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Section 1.11 Homomorphisms and isomorphisms

Let us define what we mean more precisely when we say that two groups are “the same” as abstract groups. We want to define the notion of isomorphisms for groups. But to start we define homomorphisms. In a few words, those are maps between groups that preserve the group structure. Isomorphisms are then defined as homomorphisms that are bijective.

Subsection 1.11.1 Definitions and examples

Definition 1.11.1. Group homomorphism.

Let \((G,\star)\) and \((H,\cdot)\) be groups, with corresponding group operations denoted by \(*\) and \(\cdot\) respectively. A group homomorphism \(f: G \to H\) is a map that preserves the group structure, that is

\begin{equation*} f(a \star b) = f(a) \cdot f(b) \qquad \text{for all } a,b \in G. \end{equation*}

In other words, a group homomorphism is a map between sets that is consistent with their group structures.

The most simple example of a homomorphism is called the trivial homomorphism. Take any group \((G,\star)\text{,}\) and let \(H\) be the group consisting of a single element \(\{e\}\text{.}\) Then the trivial homomorphism is given by the mapping \(f(a) = e\) for all \(a \in G\text{,}\) which trivially preserves the group structure. For any \(G\) with order \(>1\text{,}\) it is of course not an isomorphism, since the mapping is not one-to-one (all elements of \(G\) are mapped into \(e \in H\)).

While a group homomorphism is a mapping between groups that is consistent with their group structures, in general it may “lose information”, if many group elements are mapped to the same group element. This is clear by looking at the trivial group homomorphism, which maps all groups to the trivial group with a single element. So we want to define a group isomorphism as an homomorphism that “preserves all information about the group.”

Definition 1.11.3. Group isomorphism and automorphism.

An isomorphism is a group homomorphism that is bijective. We say that two groups are isomorphic, denoted by \(G \simeq H\text{,}\) if there is an isomorphism \(f:G \to H\text{.}\) An isomorphism of a group into itself is called an automorphism.

An interesting example of an isomorphism is given by the exponential map, \(\exp: (\mathbb{R},+ ) \to (\mathbb{R}_{>0}, \cdot)\text{.}\) The notation \((\mathbb{R},+ )\) denotes the group of real numbers under addition, while \((\mathbb{R}_{>0}, \cdot)\) denotes the group of positive real numbers under multiplication. The mapping \(\exp\) takes any real number and outputs its exponential, which is a positive real number. This mapping preserves the group structure (the operation in the first group is addition, while in the second group it is multiplication). What this means is that the exponential of the sum of two real numbers should be equal to the product of the exponential of these real numbers. Indeed:

\begin{equation*} \exp(a+b) = \exp(a) \cdot \exp(b) \qquad \text{for all } a,b \in \mathbb{R}. \end{equation*}

Thus \(\exp\) is a group homomorphism. Moreover, from standard properties of the real exponential it is easy to see that the homomorphism is bijective, thus it is an isomorphism.

A similar example was studied in the previous section (see Example 1.10.6). We considered the quotient group \(\mathbb{R} / \mathbb{Z}\text{,}\) and define a mapping \(f: \mathbb{R} / \mathbb{Z} \to U(1)\) by \(f(a+\mathbb{Z}) = \mathrm{e}^{2 \pi i a}\text{.}\) As in the previous example, it follows from properties of the exponential that this mapping preserves the group structure (being coset multiplication in \(\mathbb{R} / \mathbb{Z}\) and multiplication of exponentials in \(U(1)\)), and hence it is a group homomorphism. Moreover, one can see that it is bijective, therefore it is a group isomorphism. This is what we meant when we said that the two groups were “the same as abstract groups”.

Subsection 1.11.2 First isomorphism theorem

Given an homomorphism \(f:G \to H\text{,}\) we can define its kernel as being the subset of elements of \(G\) that are mapped to the identity element in \(H\text{.}\) It is straightforward to show that it is a subgroup of \(G\text{:}\)

Definition 1.11.6. The kernel of an homomorphism.

Let \(G\) and \(H\) be groups, and \(f:G \to H\) a homomorphism. The kernel of \(f\) is

\begin{equation*} \ker(f) = \{ g \in G | f(g) = e \in H \}. \end{equation*}

One can show that \(\ker(f)\) is a subgroup of \(G\text{.}\)

Prove that \(\ker(f)\) is a subgroup of \(G\text{.}\)

In fact, one of the important theorems in group theory is known as the first isomorphism theorem (also sometimes called the fundamental theorem of homomorphisms). It is quite fundamental, as it relates the structure of the kernel and image of any group homomorphism. The statement of the theorem goes as follows:

We know that \(\ker(f)\) is a subgroup of \(G\text{.}\) Let us show that it is normal. Let \(x \in \ker(f)\text{,}\) and pick any \(g \in G\text{.}\) We want to show that \(g x g^{-1} \in \ker(f)\text{.}\) Since \(f:G \to H\) is a group homomorphism, we have:

\begin{equation*} f( g x g^{-1} ) = f(g) f(x) f(g^{-1}) = f(g) e_H f(g^{-1}) = f(g) f(g^{-1}) = f( g g^{-1}) = f( e_G)= e_H, \end{equation*}

where \(e_H\) (resp. \(e_G\)) denotes the identity element in \(H\) (resp. in \(G\)). Thus we conclude that \(g x g^{-1}\) is in the kernel of \(f\text{,}\) and hence \(\ker(f)\) is normal.

The isomorphism \(\phi\) between \(G/\ker(f)\)and \(f(G)\) is simply given by sending the coset \(g \ker(f)\) generated by \(g \in G\) to the image \(f(g) \in f(G)\) of \(g\) under the group homomorphism. \(\phi\) is of course surjective, and it is a good exercise to show that it is also one-to-one, hence and isomorphism.

This theorem is quite deep. In particular, it implies the following result:

Given any group \(G\) and normal subgroup \(N \subset G\text{,}\) there is a natural group homomorphism \(\pi: G \to G/N\text{,}\) which sends each element of \(G\) to the coset to which it belongs. This homomorphism is surjective, but not one-to-one (unless \(N\) is trivial). The kernel of \(\pi\) consists of all elements of \(G\) that are mapped to the identity coset of \(N\) in \(G\text{,}\) which is just a copy of \(N\) itself. Thus one can identify \(\ker(\pi) = N\text{.}\)

The first isomorphism theorem (and the direct corollary above, which is also referred to as first isomorphism theorem) is useful for many reasons. In particular, it can be used to show that a given subgroup is normal, by realizing it as the kernel of a group homomorphism. It can also be used to construct quotient groups as images of group homomorphisms.

Consider the general linear group \(GL(n,\mathbb{C})\) and define the mapping \(\det : GL(n,\mathbb{C}) \to \mathbb{C}^*\) by taking the determinant of the matrices (zero is not in the image since \(GL(n,\mathbb{C})\) contains invertible matrices). The image \(\mathbb{C}^*\) here is the group of non-zero complex numbers under multiplication. First, let us argue that \(\det\) is a group homomorphism. That is, we must show that it respects the group structure (in the first group the operation is matrix multiplication, in the second group it is product of complex numbers). For any two matrices \(A, B \in GL(n,\mathbb{C})\text{,}\) \(\det ( A B) = \det(A) \det(B)\text{,}\) therefore \(\det\) is a group homomorphism.

Since the identity element in the multiplicative group \(\mathbb{C}^*\) is \(1\text{,}\) the kernel of \(\det\) is given by the subgroup of \(GL(n,\mathbb{C})\) consisting of matrices with unit determinant, that is \(SL(n, \mathbb{C})\text{.}\) The first isomorphism theorem then implies that \(SL(n,\mathbb{C})\) is a normal subgroup of \(GL(n,\mathbb{C})\text{.}\)

We have seen another example of group homomorphism when we discussed parity of permutations in \(S_n\) and the definition of the alternating group \(A_n\text{.}\) In this case the first isomorphism theorem can be used to show directly that \(A_n\) is always a normal subgroup of \(S_n\text{.}\)

Consider the group \(H=\{1,-1\}\) under multiplication. Consider the mapping \(\epsilon : S_n \to H\) that assigns \(1\) to even permutations and \(-1\) to odd permutations. Let us first show that \(\epsilon\) is a group homomorphism.

For any permutations \(\pi_1, \pi_2 \in S_n\text{,}\) we need to show that \(\epsilon(\pi_1 \circ \pi_2) = \epsilon(\pi_1) \epsilon(\pi_2)\text{.}\) What this means is that we must show that the composition of two even or odd permutations is even, while the composition of an even and an odd permutation is odd. But this is clear by definition of parity. If \(r_1\) (resp. \(r_2\)) is the number of transpositions in the decomposition of \(\pi_1\) (resp. \(\pi_2\)), then the number of transpositions in the decomposition of \(\pi_1 \circ \pi_2\) is \(r_1 + r_2\text{,}\) and thus if \(r_1\) and \(r_2\) are both even or odd, then \(r_1 + r_2\) is even, while if \(r_1\) is even and \(r_2\) odd, or vice-versa, then \(r_1+r_2\) is odd.

Then \(\epsilon\) is a group homomorphism, and its kernel is the alternating group \(A_n\) consisting of even permutations. It thus follows from the first isomorphism theorem that \(A_n\) is a normal subgroup of \(S_n\text{,}\) and that the quotient group \(S_n / A_n\) is isomorphic to \(H\text{.}\)

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