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3.2: Definitions of Homomorphisms and Isomorphisms

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    84806
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    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 \(C^1\text{.}\)

    1. \(\phi:\langle \mathbb{Z},+\rangle \to \langle \mathbb{Z},+\rangle\) defined by \(\phi(x)=x\text{;}\)

    2. \(\phi:\langle \mathbb{Z},+\rangle \to \langle \mathbb{Z},+\rangle\) defined by \(\phi(x)=-x\text{;}\)

    3. \(\phi:\langle \mathbb{Z},+\rangle \to \langle \mathbb{Z},+\rangle\) defined by \(\phi(x)=2x\text{;}\)

    4. \(\phi:\langle \mathbb{R},+\rangle \to \langle \mathbb{R}^+,\cdot\,\rangle\) defined by \(\phi(x)=e^x\text{;}\)

    5. \(\phi:\langle \mathbb{R},+\rangle \to \langle \mathbb{R}^*,\cdot\,\rangle\) defined by \(\phi(x)=e^x\text{;}\)

    6. \(\phi:\langle C^1,+\rangle \to \langle C^0,+\rangle\) defined by \(\phi(f)=f'\) (the derivative of \(f\));

    7. \(\phi:\langle C^0,+\rangle \to \langle \mathbb{R},+\rangle\) 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 \(\PageIndex{1}\)

    Proof

    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.

     


    This page titled 3.2: Definitions of Homomorphisms and Isomorphisms is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Jessica K. Sklar via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.