Section3.2Definitions of Homomorphisms and Isomorphisms¶ permalink
Definition3.2.1
Let \(\langle S,*\rangle\) and \(\langle S',*'\rangle\) be binary structures. A function \(\phi\) from \(S\) to \(S'\) is a homomorphism if
\begin{equation*}
\phi(a* b)=\phi(a)*'\phi(b)
\end{equation*}
for all \(a,b\in S\text{.}\) An isomorphism is a homomorphism that is also a bijection.
Intuitively, you can think of a homomorphism \(\phi\) as a “structure-preserving” map: if you multiply and then apply \(\phi\text{,}\) you get the same result as when you first apply \(\phi\) and then multiply. Isomorphisms, then, are both structure-preserving and cardinality-preserving.
Example3.2.4
For each of the following, decide whether or not the given function \(\phi\) from one binary structure to another is a homomorphism, and, if so, if it is an isomorphism. Prove or disprove your answers! For Parts 6 and 7, \(C^0\) is the set of all continuous functions from \(\R\) to \(\R\text{;}\) \(C^1\) is the set of all differentiable functions from \(\R\) to \(\R\) whose derivatives are continuous; and each \(+\) indicates pointwise addition on \(C^0\) and \(C^1\text{.}\)
\(\phi:\langle \Z,+\rangle \to \langle \Z,+\rangle\) defined by \(\phi(x)=x\text{;}\)
\(\phi:\langle \Z,+\rangle \to \langle \Z,+\rangle\) defined by \(\phi(x)=-x\text{;}\)
\(\phi:\langle \Z,+\rangle \to \langle \Z,+\rangle\) defined by \(\phi(x)=2x\text{;}\)
\(\phi:\langle \R,+\rangle \to \langle \R^+,\cdot\,\rangle\) defined by \(\phi(x)=e^x\text{;}\)
\(\phi:\langle \R,+\rangle \to \langle \R^*,\cdot\,\rangle\) defined by \(\phi(x)=e^x\text{;}\)
\(\phi:\langle C^1,+\rangle \to \langle C^0,+\rangle\) defined by \(\phi(f)=f'\) (the derivative of \(f\));
\(\phi:\langle C^0,+\rangle \to \langle \R,+\rangle\) defined by \(\phi(f)=\displaystyle{\int_0^1 f(x)\, dx}\text{.}\)
Example3.2.5
Let \(\Gdot\) be a group and let \(a\in G\text{.}\) Then the function \(c_a\) from \(G\) to \(G\) defined by \(c_a(x)=axa^{-1}\) (for all \(x\in G\)) is a homomorphism. Indeed, let \(x,y\in G\text{.}\) Then
\begin{align*}
c_a(xy)\amp =a(xy)a^{-1}\\
\amp =(ax)e(ya^{-1})\\
\amp =(ax)(a^{-1}a)(ya^{-1})\\
\amp =(axa^{-1})(aya^{-1})\\
\amp =c_a(x)c_a(y).
\end{align*}
The homomorphism \(c_a\) is called conjugation by \(a\).
Definition3.2.6
Homomorphisms from a group \(G\) to itself are called endomorphisms, and isomorphisms from a group to itself are called automorphisms.
It can be shown that conjugation by any element \(a\) of a group \(G\) is a bijection from \(G\) to itself (can you prove this?), so such conjugation is an automorphism of \(G\text{.}\) (Beware: Some texts use “conjugation by \(a\)” to refer to the function \(x\mapsto a^{-1}xa\text{.}\)) Both versions of conjugation by \(a\) in group \(G\) are automorphisms of \(G\text{.}\))
We end with a theorem stating basic facts about homomorphisms from one group to another. (Note. This doesn't apply to arbitrary binary structures, which may or may not even have identity elements.)
Theorem3.2.7
Let \(\langle G,\cdot\rangle\) and \(\langle G',\cdot'\rangle\) be groups with identity elements \(e\) and \(e'\text{,}\) respectively, and let \(\phi\) be a homomorphism from \(G\) to \(G'\text{.}\) Then:
\(\phi(e)=e'\text{;}\) and
For every \(a\in G\text{,}\) \(\phi(a)^{-1}=\phi(a^{-1})\text{.}\)
Proof
For Part 1, note that
\begin{align*}
\phi(e)\cdot'e'\amp =\phi(e)\amp \amp \text{ (by definition of \(e'\)) }\\
\amp =\phi(e\cdot e) \amp \amp \text{ (by definition of \(e\)) }\\
\amp =\phi(e)\cdot'\phi(e) \amp \amp \text{ (since \(\phi\) is a homomorphism). }
\end{align*}
Thus, by left cancellation, \(e'=\phi(e)\text{.}\) The proof of Part 2 is left as an exercise for the reader.