3.1 Homomorphisms
We start with the official definitions.
Suppose \(M\) and \(N\) are magmas. A magma homomorphism from \(M\) to \(N\) is a map \(f:M\to N\) such that, for all \(x,y\in M\), \(f(xy) = f(x)f(y)\).
Suppose \(M\) and \(N\) are monoids with identity elements \(e_M\) and \(e_N\). A monoid homomorphism from \(M\) to \(N\) is a monoid homomorphism \(f:M\to N\) with the additional property that \(f(e_M) = e_N\).
Suppose \(M\) and \(N\) are groups. A group homomorphism from \(M\) to \(N\) is a monoid homomorphism with the additional property that, for all \(x\in M\), \(f(x^{-1}) = f(x)^{-1}\).
A magma (resp. monoid, resp. group) homomorphism \(f\) is said to be a magma (resp. monoid, resp. group) isomorphism if it is one-one and onto.
A magma (resp. monoid, resp. group) homomorphism \(f:M \to M\) is called a magma (resp. monoid, resp. group) endomorphism.
A magma (resp. monoid, resp. group) isomorphism \(f:M \to M\) is called a magma (resp. monoid, resp. group) automorphism.
We are mostly going to be interested in group homomorphisms, isomorphisms and (sometimes) automorphisms. The following Proposition is helpful in this regard.
Suppose \(G\) and \(H\) are groups and \(f:G\to H\) is a magma homomorphism. Then \(f\) is a group homomorphism as well. In other words, a map \(f:G\to H\) is a group homomorphism if and only if for all \(x,y\in G\), \(f(xy) = f(x) f(y)\).
Suppose \(G\) and \(H\) are groups and \(f:G\to H\) is a magma homomorphism. Write \(e_G\) (resp. \(e_H\)) for the identity element of \(G\) (resp. \(H\)). Then \(f(e_G) = f(e_G) e_H = f(e_G) (f(e_G)f(e_G)^{-1}) = (f(e_G) f(e_g)) f(e_G)^{-1} = f(e_G) f(e_G)^{-1} = e_H\). So \(f:G\to H\) is a monoid homomorphism.
On the other hand, suppose \(g\in G\). Then
So \(f\) is a group homomorphism.
The analogue of Proposition 3.2 is not true for monoids. In other words, there are examples of monoids \(M\) and \(N\) with identity elements \(e_M\) and \(e_N\) respectively and magma homomorhisms \(f:M\to N\) such that \(f(e_M) \neq e_N\).
In fact, here’s an easy example. Let \(N = \{ 0,1\} \) with the binary operation \(*\) given by setting \(0*0 = 0*1 = 1*0 = 0\) and \(1*1 = 1\). It’s not hard to check directly that \(N\) is associative. So it is a monoid with identity element \(1\). The subset \(M =\{ 0\} \) is a submagma. In other words, it is closed under the binary operation. But, according to Definition 2.29, it is not a submonoid because it does not contain the identity element \(1\) of \(M\).
Write \(f:N\to M\) for the inclusion map given by \(f(0) = 0\). Then \(f\) is a magma homomorphism, but it is not a monoid homomorphism.
Suppose \(N\) is a submagma of a magma \(M\). Write \(i_N:N\to M\) for the map given by \(n\mapsto n\).
The map \(i_N\) is a magma homomorphism.
If \(M\) is a monoid and \(N\) is a submonoid, then \(i_N\) is a monoid homomorphism.
If \(M\) is a group and \(N\) is a subgroup, then \(i_N\) is a group homomorphism.
The only part of this that isn’t obvious is 2. For this, the important point is that, by the definition of a submonoid in Definition 2.29, \(N\) contains the idenity element \(e_M\) of \(M\). And \(e_M\) is then the identity element of \(N\). So the conditions for a monoid homomorphism in Definition 3.1 are satisfied.
We call \(i_N:N\to M\) the inclusion homomorphism.
If \(S\) is any set, we write \(\operatorname{\mathrm{id}}_S\) for the map \(\operatorname{\mathrm{id}}_S: S\to S\) given by \(\operatorname{\mathrm{id}}_S(s) = s\). This maps is called the identity map from \(S\) to itself. Recall that a map of sets \(f:S\to T\) is a bijection if and only if there exists an inverse function \(g:T\to S\) such that \(f\circ g = \operatorname{\mathrm{id}}_T\) and \(g\circ f = \operatorname{\mathrm{id}}_S\). The inverse function \(g\) is then unique and is usually written as \(f^{-1}:T\to S\). It follows directly that, in this case, \(f^{-1}:T\to S\) is also a bijection.
The following assertions hold.
If \(M\) is magma (resp. monoid, resp. group) then \(\operatorname{\mathrm{id}}_M:M\to M\) is a magma (resp. monoid, resp. group) endomorphism.
If \(M\) and \(N\) are magmas and \(f:M\to N\) is a magma isomorphism, then the inverse function \(f^{-1}:N\to M\) is also a magma isomorphism.
If \(M\) and \(N\) are monoids and \(f:M\to N\) is a monoid isomorphism, then the inverse function \(f^{-1}:N\to M\) is also a monoid isomorphism.
If \(M\) and \(N\) are groups and \(f:M\to N\) is a group isomorphism, then the inverse function \(f^{-1}:N\to M\) is also a group isomorphism.
1 This follows from Proposition 3.4.
2 Suppose \(f:M\to N\) is a bijection of magmas, and \(n_1, n_2\in N\). Set \(m_i := f^{-1}(n_i)\) for \(i=1,2\) so that \(f(m_i) = n_i\).
So \(f^{-1}:N\to M\) is a homomorphism of magmas. As it is a bijetion, it is also an isomorphism of magmas.
3 Suppose \(f:M\to N\) is a bijection of monoids with \(e_M\) (resp. \(e_N\)) the identity element of \(M\) (resp. \(N\)). Then \(f^{-1}:N\to M\) is a magma isomorphism by 2. And \(f(e_M) = e_N\), which implies that \(f^{-1}(e_N) = e_M\). So \(f^{-}:N\to M\) is also a monoid homomorphism. As it is a bijection, it is also a monoid isomorphism.
The composition \(S\circ T:G\to K\) of two group homorphisms \(T:G\to H\) and \(S:H\to K\) is a group homomorphism. (And similarly for magma and monoid homomorhisms).
Let’s prove it for group homomorphisms. So suppose \(T:G\to H\) and \(S:H\to K\) are group homomorphisms and \(x,y\in G\). Then \((S\circ T)(xy) = S(T(xy)) = S(T(x)T(y)) = S(T(x))S(T(y)) = (S\circ T)(x) (S\circ T)(y)\). So, by Proposition 3.2, we’re done.
The same proof (of course) works for magma homomorphisms, and the proof for monoid homomorphisms is easy (and left to the reader).
The composition \(S\circ T:G\to K\) of two group isomorphisms \(T:G\to H\) and \(S:H\to K\) is a group isomorphism. (And similarly for magma and monoid homomorhisms).
Follows directly from Proposition 3.7 and the fact that the composition of bijetions is a bijection.
Suppose \(G\) and \(H\) are groups. We say that \(G\) and \(H\) are isomorphic and write \(G\cong H\) if there exists an isomorphism \(f:G\to H\).
Group isomorhism is an equivalence relation on the class of groups. In other words, if \(G_1\), \(G_2\) and \(G_3\) are groups, then:
\(G_1\cong G_1\).
\(G_1\cong G_2 \Rightarrow G_2\cong G_1\).
If \(G_1\cong G_2\) and \(G_2\cong G_3\), then \(G_1\cong G_3\).