3.2: Definitions of Homomorphisms and Isomorphisms
Definition: Homomorphism and Isomorphism
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.
Note
We may omit the \(*\) and \(*'\text{,}\) as per our group conventions, but we include them here to emphasize that the operations in the structures may be distinct from one another. When we omit them and write \(\phi(st)=\phi(s)\phi(t)\text{,}\) then it is the writers' and readers' responsibility to keep in mind that \(s\) and \(t\) are being operated together using the operation in \(S\text{,}\) while \(\phi(s)\) and \(\phi(t)\) are being operated together using the operation in \(S'\text{.}\)
Remark
There may be more than one homomorphism [isomorphism] from one binary structure to another (see Example \(3.2.1\)).
Example \(\PageIndex{1}\)
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 \(\mathbb{R}\) to \(\mathbb{R}\text{;}\) \(C^1\) is the set of all differentiable functions from \(\mathbb{R}\) to \(\mathbb{R}\) whose derivatives are continuous; and each \(+\) indicates pointwise addition on \(C^0\) and \(C^1\text{.}\)
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\(\phi:\langle \mathbb{Z},+\rangle \to \langle \mathbb{Z},+\rangle\) defined by \(\phi(x)=x\text{;}\)
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\(\phi:\langle \mathbb{Z},+\rangle \to \langle \mathbb{Z},+\rangle\) defined by \(\phi(x)=-x\text{;}\)
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\(\phi:\langle \mathbb{Z},+\rangle \to \langle \mathbb{Z},+\rangle\) defined by \(\phi(x)=2x\text{;}\)
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\(\phi:\langle \mathbb{R},+\rangle \to \langle \mathbb{R}^+,\cdot\,\rangle\) defined by \(\phi(x)=e^x\text{;}\)
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\(\phi:\langle \mathbb{R},+\rangle \to \langle \mathbb{R}^*,\cdot\,\rangle\) defined by \(\phi(x)=e^x\text{;}\)
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\(\phi:\langle C^1,+\rangle \to \langle C^0,+\rangle\) defined by \(\phi(f)=f'\) (the derivative of \(f\));
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\(\phi:\langle C^0,+\rangle \to \langle \mathbb{R},+\rangle\) defined by \(\phi(f)=\displaystyle{\int_0^1 f(x)\, dx}\text{.}\)
Example \(\PageIndex{2}\)
Let \(\langle G,\;\cdot\rangle\) 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
\(\begin{array} &c_a(xy)& =a(xy)a^{-1}\\ & =(ax)e(ya^{-1})\\ & =(ax)(a^{-1}a)(ya^{-1})\\ & =(axa^{-1})(aya^{-1})\\ & =c_a(x)c_a(y). \end{array}\)
The homomorphism \(c_a\) is called conjugation by \(a\).
Definition: Endomorphism and Automorphism
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.)
Theorem \(\PageIndex{1}\)
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
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For Part 1, note that
\(\begin{array}&ϕ(e)⋅′e′&=ϕ(e) &(\text{by definition of }e′)\\ &=ϕ(e⋅e) &(\text{by definition of } e) \\&=ϕ(e)⋅′ϕ(e) &(\text{since \(ϕ\) is a homomorphism}). \end{array} \)
Thus, by left cancellation, \(e′=ϕ(e)\). The proof of Part 2 is left as an exercise for the reader.