3.10 Products and homomorphisms
Suppose \(M\) and \(N\) are magmas, and let \(M\times N\) denote the Cartesian product of \(M\) and \(N\). Define a binary operation on \(M\times N\) by setting \((m_1,n_1)(m_2,n_2) = (m_1m_2, n_1n_2)\). We call the Cartesian product \(M\times N\) with this binary operation the product magma.
Suppose \(M\) and \(N\) are magmas.
If \(M\) and \(N\) are monoids with identity element \(1\), then \(M\times N\) (with the above binary from Definition 3.52) is a monoid with identity element \((1,1)\).
Suppose \(M\) and \(N\) are groups. Then \(M\times N\) is a group, and, if \((m,n)\in M\times N\), \((m,n)^{-1} = (m^{-1}, n^{-1})\).
Exercise.
Suppose \(H\) and \(K\) are groups and \((h,k)\in H\times K\).
If either \(|h| =\infty \) or \(|k| = \infty \), then \(|(h,k)| = \infty \).
If \(|h|, |k| {\lt}\infty \), then \(|(h,k)| = \operatorname{\mathrm{lcm}}(|h|, |k|)\).
We have \((h,k)^n = 1 \Leftrightarrow h^n = k^n = 1\). In the case considered by 1 this happens only for \(n = 0\). So \(|(h,k)| =\infty \). On the other hand, in the case considered by 2, \(h^n = k^n = 1\) iff \(|h| | n\) and \(|k| | n\). And this happens if and only if \(\operatorname{\mathrm{lcm}}(|h|, |k|)| n\). So the order of \((h,k)\) is \(\operatorname{\mathrm{lcm}}(|h|, |k|)\).
As Thereom 3.54 demonstrates, products of groups are easy to think about in terms of the groups themselves. So we want to be able to identify products. One way of doing this is to try to identify homomorphisms into and out of products.
Suppose \(M\) and \(N\) are magmas. Define maps \(\pi _1:M\times N\to M\) and \(\pi _2:M\times N\to N\) by setting \(\pi _1(m,n) = m\) and \(\pi _2(m,n)= n\). Then \(\pi _1\) and \(\pi _2\) are surjective magma homomorphisms.
Basically obvious.
In the situation of Proposition 3.55, the homomorphisms \(\pi _1\) and \(\pi _2\) are called the projections on the first and second factor respectively.
Suppose \(G\), \(H\) and \(K\) are groups. Define a map
by setting \(\Pi (f) = (\pi _1\circ f, \pi _2\circ f)\). Then \(\Pi \) is a bijection.
In particular, for any pair of homomorphism \(f_1:G\to H\) and \(f_2:G\to K\) there is a unique homomorphism \(f:G\to H\times K\) such that \(f_i = \pi _i\circ f\) (\(i = 1,2\)). In fact, \(f:G\to H\times K\) is given by \(f(g) = (f_1(g), f_2(g))\).
Suppose \((f_1,f_2)\in \operatorname{\mathrm{Hom}}(G,H)\times \operatorname{\mathrm{Hom}}(G,K)\). Define \(f:G\to H\times K\) by setting \(f(g) = (f_1(g), f_2(g))\). Then it is easy to check that \(f:G\times H\times K\) is a homomorphism. (Exercise!) This shows that \(\Pi \) is onto.
On the other hand, any map \(f:G\to H\times K\) is given by \(f(g) = (f_1(g),f_2(g))\) where \(f_1:G\to H\) and \(f_2:G\to K\). So \(f\) is determined by the maps \(f_1\) and \(f_2\). This shows that \(\Pi \) is one-one.
The analogues of Theorem 3.57 are true for magmas and monoids as well.
Theorem 3.57 answer the question about how to get homomorphisms into products. But we’d like to also understand homomorphisms out of products.
Suppose \(H\) and \(K\) are groups with idenity elements written as \(1\). Define maps \(i_1:H\to H\times K\) and \(i_2:H\to H\times K\) by setting \(i_1(h) = (h,1)\) and \(i_2(k) = (1,k)\). Then \(i_1\) and \(i_2\) are both injective group homomorphisms.
Basically obvious. And it works for monoids too. But note that, for general magmas, there’s no identity element. So there’s no analogous statement.
The homomorphisms \(i_1\) and \(i_2\) in Proposition 3.59 are called the inclusion of the first and second factor respectively.
Suppose \(f:G\to H\) is an injective group homomorphism. Then the map \(h:G\to f(G)\) given by \(g\mapsto f(g)\) is an isomorphism of groups.
The map \(h\) is a group homomorphism, which is one-one and onto. So it is an isomorphism by definition.
The maps \(i_1:H\to H\times K\) and \(i_2:K\to H\times K\) induces isomorphisms from \(H\) and \(K\) to the subgroups \(H\times \{ 1\} \leq H\times K\) and \(\{ 1\} \times K\leq H\times K\) respectively.
Suppose \(G\), \(H\) and \(K\) are groups. Define a map
by \(f\mapsto (f\circ i_1, f\circ i_2)\). Then \(R\) is one-one, and the image of \(R\) is the set of pairs \((f_1,f_2)\in \operatorname{\mathrm{Hom}}(H,G)\times \operatorname{\mathrm{Hom}}(K, G)\) such that, for all \((h,k)\in H\times K\),
In particular, if \(f_1:H\to G\) and \(f_2:K\to G\) are two group homomorphisms satisfying 3.65 for every \((h,k)\in H\times K\), then there is a unique group homomorphism \(f:H\times K\to G\) such that, for all \((h,k)\in H\times K\), \(f(h,1) = f_1(h)\) and \(f(1,k) = f_2(k)\). Moreover, we have \(f(h,k) = f_1(h)f_2(k)\).
First let’s show that \(R\) is one-one. Suppose \(f:H\times K\to G\) is a group homomorphism. Then
So \(f\) is determined by the group homomorphisms \(f\circ i_1: H\to G\) and \(f\circ i_2:K\to G\). Therefore, \(R\) is one-one.
Now, suppose \(f:H\times K\to G\) is a group homomorhism and set \(f_k = f\circ i_k\) for \(k=1,2\). Suppose \((h,k)\in H\times K\). Then, we have
So 3.65 is satisfied.
On the other hand, suppose \(f_1:H\to G\) and \(f_2:K\times G\) are two group homomorhisms satisfying 3.65 for every pair \((h,k)\in H\times K\). Define \(f:H\times K\to G\) by setting \(f(h,k) = f_1(h)f_2(k)\). Then, if \((h_1,k_1), (h_2,k_2)\in H\times K\), we have
which shows that \(f:H\times K\to G\) is a group homomorphism.