Homomorphisms

Section 1.6 Homomorphisms

Definition 1.6.1. Homomorphism.

Let \(G\) and \(H\) be groups. A (group) homomorphism is a map \(\phi\colon G\rightarrow H\) satisfying

\begin{equation} \phi(g_1g_2)=\phi(g_1)\phi(g_2)\tag{1.6.1} \end{equation}

for all \(g_1,g_2\in G\text{.}\)

Remark 1.6.2. Homomorphism.

There is an important subtlety in the defining property (1.6.1): namely, there are two group operations at play. In more detail, the input \(g_1g_2\) of \(\phi\) on the left side is a product within the domain group \(G\text{;}\) the right side \(\phi(g_1)\phi(g_2)\) on the other hand is a product within the codomain group \(H\text{.}\)

🔗

A useful way of thinking of a group homomorphism is as a function between groups that respects the group structure, in the sense articulated by (1.6.1). You might ask: shouldn’t a group homomorphism also “respect” the group identities and group inverses? It turns out that the property (1.6.1) is itself enough to guarantee this.

🔗

Proposition 1.6.3. Homomorphism properties.

Let \(\phi\colon G\rightarrow H\) be a group homomorphism, and let \(e_G\) and \(e_H\) be the identity elements of \(G\) and \(H\text{,}\) respectively.

\(\displaystyle \phi(e_G)=\phi(e_H)\)
🔗

🔗
\(\displaystyle \phi(g^{-1})=\phi(g)^{-1}\)
🔗

🔗

🔗

Proof.

Since \(e_G=e_G\, e_G\) and \(\phi\) is a homomorphism, we have \(\phi(e_G)=\phi(e_G)\phi(e_G)\text{.}\) Now do some algebra in \(H\text{:}\)

\begin{align*} \phi(e_G)\amp =\phi(e_G)\phi(e_G)\amp \\ \phi(e_G)^{-1}\phi(e_G)\amp =\phi(e_G)^{-1}\phi(e_G)\phi(e_G) \amp \\ e_H \amp = \phi(e_G)\text{.} \end{align*}

🔗

🔗
Fix \(g\in G\text{.}\) We have

\begin{align*} gg^{-1}=g^{-1}g=e_G \amp \implies \phi(gg^{-1})=\phi(g^{-1}g)=\phi(e_G) \\ \amp \implies \phi(g)\phi(g^{-1})=\phi(g^{-1})\phi(g)=e_H \\ \amp \implies \phi(g^{-1})=\phi(g)^{-1} \text{,} \end{align*}

where the last implication follows from the fact that inverses in groups are unique.

🔗

🔗

🔗

Example 1.6.4. Homormorphisms.

Decide whether the given map is a group homormorphism.

Trivial homomorphism.
Let \(G\) and \(H\) be groups. Define \(\phi\colon G\rightarrow H\) as \(\phi(g)=e_H\) for all \(g\in G\text{.}\)
🔗

🔗
Identity map.
Let \(G\) be a group, and let \(\id\colon G\rightarrow G\) be the identity map as usual.
🔗

🔗
Fix \(m\in \Z\) and define \(\phi\colon \Z\rightarrow \Z\) as \(\phi(n)=mn\text{.}\)
🔗

🔗
Define \(\phi\colon \Z\rightarrow \Z\) as \(\phi(n)=n^2\text{.}\)
🔗

🔗
Determinant.
Assume \(R\in \{\Z,\Q,\R,\C\}\) or \(R=\Z/m\Z\) for some positive integer \(m\text{.}\) Fix a positive integer \(n\) and define \(\phi\colon \GL_n(R)\rightarrow R^*\) as \(\phi(A)=\det A\text{.}\)
🔗

🔗
Exponential map.
Define \(\phi\colon \R\rightarrow \R^*\) as \(\phi(x)=e^x\text{.}\)
🔗

🔗

🔗

Solution.

🔗

The following proposition serves as a useful tool for constructing homomorphisms out of \(D_n\text{.}\) You will prove it in your homework.

🔗

Proposition 1.6.5. Homomorphisms from \(D_n\).

Fix an integer \(n\geq 3\text{,}\) and let \(D=\angvec{r,s\mid r^n=s^2=e, sr=r^{-1}s}\text{.}\) Given a group \(G\) and elements \(a,b\in G\) satisfying

\begin{align*} a^n \amp =b^2=e \amp ba\amp =a^{-1}b\text{,} \end{align*}

the map \(\phi\colon D_n\rightarrow G\) defined as

\begin{equation*} \phi(r^is^j)=a^is^j, \ 0\leq i\leq n-1, 0\leq j\leq 1 \end{equation*}

is a homomorphism.

🔗

Proof.

The proof is left as a homework exercise.

🔗

Definition 1.6.6. Isomorphism.

An isomorphism is a bijective group homomorphism \(\phi\colon G\rightarrow H\text{.}\) Groups \(G\) and \(H\) are isomorphic, denoted \(G\cong H\text{,}\) if there is an isomorphism \(\phi\colon G\rightarrow H\text{.}\)

🔗

Example 1.6.7. Permutation groups.

Let \(A\) and \(B\) be nonempty sets of the same cardinality: i.e., \(\abs{A}=\abs{B}\text{.}\) Prove: \(S_A\cong S_B\text{.}\)

🔗

Solution.

By definition of cardinality, since \(\abs{A}=\abs{B}\text{,}\) there exists a bijective function \(\alpha\colon A\rightarrow B\text{.}\) Using \(\alpha\text{,}\) we define \(\phi\colon S_A\rightarrow S_B\) as

\begin{align*} \phi(\sigma) \amp =\alpha\circ \sigma\circ \alpha^{-1}\text{.} \end{align*}

Let’s first see why this is a well-defined map from \(S_A\) to \(S_B\text{.}\) From

\begin{align*} \alpha^{-1}\colon B \amp \rightarrow A\\ \sigma\colon A\amp \rightarrow A \amp \\ \alpha\colon A \amp \rightarrow B\text{,} \end{align*}

we see that

\begin{equation*} \alpha\circ\sigma\circ\alpha^{-1}\colon B\rightarrow B\text{.} \end{equation*}

Next since \(\alpha, \alpha^{-1},\) and \(\sigma\) are bijective, so is their composition. Thus \(\phi(\sigma)=\alpha\circ\sigma\circ\alpha^{-1}\in S_B\text{,}\) as desired.

🔗

Next, we show \(\phi\) is a homomorphism. We must show that \(\phi(\sigma\tau)=\phi(\sigma)\phi(\tau)\) for all \(\sigma, \tau\in S_A\text{.}\) By definition, we have (dropping the \(\circ\) symbols)

\begin{align*} \phi(\sigma)\phi(\tau) \amp =(\alpha\sigma\alpha^{-1})(\alpha\tau\alpha^{-1})\\ \amp = \alpha \sigma \id_A\tau \alpha^{-1}\amp (\alpha^{-1}\circ\alpha=\id_A)\\ \amp = \alpha \sigma \tau \alpha^{-1} \amp (\id_A\circ \tau=\tau) \\ \amp =\phi(\sigma\tau)\text{.} \end{align*}

This proves \(\phi\) is a homomorphism. To show it is bijective, and hence an isomorphism, we will show that it has an inverse function \(\phi^{-1}\colon S_B\rightarrow S_A\text{.}\) To this end, given \(\tau\in S_B\text{,}\) define

\begin{equation*} \phi^{-1}(\tau)=\alpha^{-1}\circ\tau \circ \alpha\text{.} \end{equation*}

Reasoning exactly as above, we see that \(\phi^{-1}\) is well-defined: i.e., \(\alpha^{-1}\circ\tau\circ\alpha\) is indeed an element of \(S_A\text{.}\) Lastly, computations much like the one above show that

\begin{align*} \phi^{-1}\circ\phi\, (\sigma) \amp = \sigma \text{ for all } \sigma\in S_A \\ \phi\circ\phi^{-1}\, (\tau) \amp = \tau \text{ for all } \tau\in S_B \end{align*}

and thus that the two functions are inverses of one another.

🔗

Prev Top Next