4.8 Homomorphisms and isomorphisms

Let \(G,*\) and \(H,\triangle\) be groups. A function \(f:G \to H\) doesn’t necessarily tell us anything about the relationship between G and H as groups unless we insist that it interacts in some specific way with the group operations \(*\) and \(\triangle\). We define a group homomorphism \(G \to H\) to be a function which respects the group structure on G and H in the following sense:

Definition 4.14 Let \(G,*\) and \(H,\triangle\) be groups.

  • A group homomorphism \(f:G \to H\) is a function such that for all \(x,y \in G\) we have \[ f(x*y)= f(x)\triangle f(y).\]
  • A group isomorphism is a group homomorphism which is a bijection.

We usually just say homomorphism and isomorphism instead of group homomorphism and group isomorphism.

Example 4.10

  • The identity map \(G \to G\) is an isomorphism.
  • Let \(G,*\) and \(H,\triangle\) be any groups. The map \(f: G \to H\) given by \(f(g)=e_H\) for all \(g \in G\) is a group homomorphism.
  • The map \(f: \zz \to \zz_n\) given by \(f(z)=[z]_n\) is a group homomorphism. To see this, we have to check that \(f(z+w) = f(z)+f(w)\). This is true because \(f(z+w)=[z+w]_n\) and by definition, \([z]_n+[w]_n=[z+w]_n\).
  • Let \(H = \{1, -1\}\) so that \(H,\times\) is a cyclic group of order 2. Then our results on the sign of a permutation show that \(\sgn : S_n \to H\) is a group homomorphism.
  • The map \(\det : GL(n,\rr) \to \rr^\times\) is a group homomorphism.
  • There is an isomorphism between \(S_3\) and the dihedral group of order 6.

Lemma 4.12 Let \(f:G \to H\) be a group homomorphism. Then

  • \(f(e_G)=e_H\), and
  • for any \(g \in G\) we have \(f(g^{-1})=f(g)^{-1}\).

Proof.

  • We have \[\begin{align*} f(e_G) &= f(e_G*e_G) & \text{as } e_G*e_G=e_G \\& = f(e_G) \triangle f(e_G) & \text{as } f \text{ is a homomorphism.} \end{align*}\] Now multiply both sides on the left by the inverse of \(f(e_G)\), to get \[\begin{equation*} e_H = f(e_G). \end{equation*}\]
  • By the previous part, \(e_H = f(g*g^{-1}) = f(g) \triangle f(g^{-1})\). Multiply on the left by \(f(g)^{-1}\) to get \[\begin{equation*} f(g)^{-1} = f(g^{-1}) \end{equation*}\]

Definition 4.15 Let \(f:G \to H\) be a group homomorphism.

  • \(\ker f\), the kernel of f, is defined to be \(\{ g \in G: f(g) = e_H\}\).
  • \(\im f\), the image of f, is defined to be \(\{ f(g): g \in G\}\).

The latter isn’t a new definition — it’s just the image of the function f as we already defined it.

Proposition 4.8 Let \(f:G \to H\) be a group homomorphism. Then \(\ker f\) is a subgroup of G and \(\im f\) is a subgroup of H.

Proof. The kernel contains \(e_G\) because \(f(e_G) = e_H\) by the previous Lemma. If \(g \in \ker f\) then \[\begin{align*} f(g^{-1}) &= f(g)^{-1} & \text{by the lemma} \\ &= e_H^{-1} & \text{as } g \in \ker f \\ &= e_H \end{align*}\] so \(g^{-1} \in \ker f\) which is therefore closed under inverses, and if \(g,h \in \ker f\) then \[\begin{align*} f(g*h) & =f(g)\triangle f(h) & \text{as } f \text{ is a homomorphism} \\ &= e_H \triangle e_H & \text{as } g,h \in \ker f \\ &= e_H \end{align*}\] so \(g*h \in \ker f\). It follows \(\ker f\) is a subgroup.

Now the image. The previous lemma shows \(e_h \in \im f\), and that \(\im f\) is closed under inverses. If \(f(g),f(h) \in \im f\) then \(f(g)\triangle f(h) = f(gh) \in \im f\) as f is a homomorphism, so \(\im f\) is a subgroup.