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% 21/01/2023 12PM

\begin{definition*}
    An isomorphism of groups is a group
    homomorphism that is also a bijection. We say
    $G$ and $H$ are isomorphic (written $G \cong
    H$) if $\exists$ isomorphism $\phi \colon G
    \to H$. (Exercise: Check $\phi^{-1} \colon H
    \to G$ is a group homomorphism).
\end{definition*}

\begin{theorem*}[First Isomorphism Theorem]
    Let $\phi \colon G \to H$ be a group
    homomorphism. Then $\Ker(\phi) \normalsub G$
    and $G / \Ker(\phi) \cong \Im(\phi)$.
\end{theorem*}

\begin{proof}
    Let $K = \Ker(\phi)$. Already checked $K$ is
    normal. Define $\Phi \colon G / K \to
    \Im(\phi)$, $gK \mapsto \phi(g)$. Check $\Phi$
    is well-defined and injective:
    \begin{align*}
        g_1 K = g_2 K
        &\iff g_2^{-1} g_1 \in K \\
        &\iff \phi(g_2^{-1} g_1) = 1 \\
        &\iff \phi(g_2) = \phi(g_1)
    \end{align*}
    Check $\Phi$ is a group homomorphism:
    \begin{align*}
        \Phi(g_1 K g_2 K)
        &= \Phi(g_1 g_2 K) \\
        &= \phi(g_1 g_2) \\
        &= \phi(g_1) \phi(g_2) \\
        &= \Phi(g_1 K) \Phi(g_2 K)
    \end{align*}
    $\Phi$ is surjective: Let $x \in \Im(\phi)$,
    say $\phi(g) = x$ for some $g \in G$. Then $x
    = \Phi(gK) \in \Im(\Phi)$.
\end{proof}

\begin{example*}
    $\phi \colon \CC \to \CC^\times = \{x \in \CC
    \mid x \neq 0\}$, $z \mapsto e^z$. Since $e^{z
    + w} = e^z e^w$, this is a group homomorphism
    from $(\CC, +)$ to $(\CC^\times, x)$.
    \[ \Ker(\phi) = \{z \in \CC \mid e^z = 1\} =
    2\pi i \ZZ \]
    \[ \Im(\phi) = \CC^\times \qquad \text{(by
    existence of log)} \]
    therefore $\CC / 2\pi i \ZZ \cong \CC^\times$.
\end{example*}

\begin{flashcard}[second-iso-thm]
\begin{theorem*}[Second Isomorphism Theorem]
    Let $H \cloze{\le} G$, and $K
    \cloze{\normalsub} G$. Then $\cloze{HK
    = \{hk \colon h \in H, k \in K\} \le G}$
    and $\cloze{H \cap K \normalsub H}$. Moreover
    \[ \cloze{HK / K \cong H / H \cap K} \]
\end{theorem*}
\end{flashcard}

\begin{flashcard}[second-iso-thm-proof]
\prompt{Proof of second isomorphism theorem? \\}
\begin{proof}
    \cloze{
    Let $h_1 k_1, h_2 k_2 \in HK$ (so $h_1, h_2
    \in H$, $k_1, k_2 \in K$). Then
    \[ h_1 k_1 (h_2 k_2)^{-1}
    = \ub{h_1 h_2^{-1}}_{\in H}
    \ub{h_2 k_1 k_2^{-1} h_2^{-1}}_{\in K}
    \in HK \]
    Thus $HK \le G$ (by Remark from last lecture).

    \myskip
    Let $\phi \colon H \to G / K$, $h \mapsto h
    \to hK$. This is the composite of $H \injto G$
    and the quotient map $G \to G / K$, hence
    $\phi$ is a group homomorphism.
    \[ \Ker(\phi) = \{h \in H \mid hK = k\} = H
    \cap K \normalsub H \]
    \[ \Im(\phi) = \{hK \mid h \in H\} = HK / K \]
    First isomophism theorem implies $H / H \cap K
    \cong HK / K$.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    Suppose $K \normalsub G$. There is a bijection
    \[ \{\text{subgroups of $G / K$}\}
    \leftrightarrow \{\text{subgroups of $G$
    containing $H$}\} \]
    defined by $X \mapsto \{g \in G : gK \in X\}$
    and $H / K \mapsfrom H$. This restricts to a
    bijection
    \[ \{\text{normal subgroups of $G / K$}\}
    \leftrightarrow \{\text{normal subgroups of
    $G$ containing $K$}\} \]
\end{remark*}

\begin{flashcard}[third-iso-thm]
\begin{theorem}[Third Isomorphism Theorem]
    Let $K \normalsub H \normalsub G$ be normal subgroups of
    $G$. Then
    \[ \cloze{\frac{G / K}{H / K}} \cong \cloze{G
    / H} \]
\end{theorem}
\end{flashcard}

\begin{flashcard}[third-iso-thm-proof]
\prompt{Proof of third isomorphism theorem? \\}
\begin{proof}
    \cloze{
    Let $\phi : G / K \to G / H$, $gK \mapsto gH$.
    If $g_1 K = g_2 K$, then $g_2^{-1} g_1 \in K
    \le H \implies g_1H = g_2 H$. Thus $\phi$
    well-defined. $\phi$ is surjective group
    homomorphism with kernel $H / K$.
}
\end{proof}
\end{flashcard}

\myskip
If $K \normalsub G$ then studying the groups $K$
and $G / K$ gives some information about $G$. This
is not always available.

\begin{definition*}
    A group $G$ is \emph{simple} if $1$ and $G$
    are its only normal subgroups, except if $G$
    is the trivial group (convention).
\end{definition*}

\begin{lemma}
    Let $G$ be an abelian group. $G$ is simple if
    and only if $G \cong C_p$ for some prime $p$.
\end{lemma}

\begin{proof}
    \begin{enumerate}
        \item[$\Leftarrow$] Let $H \le C_p$. By
            Lagrange's Theorem, $|H| ~\big|~
            |C_p| = p$. So $|H|$ is 1 or $p$, i.e.
            $H = \{1\}$ or $H = C_p$. Thus $C_p$
            is simple.
        \item[$\Rightarrow$] Let $1 \neq g \in G$.
            $G$ contains the subgroup $\langle g
            \rangle = \langle \ldots, g^{-2},
            g^{-1}, 1, g, g^2, \ldots \rangle$ -
            normal in $G$ since $G$ is abelian.
            Since $G$ is simple, $\langle g
            \rangle = G$. If $G$ is infinite, $G
            \cong (\ZZ, +)$ and $2\ZZ \le \ZZ$,
            contradiction. Otherwise $G \cong C_n$
            for some $n$, so $g^n = 1$. If $m \mid
            n$, then $g^{n / m}$ generates a
            subgroup of order $m$ inside $G$. So
            $G$ is simple $\implies$ only factors
            of $n$ are 1 and $n$, so $n$ is prime.
    \end{enumerate}
\end{proof}

\begin{lemma}
    If $G$ is a finite group, then $G$ has a
    composition series
    \[ 1 = G_0 \normalsub G_1 \normalsub \cdots
    \normalsub G_{m - 1} \normalsub G_m = G \]
    with each quotient $G_i / G_{i - 1}$ simple.
\end{lemma}

\begin{warning*}
    $G_i$ need not be normal in $G$; we only
    necessarily know that $G_i$ is normal in $G_{i
    + 1}$.
\end{warning*}

\begin{proof}
    Induct on $|G|$. Case $|G| = 1$. If $|G| > 1$,
    let $G_{m - 1}$ be a normal subgroup of
    largest possible order $\neq |G|$. By earlier
    Remark, $G / G_{m - 1}$ must be simple. Apply
    induction to $G_{m - 1}$.
\end{proof}