CHAPTER 04 · MA-0561

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Group Homomorphisms

Elementary definitions

Definition (Group homomorphism)

Let \((G,\ast_{G}), (H,\ast_{H})\) be groups. A group homomorphism between \(G\) and \(H\) is a function \(f:G\to H\) such that

\[\forall a,b, \in G \quad \Big( f(a \ast_{G} b) = f(a) \ast_{h} f(b)\Big).\]

Examples (Basic homomorphisms)

  1. Let \(G\) be a group and \(f:G\to G\) such that \(f(x) = 1_{G}\) for all \(x \in G\). Note that

    \[f(x \ast y) = 1_{G} = 1_{G} \ast 1_{G} = f(x) \ast f(g)\]

  2. Consider \(f:(\mathbb{R},+) \to (\mathbb{R}^{\ast}, \cdot)\) such that \(f(x) = e^{x}\). Note that

    \[f(x+y) = e^{x+y} = e^{x} \cdot e^{y} = f(x) \cdot f(y).\]

Proposition (Properties of homomorphisms)

Let \((G,\ast_{G}), (H,\ast_{H})\) be groups and \(f:G\to H\) a homomorphism. Then:

  1. \(f(1_{G}) = 1_{H}\),
  2. \(\forall g \in G \quad \Big( f(g^{-1}) = \big(f(g)\big)^{-1}\Big)\),
  3. \[\forall n \in \mathbb{Z} \quad \Big(f(g^{n}) = \big(f(g)\big)^{n}\Big)\]
  4. If \(g \in G\) and \(\lvert g \rvert = n\), then \(\lvert f(g) \rvert \Big\lvert n\).

Proof: For (1), note that

\[f(1_{G}) = f(1_{G} \ast_{G} 1_{G}) = f(1_{G}) \ast_{H} f(1_{G}),\]

from which we conclude that \(f(1_{G}) = 1_{H}\). For (2), note that

\[g \ast_{G} g^{-1} = 1_{G} \implies f(g) \ast_{H} f(g^{-1}) = 1_{H} \implies f(g^{-1}) = \big(f(g)\big)^{-1}.\]

For (3), in the case \(n\geq 0\), we proceed by induction. The base case \(n=0\) was proved in (1). Suppose as inductive hypothesis that \(f(g^{n})=\big(f(g)\big)^{n}\) for some \(n \in \mathbb{N}\). Note that

\[\begin{aligned} f(g^{n+1}) = f(g^{n} \ast_{G} g) &=f(g^{n}) \ast_{H} f(g)\\ &=\big(f(g)\big)^{n} \ast_{H} f(g) \\ &= \big(f(g)\big)^{n+1}. \end{aligned}\]

Now, for \(n<0\), note that \(-n > 0\). Thus, \(g^{n} = (g^{-n})^{-1}\). Thus,

\[f(g^{n}) = f\big((g^{-n})^{-1}\big) = \big(f(g^{-n})\big)^{-1} = \Big(\big(f(g)\big)^{-n}\Big)^{-1} = \big(f(g)\big)^{n}.\]

Finally, for (4), since \(\lvert g \rvert = n\),

\[g^{n} = 1_{G} \implies f(g^{n}) = \big(f(g)\big)^{n} = 1_{H},\]

from which we deduce that \(\lvert f(g) \rvert \Big\lvert n\).

Example (Projection and homomorphisms into \(\mathbb{Z}\))

  1. Consider \(f:(\mathbb{Z}, +)\to(\mathbb{Z}_{n}, +)\) such that \(x \mapsto [x]\). Note that

    \[f(x+y) = [x+y] = [x] + [y] = f(x) + f(y),\]

from which we conclude that \(f\) is a homomorphism.

  1. Is there a non-trivial homomorphism \(f:\mathbb{Z}_{n} \to \mathbb{Z}\)? If there were, note that \(\lvert [1] \rvert = n\) and hence \(\lvert f[1] \rvert \Big\lvert n\), a contradiction, since the order of the elements of \(\mathbb{Z}\) is infinite (except for zero).

Proposition (Composition of homomorphisms)

Let \(G,H,K\) be groups and \(f:G\to H\), \(g:H\to K\) homomorphisms. Then \(g \circ f:G \to K\) is a homomorphism.

Proof: Exercise

Definition (Kernel and image of a homomorphism)

Let \(G,H\) be groups and \(f:G\to H\) a homomorphism. Define

\[\begin{aligned} \operatorname{Ker}(f) &= \{ x \in G: f(x) = 1_{H} \} \subseteq G \\ \operatorname{Im}(f) &= \{ f(x):x \in G \} \subseteq H. \end{aligned}\]

Theorem (Kernel and image are subgroups)

Let \(G,H\) be groups and \(f:G\to H\) a homomorphism. Then,

  1. \[\operatorname{Ker}(f) \leq G\]
  2. \[\operatorname{Im}(f) \leq H\]

Proof: For (1), note that \(1_{G} \in \operatorname{Ker}(f)\), so \(\operatorname{Ker}(f) \neq \emptyset\). Let \(x,y \in \operatorname{Ker}(f)\).

\[f(x \ast_{G} y^{-1}) = f(x) \ast_{H} f(y^{-1}) =1_{H} \implies x \ast_{G} y \in \operatorname{Ker}(f).\]

For (2), note that \(f(1_{G}) = 1_{H} \in \operatorname{Im}(f)\), so \(\operatorname{Im}(f) \neq \emptyset\). Let \(a,b \in \operatorname{Im}(f)\). Hence, there exist \(x,y \in G\) such that \(f(x) = a\) and \(f(y) = b\). By the properties of homomorphisms, \(b^{-1} = f(y^{-1})\). Thus,

\[f(x \ast_{G} y^{-1}) = f(x) \ast_{H} \big(f(y)\big)^{-1} = a \ast_{H} b^{-1} \in \operatorname{Im}(f).\]

Example (Projection \(\mathbb{Z} \to \mathbb{Z}_n\))

Let \(\pi:\mathbb{Z} \to \mathbb{Z}_{n}\) be such that \(x \mapsto [x]\). We have that

\[\begin{aligned} \operatorname{Im}(\pi) &= \{ \pi(x):x \in \mathbb{Z} \} \\ &=\{ [x]: x \in \mathbb{Z} \} \\ &=\mathbb{Z}_{n},\\ \operatorname{Ker}(\pi) &=\{ x:\pi(x) = [0] \} \\ &=\{ x: [x] = 0 \} \\ &=n \cdot \mathbb{Z}. \end{aligned}\]

Definition (Monomorphisms, epimorphisms and isomorphisms)

Let \(f:G\to H\) be a group homomorphism:

  1. if \(f\) is injective, we say that \(f\) is a monomorphism,
  2. if \(f\) is surjective, we say that \(f\) is an epimorphism,
  3. if \(f\) is bijective, we say that \(f\) is an isomorphism.

Theorem (Existence of the inverse isomorphism)

If \(f:G\to H\) is an isomorphism, then \(f^{-1}:H\to G\) (as a function) is an isomorphism.

Proof: Exercise.

Theorem (Monomorphisms and kernel)

Let \(f:G\to H\) be a homomorphism. Then,

\[f \text{ is a monomorphism} \iff \operatorname{Ker}(f) = \{ 1_{G} \}\]

Proof: exercise. Note that \(f(x)=f(y) \implies f(x \ast y^{-1}) = 1_{H}\).

Theorem (Inverse image and homomorphisms)

Let \(f:G\to H\) be a homomorphism and let \(H'\leq H\). Then \(f^{-1}(H') \leq G\)

Proof: Note that \(1_{G} \in f^{-1}(H')\) since \(f(1_{G}) = 1_{H} \in H \implies f^{-1}(H') \neq \emptyset\). Let \(x,y \in f^{-1}(H')\). We must show that \(x \ast_{G} y^{-1} \in f^{-1}(H')\). Since \(f(x) \in H', f(y) \in H'\), we have that \(\big(f(y)\big)^{-1} = f(y^{-1}) \in H'\). Then \(f(x \ast_{G} y^{-1}) = f(x)\ast_{H} f(y^{-1}) \in H'\).

Definition (Automorphism and automorphism group)

Let \(G\) be a group. An automorphism of \(G\) is an isomorphism from \(G\) to \(G\). Define \(\mathrm{Aut} \hspace{2pt}(G) = \{ f:G\to G \text{ isomorphism}\}\) as the set of automorphisms of \(G\). Note that \(\mathrm{Aut} \hspace{2pt}(G) \subseteq S_{G} = \{ \phi: G\to G \text{ bijective}\}\).

Theorem (Subgroup of automorphisms)

\(\mathrm{Aut} \hspace{2pt}(G) \leq S_{g}\).

Proof: Exercise

Definition (Isomorphic groups)

We say that two groups \(G\) and \(H\) are isomorphic if there exists an isomorphism \(f:G\to H\). We write \(G \cong H\).

Example (Isomorphism \(\mathbb{Z}_4 \cong G\))

Consider \(G = (\{ 1,i,-1,-i \}, \ast)\).

\(Z_{4} \cong G\) under the isomorphism

\[\begin{aligned} 0 &\mapsto 1 \\ 1 &\mapsto i \\ 2 &\mapsto -1 \\ 3 &\mapsto -i \end{aligned}\]

Note (Being isomorphic is an equivalence relation)

Note that the relation \(G \cong H\) is reflexive, transitive and symmetric (this is easy to prove). Hence it is an equivalence relation.

Theorem (Isomorphisms of cyclic groups)

If \(G = \langle g \rangle\), then:

  1. If \(\lvert g \rvert = \infty \implies G \cong \mathbb{Z}\).
  2. If \(\lvert g \rvert = n \implies G \cong \mathbb{Z}_{n}\).