H is a Normal Subgroup of G IFF H is the Kernel of a Homo. on G

H is a Normal Subgroup of G IFF H is the Kernel of a Homomorphism on G

Lemma 1: Let $G$ and $H$ be groups. If $\varphi : G \to H$ is a group homomorphism then for all $h \in H$, we have that if $g \in \varphi^{-1}(h)$ then:
a) For all $g' \in \varphi^{-1}(h)$ we have that $\varphi^{-1}(h) = g' \ker (\varphi)$.
b) For all $g' \in \varphi^{-1}(h)$ we have that $\varphi^{-1}(h) = \ker (\varphi) g'$
  • Proof of a) Let $h \in H$. Suppose that $g \in \varphi^{-1}(h)$. Then $\varphi(g) = h$.
  • Let $g' \in \varphi^{-1}(h)$. Then $\varphi(g') = h$.
  • So let $g_1 \in \varphi^{-1}(h)$. Then $\varphi(g_1) = h$. Let $t = g'^{-1}g_1$. Then since $\varphi(g') = h$ and since $\varphi(g_1) = h$ we have that:
\begin{align} \quad \varphi(g'^{-1}g_1) = \varphi(g'^{-1}) \varphi(g_1) = \varphi(g')^{-1} \varphi(g_1) = h^{-1}h = 0 \end{align}
  • So $t \in \ker (\varphi)$. Furthermore, we have that $g_1 = g'(g'^{-1}g_1) = g't \in g' \ker (\varphi)$. Thereore:
\begin{align} \quad \varphi^{-1}(h) \subseteq g' \ker (\varphi) \quad (*) \end{align}
  • Now let $g_1 \in g' \ker (\varphi)$. Then $g_1 = g't$ for some $t \in \ker (\varphi)$. Therefore:
\begin{align} \quad \varphi(g_1) = \varphi(g't) = \varphi(g') \varphi(t) = \varphi(g') 1 = h \end{align}
  • So $g_1 \in \varphi^{-1}(h)$. Thus:
\begin{align} \quad \varphi^{-1}(h) \supseteq g' \ker (\varphi) \quad (**) \end{align}
  • From $(*)$ and $(**)$ we conclude that $\varphi^{-1}(h) = g' \ker (\varphi)$. $\blacksquare$
Theorem 2: Let $G$ be a group and let $H$ be a subgroup of $G$. Then $H$ is a normal subgroup of $G$ if and only if there exists a homomorphism $\varphi$ on $G$ such that $H = \ker (\varphi)$.
  • Proof: $\Rightarrow$ Suppose that $H$ is a normal subgroup of $G$. Then the set of left cosets $G/H = \{ gH : g \in G \}$ with the operation defined for all $g_1H, g_2H \in G/H$ by:
\begin{align} \quad (g_1H)(g_2H) = (g_1 \cdot g_2H) \end{align}
  • is well-defined and forms a group whose identity is $1H$ and where for each $g \in G$, the inverse of $gH$ in $G/H$ is $g^{-1}H$.
  • Let $\varphi : G \to G/H$ be the natural homomorphism defined for all $g \in G$ by:
\begin{align} \quad \varphi(g) = gH \end{align}
  • Indeed, $\varphi$ is a homomorphism since for all $g_1, g_2 \in G$ we have that:
\begin{align} \quad \varphi(g_1 \cdot g_2) = (g_1 \cdot g_2)H = (g_1H)(g_2H) = \varphi(g_1) \varphi(g_2) \end{align}
  • We claim that $H = \ker (\varphi)$.
  • Let $h \in H$. Then we have that $\varphi(h) = hH = H = 1H$, and so indeed, $h \in \ker (\varphi)$, so $H \subseteq \ker (\varphi)$. Now let $g \in \ker (\varphi)$. Then $\varphi(g) = 1H$, that is, $gH = 1H = H$. Since $H$ is a subgroup, $1 \in H$. So the equality $gH = H$ tells us that $g = g \cdot 1 = g \in gH = H$. Thus $\ker (\varphi) \subseteq H$. Hence we conclude that $H = \ker (\varphi)$.
  • $\Leftarrow$ Suppose that there exists a homomorphism $\varphi : G \to S$ such that $H = \ker (\varphi)$. By Lemma 1 we have that for all $g' \in \bigcup_{s \in S} \varphi^{-1}(s)$ that $g' \ker(\varphi) = \ker (\varphi)g'$. But this just says that for all $g' \in \varphi^{-1}(S) = G$ we have that $g'H = Hg'$. So $H$ is a normal subgroup of $G$. $\blacksquare$
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