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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 1.5 Group Homomorphisms

Subsection Homomorphism? I Barely Know ’em!

“In the end, the character of a civilization is encased in its structures.”
―Frank Gehry
The Greek roots "homo" and "morph" together mean "same shape".

Definition 1.59. Group Homomorphism.

If \(G\) and \(H\) are groups, a homomorphism from \(G\) is \(H\) is a function \(\varphi: G \to H\) such that
\begin{equation*} \varphi(x *_G y) = \varphi(x) *_H \varphi(y)\text{,} \end{equation*}
where \(*_G\) and \(*_H\) denote the multiplication rules for \(G\) and \(H\text{,}\) respectively.

Convention 1.60. Homomorphism Notation: \(\varphi\) vs. \(f\).

We use \(\varphi\) to denote an arbitrary homomorphism instead of \(f\) to differentiate between homormorphisms and ordinary functions.
Some folk use \(f\) for homomorphisms as well as functions that are not homomorphisms. This is usually harmless, we avoid doing so here only to remove as much ambiguity as possible.
Intuitively, a group homomorphism preserves the algebraic structure of the group, while allowing us to compare and relate different groups. In this way, group homomorphisms allow us to study the properties of groups by comparing them to other groups that we already understand well.

Example 1.61. Homormophism Examples.

  1. The identity map is a group homomorphism for any group \(G\text{,}\) this is known as the trivial homomorphism.
  2. The zero map is a group homomorphism from \(\C\to\C\text{.}\)
  3. The sign homomorphism \(\sign: S_n\to \{-1,1\}\) that sends even permuations to \(1\) and odd permutations to \(-1\) is a homomorphism.
Answer.
  1. Let \(x,y\in G\text{,}\) and let \(\id_G\) denote the identity map on \(G\text{.}\) Then
    \begin{align*} \id_G(xy) &= xy && \text{Definition of }\id_G \text{ on } xy\\ &= \id_G(x)\id_G(y) && \id_G(x)=x \text{ and } x \id_G(y)=y \end{align*}
    Thus \(\id_G\) is operation preserving, making it a group homomorphism for any group \(G\text{.}\)
  2. Let \(x,y\in \C\text{,}\) and let \(0_\C\) denote the zero map. As \(0_\C(xy) = 0 = 0_\C(x)0_\C(y)\text{,}\) \(0_\C\) is indeed a group homomorphism.
  3. First we note that \(\{-1,1\}\) is indeed a group, as \(\{-1,1\}=\Z^\times\text{,}\) as seen in Example 1.16. Let \(\sigma,\tau\in S_n\text{.}\) We proceed via cases.
    First, suppose both \(\sigma\) and \(\tau\) are even permutations. Thus \(\sigma\tau\) is an even permutation by Proposition 1.48, and hence
    \begin{equation*} \sign(\sigma)\sign(\tau) = (1)(1) = (1) = \sign(\sigma\tau). \end{equation*}
    Now suppose \(\sigma\) is even and \(\tau\) is odd. Thus \(\sigma\tau\) is an odd permutation by Proposition 1.48, and hence
    \begin{equation*} \sign(\sigma)\sign(\tau) = (1)(-1) = (-1) = \sign(\sigma\tau). \end{equation*}
    Next, suppose \(\sigma\) is odd and \(\tau\) is even. Thus \(\sigma\tau\) is an odd permutation by Proposition 1.48, and hence
    \begin{equation*} \sign(\sigma)\sign(\tau) = (-1)(1) = (-1) = \sign(\sigma\tau). \end{equation*}
    Finally, suppose both \(\sigma\) and \(\tau\) are odd permutations. Thus \(\sigma\tau\) is an even permutation by Proposition 1.48, and hence
    \begin{equation*} \sign(\sigma)\sign(\tau) = (-1)(-1) = (1) = \sign(\sigma\tau). \end{equation*}
    This shows \(\sign\) is a homomorphism in all cases.

Exercise 1.62.

  1. The exponential map \(\exp: ({\mathbb{R}}, +) \to (\mathbb{R}^\times, \cdot)\) is a homomorphism.
  2. The natural log map \(\ln: (\mathbb{R}^\times, \cdot) \to ({\mathbb{R}}, +)\) is a homomorphism.
  3. For any positive integer \(n\text{,}\) the map determinant map \(\det: \operatorname{GL}_n(\C) \to (\C^\times, \cdot)\) is a group homomorphism.
Specifically, homomorphisms will preserve the additive structure, such as the identity element and inverse elements.

Proof.

Let \(\varphi: G \to H\) be a homomorphism of groups.
  1. Observe:
    \begin{align*} \varphi(e_G) &= \varphi(e_Ge_G)\\ &= \varphi(e_G)\varphi(e_G) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}}\\ \varphi(e_G)\varphi(e_G)\inv &= \varphi(e_G)\varphi(e_G)\varphi(e_G)\inv &&\text{multiply by }\varphi(e_G)\inv\\ e_H &= \varphi(e_G) \end{align*}
  2. Let \(x\in G\text{.}\) I claim that \(\varphi(x\inv)\) is the inverse of \(\varphi(x)\) in \(H\text{:}\)
    \begin{align*} \varphi(x)\varphi(x\inv) &= \varphi(xx\inv) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}}\\ &= \varphi(e_G)\\ &= e_H && \knowl{./knowl/xref/gp-ident-pres.html}{\text{Part 1}} \end{align*}
    Thus \(\varphi(x\inv)\) and \(\varphi(x)\inv\) are both inverses of \(\varphi(x)\) in \(H\text{,}\) and so \(\varphi(x^{-1}) = \varphi(x)\inv\text{,}\) since inverses are unique.

Exercise 1.64.

Let \(\varphi: G \to H\) be a homomorphism of groups, and let \(g\in G\text{.}\) Then \(\varphi(g^n)=\varphi(g)^n\text{.}\)

Proof.

Suppose \(\varphi:G\to H\) and \(\phi:H\to K\) be group homomorphisms, and let \(x,y\in G\text{.}\)
\begin{align*} \phi\circ\varphi(xy) &= \phi\circ\left(\varphi(x)\varphi(y)\right) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}} \text{ applied to } \varphi \\ &= \phi\left(\varphi(x)\varphi(y)\right) && \knowl{./knowl/xref/def-composition.html}{\text{Definition A.20}} \\ &= \phi\left(\varphi(x)\right)\phi\left(\varphi(y)\right) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}} \text{ applied to } \phi\\ &= \left(\phi\circ\varphi(x)\right)\left(\phi\circ\varphi(y)\right) && \knowl{./knowl/xref/def-composition.html}{\text{Definition A.20}} \end{align*}
Thus \(\phi\circ\varphi:G\to K\) is a group homomorphism.

Exercise 1.66. \(\Hom\) Sweet \(\Hom\).

Let \(G\) and \(H\) be abelian groups. Then the set \(\Hom_{Ab}(G,H)\) of all group homomorphisms from \(G\) to \(H\) is itself an abelian group.

Definition 1.67. Kernel.

Let \(\varphi: G \to H\) be a homomorphism of groups. The kernel of \(\varphi\) is the set
\begin{equation*} \ker(\varphi) := \{g\in G \mid \varphi(g)=e_H\}. \end{equation*}

Proof.

  1. \((\Rightarrow)\) Suppose \(\varphi: G \to H\) is injective. By Theorem 1.63 we know \(\varphi(e_G) = e_H\text{.}\) Thus \(e_G\in\ker(\varphi)\) and \(\{e_G\}\sse\ker(\varphi)\text{.}\)
    Let \(x\in\ker(\varphi)\text{.}\) Thus \(\varphi(x)=e_H\text{.}\) As \(\varphi\) is injective, this means \(x=e_G\text{.}\) Hence \(\ker(\varphi)\sse\{e_G\}\text{,}\) making the two sets equal.
    \((\Leftarrow)\) Suppose \(\operatorname{im}(\varphi) = H\text{.}\) Suppose further there exist \(x,y\in G\) such that \(\varphi(x) = \varphi(y)\text{.}\)
    I claim \(x\inv\) is the inverse of \(y\text{.}\) Observe
    \begin{align*} \varphi(x^{-1}y) &= \varphi(x^{-1})\varphi(y) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}}\\ &= \varphi(x)^{-1}\varphi(y) && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}}\\ &= \varphi(x)^{-1}\varphi(x) && \varphi(x) = \varphi(y)\\ &= e_H \end{align*}
    Thus \(x\inv\) is the inverse of \(y\text{,}\) making \(x=y\) as \(e\) inverses are unique.
  2. This proof is identical to that of Exercise A.19; the homomorphism aspect adds nothing of interest.
     3 
    Though one could argue that adding nothing of interest is interesting, causing it to add something of interest.

Exercise 1.69. Homomorphisms and Order.

If \(\varphi: G \to H\) is a homomorphism of groups and \(|g|\) is finte, then \(|\varphi(g)|\big||g|\text{.}\)

Subsection Isomorphism? I Know ’em!

“The test of a first-rate intelligence is the ability to hold two opposed ideas in mind at the same time and still retain the ability to function.”
―F. Scott Fitzgerald

Definition 1.70. Group Ismorphism.

A homomorphism \(\varphi: G \to H\) is called an isomorphism if there exists a homomorphism \(\phi: H \to G\) such that \(\varphi \circ \phi = \operatorname{id}_H\) and \(\phi \circ \varphi = \operatorname{id}_G\text{.}\)
If \(\varphi:G\to H\) is an isomorphism, \(G\) and \(H\) are called isomorphic, written \(G\cong H\text{.}\)
Intuitively, a group isomorphism establishes a one-to-one correspondence between the elements of two groups, such that they have exactly the same algebraic structure. This means that the groups are essentially the same, up to a relabeling of their elements.

Proof.

\((\Rightarrow)\) Suppose \(\varphi\) an isomorphism. Then there exists a homomorphism \(\phi: H \to G\) such that \(\varphi \circ \phi = \operatorname{id}_H\) and \(\phi \circ \varphi = \operatorname{id}_G\text{.}\) Thus \(\varphi\) is invertible. By Theorem A.21 \(\varphi\) is bijective.
(\(\Leftarrow\)) Suppose \(\varphi\) is bijective homomorphism. By Theorem A.21 \(\varphi\) is invertible, meaning there exists some function \(f: H \to G\)
 4 
Notice we use \(f\text{,}\) not \(\phi\text{!}\) We are not able to assume \(f\) is a homomorphism, we must prove this!
such that \(\varphi \circ f = \operatorname{id}_H\) and \(f \circ \varphi = \operatorname{id}_G\text{.}\)
I claim \(f\) is actually a homomorphism: For \(x,y \in H\) we have
\begin{align*} \varphi(f(xy)) &= xy && \varphi\circ f = \operatorname{id}_H\\ &= \varphi(f(x))\varphi(f(y)) && \varphi\circ f = \operatorname{id}_H\\ &= \varphi(f(x)f(y)) && \varphi \text{ is a } \knowl{./knowl/xref/def-ghom.html}{\text{homomorphism}} \end{align*}
Thus \(\varphi(f(xy))=\varphi(f(x)f(y))\) Since \(\varphi\) is bijective it is injective, and thus \(f(xy) = f(x)f(y)\text{,}\) making \(f\) a homomorphism.

Remark 1.72.

Moving forward, we will essentially treat Theorem 1.71 as the de-facto definition of a group isomorphism. Rather than end every proof with a reference to Theorem 1.71, we will instead enshrine its monumental achievements of Theorem 1.71 within this remark. To alleviate some of the corresponding guilt asociated with this decision I am referencing Theorem 1.71 as many times as possible, as Theorem 1.71 is important and Theorem 1.71 should not be forgotten.

Example 1.73. Isomorphism Examples.

  1. The identity map is a group isomomorphism for any group \(G\text{.}\)
  2. \(S_3\cong D_6\text{.}\)
  3. \((\Z,+)\cong(\E,+)\text{,}\) where \((\E,+)\) is as defined in Exercise 1.20.

Exercise 1.74. More Isomorphisms.

  1. Let \(G\) be a group and \(G^{op}\) the of \(G\text{.}\) Then \(G\cong G^{op}\text{.}\)
  2. \(\displaystyle (\R,+)\cong(\R^\times,\cdot)\)
  3. The exponential and natural logarithm maps from Exercise 1.62 are inverses, making them isomorphisms as well.

Proof.

Let \(\varphi:G\to H\) be an isomorphism of groups with inverse \(\varphi\inv\text{.}\)
  1. By Theorem 1.71 \(\varphi\) is a bijective function between \(G\) and \(H\text{.}\) Thus \(|G|=|H|\text{.}\)
  2. Let \(g\in G\) and \(h\in H\text{,}\) and let \(m,n\) denote the orders of \(g,h\text{,}\) respectively. Thus
    \begin{align*} \varphi(g)^m &= \varphi(g^m) && \knowl{./knowl/xref/exe-grp-hom-order-split.html}{\text{Exercise 1.64}}\\ &= \varphi(e_G) && m \text{ is the order of } g\\ &= e_H && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}} \end{align*}
    and
    \begin{align*} \varphi\inv(h)^n &= \varphi\inv(h^n) && \knowl{./knowl/xref/exe-grp-hom-order-split.html}{\text{Exercise 1.64}}\\ &= \varphi\inv(e_H) && n \text{ is the order of } h\\ &= e_G && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}} \end{align*}
    Thus \(\varphi(g)^m=e_H\) and \(\varphi\inv(h)^n=e_G\text{.}\)
    I claim \(m\) and \(n\) are the smallest positive integers where this holds. Suppose by way of contradiction there exist \(m'<m\) and \(n'<n\) such that \(\varphi(g)^{m'}=e_H\) and \(\varphi\inv(h)^{n'}=e_G\text{.}\) But then
    \begin{align*} g^{m'} &= \varphi\inv \varphi(g^{m'}) \\ &= \varphi\inv (e_H) && \varphi(g)^{m'}=e_H\\ &= e_G && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}} \end{align*}
    and
    \begin{align*} h^{n'} &= \varphi\varphi\inv(h^{n'}) \\ &= \varphi(e_G) && \varphi\inv(h)^{n'}=e_G\\ &= e_H && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}} \end{align*}
    As \(m'<m\) and \(n'<n\text{,}\) this contradicts \(|g|=m\) and \(|h|=n\text{.}\)
  3. \((\Rightarrow)\) Suppose \(G\) is abelian, and let \(x,y\in H\text{.}\)
    \begin{align*} xy &= \varphi\varphi\inv(xy)\\ &= \varphi\varphi\inv(x)\varphi\inv(y) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}} \\ &= \varphi\varphi\inv(y)\varphi\inv(x) && G \text{ abelian }, \varphi\inv(x), \varphi\inv(y)\in G, \\ &= \varphi\varphi\inv(yx) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}} \\ &= yx \end{align*}
    Thus every element two elements in \(H\) commute, making \(H\) abelian.
    \((\Leftarrow)\) Suppose \(H\) is abelian, and let \(x,y\in G\text{.}\)
    \begin{align*} xy &= \varphi\varphi\inv(xy)\\ &= \varphi\varphi\inv(x)\varphi\inv(y) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}} \\ &= \varphi\varphi\inv(y)\varphi\inv(x) && H \text{ abelian }, \varphi\inv(x), \varphi\inv(y)\in H, \\ &= \varphi\varphi\inv(yx) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}} \\ &= yx \end{align*}
    Thus every element two elements in \(G\) commute, making \(G\) abelian.

Exercise 1.76. Something’s Missing.

Justify why the following pairs of groups are not isomorphic.
  1. \(Q_8\) and \(D_8\)
  2. \(\Z/6\) and \(S_3\)
  3. \(D_{24}\) and \(S_4\)
  4. \(\Q\) and \(\Q^\times\)

Exercise 1.77. There Can Only Be One....

All trivial groups are (canonically) isomorphic. Hence, we usually speak of the trivial group.

Proof.

  1. The identity map from \(G\to G\) is an isomorphism
  2. Suppose \(\varphi:G\to H\) is an isomorphism. Then there exists a homomorphism \(\phi: H \to G\) such that \(\varphi \circ \phi = \operatorname{id}_H\) and \(\phi \circ \varphi = \operatorname{id}_G\text{.}\) Thus \(\phi\) is an isomorphism from \(H\) to \(G\text{,}\) and so \(H\cong G\text{.}\)
  3. Let \(\varphi: G\to H\) and \(\phi: H\to K\) be isomorphisms. From Theorem 1.65 we see \(\phi\circ\varphi\) is a homomorphism of groups. As both \(\varphi\) and \(\phi\) are injective, \(\phi\circ\varphi\) is injective by Theorem A.21. Similarly, as both \(\varphi\) and \(\phi\) are surjective, \(\phi\circ\varphi\) is surjective by Theorem A.21. Thus \(\phi\circ\varphi\) is a bijective homomorphism from \(G\) to \(K\text{,}\) we see \(G\cong K\) by Theorem 1.71

Remark 1.79.

The quotation marks in the title of this result are important. Equivalence relations as we know them are only defined on sets, but there is no set of all groups, much as there is no set of all sets. This dips into the realm of Russel’s Paradox and higher category theory, which we will steer clear of for the time being.

Subsection Automorphism? I’m Am ’em!

“Self preservation is the first law of nature.”
―Samuel Butler

Definition 1.80. Group Automorphism.

Let \(G\) be a group. An isomorphism \(\varphi:G\to G\) is called an automorphism of \(G\text{.}\)

Example 1.81.

In Example 1.73 we saw that the identity map is a group isomomorphism from a group to itself, making \(\id\) an automorphism.

Definition 1.82. Automorphism Group.

The set of automorphisms of a group \(G\) is called the automorphism group of \(G\) and denoted \(\Aut(G)\text{.}\)
Let’s verify that this group is indeed what it claims to be.
 5 
The more pedantic reader might object to a group claiming anything, as it is an abstract concept. However, the more sassy author would politely tell them to keep it to themselves.

Proof.

Let \(\varphi, \phi, \psi\in\Aut(G)\text{.}\) As function composition is associative, we see \((\varphi\circ\phi)\circ\psi = \varphi\circ(\phi\circ\psi)\text{.}\)
Recall from Example 1.81 that \(\id:G\to G\) is an automorphism. I claim \(\id\) is the identity element of \(\Aut(G)\text{.}\) Let \(\varphi\in\Aut(G)\text{.}\) We will show \(\varphi\circ \id=\varphi=\id\circ \varphi\text{.}\) As all functions share the same domain and co-domain, we need only show that the functions map elements equally, Let \(x \in G\text{,}\) and observe:
\begin{align*} \varphi\circ \id(x) &= \varphi(\id(x)) \\ &= \varphi(x) \\ &= \id(\varphi(x)) \\ &= \id\circ\varphi(x) \end{align*}
Thus \(\id\) is indeed the identity element in \(\Aut(G)\text{.}\)
Let \(\varphi\in\Aut(G)\text{.}\) As \(\varphi\) is an isomorphism, there exists a homomorphism \(\phi: G \to G\) such that \(\varphi \circ \phi = \operatorname{id}\) and \(\phi \circ \varphi = \operatorname{id}\text{.}\) Thus \(\phi\in\Aut(G)\text{,}\) making it the inverse of \(\varphi\text{.}\)

Exercise 1.84. Complex Conjugation Automorphism.

The function \(\phi:\C\to\C\) given by \(\phi(a+bi)=a-bi\) is an automorphism.

Exercise 1.85. \(\Aut(\Z)\).

\(\Aut(\Z)\cong\Z/2\)

Remark 1.86.

Notice that we have already shown that the composition of automorphisms is an automorphism, both through Theorem 1.78 (as automorphisms are isomorphisms) and Proposition 1.83 (as groups are closed under their operation)

Definition 1.87. Inner Automorphism.

Let \(G\) be a group and \(x\in G\text{.}\) The function \(\psi_x:G\to G\) defined by \(\psi_x(a)=xax^{-1}\) is called the inner automorphism of \(G\) induced by \(x\text{.}\) The set of inner automorphisms of \(G\) is denoted \(\Inn(G)=\{\psi_x\mid x\in G\}\text{.}\)
Let’s make sure this is indeed an automorphism to avoid any potential awkwardness down the line.

Proof.

Let \(G\) be a group and \(g\in G\text{.}\) For any \(x,y\in G\text{,}\) observe
\begin{align*} \psi_g(xy) &= g(xy)g^{-1} && \knowl{./knowl/xref/def-inner-automorphisms.html}{\text{Definition 1.87}}\\ &= gx(g^{-1} g)yg^{-1} \\ &= \psi_g(x) \psi_g(y) && \knowl{./knowl/xref/def-inner-automorphisms.html}{\text{Definition 1.87}} \end{align*}
proving that \(\psi_g\) is a homomorhism from \(G\) to itself.
I claim \(\psi_g \circ \psi_h = \psi_{gh}\) holds for any \(g, h \in G\text{.}\) Observe
\begin{align*} \psi_g(\psi_h(x)) &= \psi_g(hxh^{-1}) && \text{apply }\knowl{./knowl/xref/def-inner-automorphisms.html}{\text{Definition 1.87}} \text{ for } \psi_h \\ &= ghxh^{-1} g^{-1} && \text{apply } \knowl{./knowl/xref/def-inner-automorphisms.html}{\text{Definition 1.87}} \text{ for } \psi_g \\ &= ghx(gh)^{-1} && \knowl{./knowl/xref/thm-group-inverses.html}{\text{Theorem 1.21}}\\ &= \psi_{gh}(x) && \text{apply }\knowl{./knowl/xref/def-inner-automorphisms.html}{\text{Definition 1.87}} \text{ for } \psi_{gh} \end{align*}
Additionally, notice that \(\psi_e(x) = exe = x\) for all \(x\in G\text{,}\) and thus \(\psi_e = \operatorname{id}_G\text{.}\)
As \(\psi_g \circ \psi_h = \psi_{gh}\text{,}\) it follows \(\psi_g \circ \psi_{g^{-1}} = \operatorname{id}_G\) and \(\psi_{g^{-1}} \circ \psi_g = \operatorname{id}_G\text{.}\) This proves \(\psi_g\) has a two-sided inverse and hence is an isomorphism.
 6 
In fact, it shows that \((\psi_g)^{-1} = \psi_{g^{-1}}\text{.}\)
Thus \(\psi_g\in \operatorname{Aut}(G)\) for all \(g\in G\text{.}\)

Convention 1.89.

In this text \(\psi_g\) is primarily reserved for conjugation automorphisms, where the subscript identifies which element is doing the conjugating.

Remark 1.90.

There is a notion of “outer automorphisms” as well, though we currently lack the tools to define them rigorously.
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