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Abstract algebra A Kursobjekt for Math 331

Aaron Greicius

Section 1.14 Isomorphism theorems: second and third

Proof.

  1. Consider the map of sets
    \begin{equation*} f\colon H\times K\rightarrow HK \end{equation*}
    defined by \(\phi(h,k)=hk\text{.}\) This map is surjective by definition of \(HK\text{.}\) Given any \(g=hk\in HK\text{,}\) we have
    \begin{align*} f\left((h',k')\right)=hk \amp \iff h'k'=hk \\ \amp \iff h^{-1}h'=k(k')^{-1}\text{.} \end{align*}
    Let \(t=h^{-1}h'=k(k')^{-1}\text{,}\) and notice that \(t\in H\cap K\text{.}\) It follows easily that for all \(g=hk\in HK\text{,}\) we have
    \begin{equation*} f^{-1}(g)=\{(ht,t^{-1}k)\mid t\in H\cap K\}\text{.} \end{equation*}
    Furthermore, the map
    \begin{align*} H\cap K \amp \rightarrow f^{-1}(hk)\\ t \amp\longmapsto (ht,t^{-1}k) \end{align*}
    can be shown to be a bijection.
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    Now, if \(H\) and \(K\) are finite, then \(HK\) and \(H\cap K\) are finite, and we have
    \begin{align*} \abs{H}\times \abs{K} \amp =\abs{H\times K}\\ \amp =\abs{\bigcup_{g\in HK}f^{-1}(g)}\\ \amp =\sum_{g\in HK}\abs{f^{-1}(g)}\\ \amp = \sum_{g\in HK}\abs{H\cap K}\\ \amp =\abs{HK}\cdot \abs{H\cap K}\text{,} \end{align*}
    as desired.
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    Moreover, if either \(H\) or \(K\) is infinite, then so is \(HK\text{,}\) in which case the equality \(\abs{HK}\abs{H\cap K}=\abs{H}\abs{K}\) still holds.
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  2. Left as an exercise.
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  3. Left as an exercise.
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Proof.

First we prove \(K\normalin HK\text{.}\) Given \(k\in K\) and \(g=hk'\in HK\text{,}\) we have
\begin{align*} gkg^{-1} \amp = hk'k(k')^{-1}h^{-1}\\ \amp =hk''h^{-1} \text{ for some } k''\in K \amp (K \text{ subgroup}) \\ \amp =k''' \text{ for some } k'''\in K \amp \amp (H\subseteq N_G(K)) \text{.} \end{align*}
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To prove the stated isomorphism, we start with the quotient map
\begin{equation*} \pi\colon HK\rightarrow HK/K \end{equation*}
and let \(\psi=\pi\vert_H\text{,}\) its restriction to the subgroup \(H\leq HK\text{.}\) We have
\begin{align*} \ker \psi \amp = H\cap \ker\pi \amp (\psi=\pi\vert_H)\\ \amp =H\cap K \amp (\ker\pi=K)\text{.} \end{align*}
By the first isomorphism theorem, we have \(H/H\cap K\cong \im\psi\text{.}\) It remains to show that \(\im\psi=\im\pi=HK/K\text{.}\) Given any \(gK\in HK/K\text{,}\) we have \(g=hk\) for some \(h\in H\) and \(k\in K\text{,}\) in which case
\begin{equation*} gK=hkK=hK\text{.} \end{equation*}
It follows that
\begin{equation*} gK=\pi(h)=\psi(h)\text{.} \end{equation*}
This proves \(\psi\) is surjective, and hence that \(\im\psi=HK/K\text{.}\)
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Remark 1.14.3. Second isomorphism theorem.

The lattice diagram below nicely summarizes the second isomorphism theorem. For this reason, this isomorphism theorem is sometimes called the diamond isomorphism theorem.
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Lattice diagram of second isomorhphism theorem
Figure 1.14.4. Lattice diagram of second isomorhphism theorem
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Theorem 1.14.2 further tells us that \(H/H\cap K\cong HK/K\text{.}\)
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Proof.

Notation 1.14.6. Images under quotients.

Let \(N\) be a normal subgroup of \(G\text{.}\) Given any subset \(A\subseteq G\text{,}\) we denote by \(A/N\) its image under the quotient map \(\pi\colon G\rightarrow G/N\text{.}\)
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