First isomorphism theorem

Section 1.13 First isomorphism theorem

Theorem 1.13.1. Quotient map: universal property.

Let $N$ be a normal subgroup of $G\text{,}$ and let $\pi\colon G\rightarrow G/N$ be the corresponding quotient map. Given any group homomorphism $\phi\colon G\rightarrow H$ satisfying $N\leq \ker \phi\text{,}$ there exists a unique group homomorphism $\overline{\phi}\colon G/N\rightarrow H$ satisfying

\begin{equation} \overline{\phi}\circ \pi=\phi\text{.}\tag{1.13.1} \end{equation}

Moreover, $\overline{\phi}$ is defined as

\begin{equation} \overline{\phi}(gN)=\phi(g)\text{.}\tag{1.13.2} \end{equation}

for all $gN\in G/N\text{.}$

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In short, there is a unique homomorphism $\overline{\phi}$ making the diagram below commutative.

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Figure 1.13.2. Universal property of quotient map
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Proof.

First consider the uniqueness claim. If $\overline{\phi}$ satisfies

\begin{equation*} \overline{\phi}\circ\pi=\phi\text{,} \end{equation*}

then since $gN=\pi(g)$ for all $g\in G\text{,}$ we must have

\begin{equation*} \overline{\phi}(gN)=\overline{\phi}\circ \pi (g)=\phi(g)\text{.} \end{equation*}

Thus there is at most one such homomorphism $\overline{\phi}\text{,}$ in fact we must define $\overline{\phi}(gN)=\phi(g)\text{.}$

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We now show that this formula gives rise to a well-defined map $\overline{\phi}\colon G/N\rightarrow H\text{.}$ We must show that if $gN=g'N\text{,}$ then $\phi(g)=\phi(g')\text{.}$ We have

\begin{align*} gN=g'N \amp \implies g'=gn \text{ for some } n\in N \\ \amp \implies \phi(g')=\phi(gn) \\ \amp \implies \phi(g')=\phi(g)\phi(n)\\ \amp \implies \phi(g')=\phi(g) \amp (N\leq \ker\phi\implies \phi(n)=e)\text{.} \end{align*}

Having shown $\overline{\phi}$ is well defined, it is easy to how that it is a group homomorphism. We have

\begin{align*} \overline{\phi}(gN\, g'N) \amp =\overline{\phi}(gg'N)\\ \amp =\phi(gg') \amp\\ \amp =\phi(g)\phi(g')\\ \amp =\overline{\phi}(gN)\overline{\phi}(g'N)\text{.} \end{align*}

Lastly, $\overline{\phi}$ satisfies (1.13.1) essentially by definition, since

\begin{equation*} \overline{\phi}(\pi(g))=\overline{\phi}(gN)=\phi(g) \end{equation*}

for all $g\in G\text{.}$

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Theorem 1.13.3. First isomorphism theorem.

Let $\phi\colon G\rightarrow H$ be a group homomorphism, and let $\pi\colon G \rightarrow G/\ker\phi$ be the quotient map.

$G/\ker\phi\cong \im\phi\text{.}$
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In more detail, letting $N=\ker\phi \text{,}$ the map $\overline{\phi}\colon G/N\rightarrow \im\phi$ defined as

\begin{equation} \overline{\phi}(gN)=\phi(g)\tag{1.13.3} \end{equation}

for all $gN\in G/\ker\phi$ is an isomorphism, and is the unique homomorphism satisfying

\begin{equation*} \overline{\phi}\circ\pi=\phi\text{.} \end{equation*}

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Proof.

Assume $\phi\colon G\rightarrow H$ is a group homomorphism. Setting $N=\ker\phi\text{,}$ Theorem 1.13.1 implies that the map $\overline{\phi}\colon G/N\rightarrow H$ is a homomorphism. Furthermore, it is easy to see that $\im\overline{\phi}=\im\phi$ and hence that

\begin{equation*} \overline{\phi}\colon G/N\rightarrow \im\phi \end{equation*}

is surjective. We claim that $\overline{\phi}$ is injective and hence that the map above is an isomorphism, showing $G/N\cong \im \phi\text{.}$

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According to Lemma 1.13.4, it suffices to show that $\ker\overline{\phi}=\{e_{G/N}\}\text{.}$ By definition, we have

\begin{align*} gN\in \ker\overline{\phi} \amp \iff \overline{\phi}(gN)=e_H\\ \amp \iff \phi(g)=e_H\\ \amp \iff g\in \ker\phi=N\\ \amp \iff gN=N\\ \amp \iff gN=e_{G/N}\text{,} \end{align*}

as desired.

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Lemma 1.13.4. Kernel and injectivity.

Let $\phi\colon G\rightarrow H$ be a group homomorphism.

Given $h=\phi(g)$ we have $\phi^{-1}(h)=g\ker\phi\text{.}$

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$\phi$ is injective if and only if $\ker\phi=\{e\}\text{.}$
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Proof.

We have

\begin{align*} \phi(g')=h=\phi(g) \amp \iff \phi(g)=\phi(g')\\ \amp \iff (\phi(g))^{-1}\phi(g')=e_H\\ \amp \iff \phi(g^{-1}g')=e_H\\ \amp \iff g^{-1}g'\in \ker\phi\\ \amp \iff g'\in g\,\ker\phi\text{.} \end{align*}

Thus $\phi^{-1}(h)=g\, \ker\phi\text{.}$

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If $\phi$ is injective, then $\phi(g)=e_H$ if and only if $g=e_G\text{,}$ showing that $\ker\phi=\{e_G\}\text{.}$ Inversely, if $\phi$ is not injective, then we can find $g\ne g'$ such that $\phi(g)=\phi(g')=h\text{.}$ But then $\abs{h^{-1}(h)}=\abs{g\ker\phi}> 1\text{.}$ Since $\abs{g\ker\phi}=\abs{\ker\phi}$ it follows that $\abs{\ker\phi}> 1$ and hence $\ker\phi\ne \{e_G\}\text{.}$
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Example 1.13.5. First isomorphism theorem.

Let $N=\{e,(12)(34),(13)(24),(14)(23)\}\text{,}$ a normal subgroup of $S_4\text{.}$ Use the first isomorphism theorem to identify $S_4/N$ with a familiar group.

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Solution.

We convinced ourselves earlier that $S_4/N$ was isomorphic to $D_3\text{,}$ which is isomorphic to $S_3\text{.}$ That argument involved finding appropriate generators of the quotient group that gave us the $D_3$ relations. Let’s see how the first isomorphism theorem can clean things up a bit.

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Consider the set $D=\{D_1,D_2,D_3\}$ where

\begin{align*} D_1 \amp =\{\{1,2\},\{3,4\}\}\\ D_2 \amp =\{\{1,3\},\{2,4\}\}\\ D_3 \amp =\{\{1,4\},\{2,3\}\}\text{.} \end{align*}

In other words, $D$ is the set of all partitions of $\{1,2,3,4\}$ into two disjoint sets of cardinality 2. It is easy to see that $S_4$ acts on $D$ as

\begin{equation*} \sigma\cdot \{\{a,b\},\{c,d\}\}=\{\{\sigma(a),\sigma(b)\},\{\sigma(c),\sigma(d)\}\}\text{.} \end{equation*}

As such we get a homomorphism $\phi\colon S_4\rightarrow S_3$ that associates to each $\sigma\in S_4$ its corresponding permutation of the elements of $D\text{.}$ The first isomorphism theorem tells us that $S_4/\ker\phi\cong \im \phi\text{.}$ We will show that $\ker\phi=N$ and $\im\phi=S_A\text{,}$ from whence it follows that

\begin{equation*} S_4/N\cong S_D\cong S_3\text{,} \end{equation*}

where the last isomorphism follows from the fact that $\abs{D}=3\text{.}$

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Let’s see why $\ker\phi=N\text{.}$ Note that by definition for an element $\sigma$ to lie in $\ker\phi\text{,}$ we need

\begin{align*} \sigma\cdot D_1 \amp =D_1 \amp \sigma\cdot D_2\amp=D_2 \amp \sigma\cdot D_3\amp= D_3\text{.} \end{align*}

An argument similar to the ones we used to compute normalizers and centralizers in $S_4$ shows that given $d=\{\{a,b\},\{c,d\}\}\in D\text{,}$ we have

\begin{equation*} \sigma\cdot d=d \iff \sigma\in \angvec{(a\ c\ b\ d),(a\ b)}=\{e, (ab),(cd),(ac)(bd),(ad)(bc),(acbd),(ab)(cd),(adbc)}\text{.} \end{equation*}

It follows that $\sigma\cdot D_i=D_i$ for all $1\leq i\leq 3$ if and only if $\sigma\in N\text{.}$ Thus $\ker\phi=N\text{.}$

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Having proved that $N=\ker\phi\text{,}$ the first isomorphism tells us that $S_4/N\cong \im\phi\text{,}$ and in particular

\begin{equation*} \abs{\im\phi}=\abs{S_4/N}=\frac{24}{4}=6\text{.} \end{equation*}

Since $\abs{S_A}=6\text{,}$ we conclude that $\im\phi=S_A\text{,}$ and hence that $S_4/N\cong S_A\text{,}$ as claimed.

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Example 1.13.6. First isomorphism theorem: $\SL_n(R)$.

Let $R$ be one of our familiar rings: i.e., $R=\Z,\Q,\R,\C$ or $R=\Z/m\Z$ for some positive integer $m\text{.}$ Fix a positive integer $n$ and define

\begin{equation*} \SL_n(R)=\{A\in \GL_n(R)\mid \det A=1\}\text{.} \end{equation*}

Show that $\SL_n(R)$ is normal in $\GL_n(R)$ and identify the quotient $\GL_n(R)/\SL_n(R)$ with a familiar group.

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Solution.

The determinant map $\det \GL_n(R)\rightarrow R^*$ is a group homomorphism, and it is easy to see that it is surjective: given $r\in R^*$ the diagonal matrix $A$ with $r$ in the first diagonal entry and 1’s elsewhere satisfies $\det A=r\text{.}$

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By definition we have $\SL_n(R)=\ker \det\text{.}$ It follows that $\SL_n(R)\normalin \GL_n(R)$ and we have

\begin{equation*} \GL_n(R)/\SL_n(R)\cong \im \det =R^* \end{equation*}

by the first isomorphism theorem.

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Example 1.13.7. First isomorphism theorem: $\R/\Z$.

Let $S_1=\{z\in \C\mid \abs{z}=1\}\text{,}$ and let $U=\{z\in \C\mid z^n=1 \text{ for some } n\in \Z_{> 0}\}$

Prove that $S^1$ is a subgroup of $\C^*$ and that $U$ is a subgroup of $S_1\text{.}$
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Prove that $\R/\Z\cong S^1\text{.}$
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Prove that $\Q/\Z\cong U\text{.}$
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Solution.

This example relies on the following notions from complex arithmetic.

Given $z=a+bi\in \C\text{,}$ with $a,b\in \R\text{,}$ we define its modulus as $\abs{z}=\sqrt{a^2+b^2}\text{.}$ Geometrically, this is just the length of the vector $v=(a,b)\in \R^2\text{.}$
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Given $z=a+bi\text{,}$ if the point $(a,b)$ has polar coordinates $(r,\theta)\text{,}$ then we have $z=r\cos\theta+r\sin\theta\, i=r(\cos\theta+\sin\theta\, i)\text{.}$ Defining
\begin{equation*} e^{i\theta}=\cos\theta+i\sin\theta\text{,} \end{equation*}
we have $z=re^{i\theta}\text{.}$ If we specify that $r\geq 0\text{,}$ then we have
\begin{equation*} r=\sqrt{a^2+b^2}=\abs{z}\text{.} \end{equation*}

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Given $z=re^{i\theta}$ and $w=se^{i\psi}\text{,}$ we have

\begin{equation*} zw=(rs)e^{i(\theta+\psi)}\text{.} \end{equation*}

In other words, geomtrically speaking, two multiply two complex numbers, we (a) multiply their moduli, and (b) add their angles.

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Using these properties, we see that the map $\phi\colon \R\rightarrow S^1$ defined as $\phi(t)=e^{2\pi i\ t}$ is a group homomorphism with kernel $\Z\text{.}$ We leave the details to discussion section.

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