They satisfy the incorrect relation because \(2(3)+3(-2)=6-6=0\). But they do not agree on \(C\): the images of \(c\) are \(2(3)=6\) and \(3(-2)=-6\), which are unequal. So these maps cannot factor through the required amalgamated product. The corrected relation \(a^2=b^3\) tests precisely the agreement condition.
The construction is useful because it separates two tasks. First, the free product retains the multiplication rules internal to each group. Then the quotient imposes the identifications specified by the maps from \(C\). In a van Kampen computation, getting those maps and their directions right is what determines which relations the intersection contributes.
Check Your Understanding
For each question, identify the maps from the common group and write the resulting relations before simplifying.
- If \(C\) is trivial, what relations does the amalgamation add to \(A*B\)?
- Suppose \(C=\mathbb{Z}\) maps to \(A=\langle a\rangle\cong\mathbb{Z}\) by \(1\mapsto a^4\) and to \(B=\langle b\rangle\cong\mathbb{Z}\) by \(1\mapsto b^5\). What is the defining identification relation?
- State the condition on homomorphisms \(f:A\to H\) and \(g:B\to H\) that allows them to define a homomorphism from \(A *_C B\) to \(H\).
- Let \(A=\langle a\mid a^3=1\rangle\), \(B=\langle b\mid b^5=1\rangle\), and let \(C=\mathbb{Z}\) map its generator to \(a\) and \(b\). Simplify the resulting presentation.
- Explain why the relation equating two images is written as a product with an inverse, rather than as their product equal to the identity.