1.3. Homomorphism and Isomorphisms.
{\color{BlueViolet}As with all mathematical objects, now that we have a well defined abstract concept (i.e., a group) we'll now be interested attempting to understand mappings between different groups. Mappings of abstract concepts simply helps mathematicians get a better sense of what they're dealing with, and most often provides new insight into understand their objects.
The most important utility of the following definition is that it not only leads one to have a better understanding of groups, but it also helps us understand when two groups are equivalent. For example, \(D_3\) and \(S_3\) equivalent, since one could view \(D_3\) as simply all the permutations of 1, 2, and 3, if we assigned these numbers to the vertices of a triangle. }
Let \((G, \cdot)\) and \((G', *)\) be groups. A homomorphism is a mapping \(\phi: G \to G'\) such that, for all \(a, b \in G\),
{\color{red}Again, here \(*\) is the group operation of \(G'\).}
Example. Consider the two groups \(GL_n(\mathbb{R})\) and \(\mathbb{R}\setminus\{0\}\). If we define \(\phi\) such that, for \(A \in GL_n(\mathbb{R})\)
then \(\phi\) defines a homomorphism.
Recall that for for any \(n \times n\) matrices \(A, B\) that \(\det(AB) = \det(A)\det(B)\). Therefore
Since \(\phi(AB) = \phi(A)\phi(B)\), we see that \(\phi\) satisfies the condition to be a homomorphism.
Let \(\phi: G \to G'\) be a homomorphism. Then all of the following hold.
- 1. If \(e_G\) is the identity of \(G\) and \(e_{G'}\) is the identity of \(G'\), then \(\phi(e_G) = e_{G'}\).
- 2. For all \(g \in G\), \(\phi(g^{-1}) = \phi(g)^{-1}\).
- 3. For \(g_1, g_2, \dots, g_n \in G\), then \(\phi(g_1 \cdot g_2 \cdot \dots \cdot g_n) = \phi(g_1)\phi(g_2)\cdots\phi(g_n)\). Consequently, if \(g = g_1 = g_2 = \cdots = g_n\), then \(\phi(g^n) = \phi(g)^{n}\).
Let \(g \in G\), and suppose \(\phi: G \to G'\) is a homomorphism.
- 1. Since \(e_G = e_G \cdot e_G\), we have that
We also know that \(\phi(e_G) \in G'\), and becuase \(G'\) is a group, there exists an inverse \(\phi(e_G)^{-1} \in G\) of \(\phi(e_G)\). Multiplying this on the left (or right) yields
as desired. * 2. Since \(gg^{-1} = e_G\), and by (1.) we know that \(\phi(e_G) = e_{G'}\). Hence
Again, \(\phi(g) \in G'\), and since \(G'\) is a group there exist an inverse \(\phi(g)^{-1} \in G\) of \(\phi(g)\). Multiplying on the left by this inverse, we get
as desired. * 3. This is just repeated application of the homomorphism property. For \(g_1, g_2, \dots g_n \in G\), \(g_1 \cdot g_2 \cdot \hspace{0.01mm} \dots \hspace{0.01mm} \cdot g_n = g_1 \cdot (g')\) where \(g' = g_2 \cdot g_3 \cdot \hspace{0.01mm} \dots \hspace{0.01mm} \cdot g_n\). Applying the homomorphism property,
Repeatedly applying the same idea, starting again with the product \(g_2 \cdot g_3 \cdot \hspace{0.01mm} \dots \hspace{0.01mm} \cdot g_n\) yields the result. The fact that \(\phi(g^n) = \phi(g)^n\) is follows immediately.
{\color{Plum} If \(\phi\) is a bijective homomorphism (i.e., one-to-one and onto) then we say that \(\phi\) is an isomorphism. Furthermore, if there exists an isomorphism between two spaces \(G\) and \(G'\), then we say these spaces are isomorphic and that \(G \cong G'\). As we'll soon see, isomorphisms gives us really nice results (hence the special terminology and notation). In addition, it can sometimes be difficult to tell when two groups \(G\) and \(G'\) are the same or different. Isomorphisms can help determine when there isn't such an equivalence.
As we'll see, the concept of an isomorphism is very powerful. However, proving it may not be that simple, and in ceratin cases the following theorem will be very useful. }
Let \(G\) and \(H\) be groups. The homomorphism \(\phi: G \to H\) is an isomorphism if and only if there exists a homomorphism \(\psi: H \to G\) such that \(\psi \circ \phi\) is the identity map on \(G\) and \(\phi \circ \psi\) is the identity map on \(H\).
(\(\implies\)) Suppose \(\phi: G \to H\) is an isomorphism. Since \(\phi\) is bijective, define the inverse map \(\phi^{-1}: H \to G\) such that if \(\phi(g) = g'\) then \(\phi^{-1}(g') = g\).
Note that this is a well defined map due to the surjectivity and injectivity of \(\phi\). To show it is a homomorphism, we need to demonstrate that \(\phi^{-1}(h_1\cdot h_2) = \phi^{-1}(h_1)\phi^{-1}(h_2)\). Thus observe that for \(h_1, h_2 \in H\) there exist \(g_1, g_2 \in G\) such that \(\phi(g_1) = h_1\) and \(\phi(g_2) = h_2\). Therefore
Thus \(\phi^{-1}\) is a homomorphism.
Now observe that for all \(g \in G\) we have that \(\phi^{-1} \circ \phi(g) = g\) and for all \(h \in H\), \(\phi \circ \phi^{-1}(h) = h.\) Thus \(\phi^{-1} \circ \phi\) is the identity on \(G\) while \(\phi \circ \phi^{-1}\) is the identity on \(H\), which proves this direction.
(\(\impliedby\)) Now suppose \(\phi: G \to H\) is a homomorphism and that there exists a homomorphism \(\psi: H \to G\) such that \(\psi \circ \phi\) is the identity map on \(G\) and \(\phi \circ \psi\) is the identity map in \(H\). In other words, \(\psi\) and \(\phi\) are inverses of each other. Thus \(\phi\) is a bijection function from \(G \to H\), which implies that \(\phi\) is an isomorphism.
We also introduce the following criteria which is frequently used to evaluate if a homomorphism is one-to-one and/or onto.
Let \(\phi: G \to G'\) be a homomorphism. Then
- 1. \(\phi\) is one-to-one if and only if \(\mbox{ker}(\phi)\) is trivial. That is, \(\mbox{ker}(\phi) = \{e_G\}\), where \(e_G\) is the identity of \(G\).
- 2. \(\phi\) is onto if and only if \(\mbox{im}(\phi) = G'\).
Therefore, \(\phi\) is an isomorphism if and only if (1) and (2) hold.
- 1. Suppose \(\phi\) is one-to-one. By proposition 1.1.1, we know that \(\phi(e_G) = e_{G'}\). But since \(\phi\) is injective we know \(e_G\) is the only element in \(G\) which is mapped to \(e_{G'}\). Therefore \(\mbox{ker}(\phi) = \{e_G\}\).
Now suppose \(\mbox{ker}(\phi) = \{e_G\}\). To show \(\phi\) is one-to-one, consider \(g, h \in G\) such that
Multiplying both sides by \(\phi(h)^{-1}\) we get
By proposition 1.1.2, we know that \(\phi(h)^{-1} = \phi(h^{-1})\). Since \(\phi\) is a homomorphism, we can then combine the terms to get
Since \(\mbox{ker}(\phi) = \{e_G\}\), we see that
Therefore \(\phi\) is one to one. * 2. Suppose \(\phi\) is onto. Then \(\mbox{im}(\phi) = G'\) is just another way of stating this fact.
Suppose \(\mbox{im}(\phi) = G'\). Then for every element \(g' \in G'\), there exists \(g \in G\) such that \(\phi(g) = g'\). That is, \(\phi\) covers every value in \(G'\) so that it is onto.
Thus, we have that a function is isomorphic if and only if it is one to one and onto. Hence, it is isomorphic if and only if (1) and (2) hold.
We also make two common definitions for special homomorphisms.
Let \(G\) be a group.
- 1. If \(\phi: G \to G\) is a group homomorphism, then we say that \(\phi\) is a endomorphism.
- 2. If \(\phi\) is a bijective endomorphism (an isomophic endomorphism) then we say that \(\phi\) is an automorphism.
The set of all automorphisms of a group \(G\), denoted as \(\text{Aut}(G)\), forms a group with an operation \(\circ\) of function composition.
We can prove this directly.
- Closure. Let \(\phi\) and \(\psi\) be automorphisms. Then \(\phi \circ \psi\) is (1) a homomorphism from \(G \to G\) and (2) a bijection (as the composition of bijections is a bijetion).
- Associativity. In general, function composition is associative.
- Identity. Let \(i:G \to G\) be the identity map. The (1) \(i\) is a group homomorphism and (2) a bijection. Therefore \(i \in \text{Aut}(G)\) and we can set \(i\) as the identity of the group. Note that
for any \(\phi \in \text{Aut}(G)\). * Inverse. Let \(\phi \in \text{Aut}(G)\). Construct the function \(\phi^{-1}\) as follows. If \(\phi(g) = g'\) for some \(g, g' \in G\), then write \(\phi^{-1}(g') = g\). Such an assignment is well-defined since \(\phi\) is a bijection. Hence we see that
Finally, observe that \(\phi^{-1}\) is (1) a homomorphism and (2) a bijection, so we see that \(\phi^{-1} \in \text{Aut}(G)\). Therefore this forms a group.