Exercise 3.3.3 (Application of the Second Isomorphism Theorem)

Question

Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K \leq G$ either
\begin{enumerate}
\item[{\bf(i)}] $K \leq H$ or
\item[{\bf(ii)}] $G = HK$ and $|K : K \cap H | = p$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\begin{proof} Suppose that $H \unlhd G$ and that $|G:H| = p$ is prime. Moreover, we suppose that (i) is not true, so that $K 
ot \subseteq H$. Then $K 
e HK \cap H$, thus
$$|K : K \cap H | >1.$$

\bigskip
\begin{center}
\begin{tikzpicture}
    
ode (n0) at (3,0) {$K \cap H$};
    
ode (n1) at (1.5,1.5) {$H$};
    
ode (n4) at (4.5,1) {$K$};
    
ode (n5) at (3,2.5) {$HK$};
    
ode (n6) at (3, 3.5) {$G$};
     \draw (n0) edge (n1)  edge (n4)  ;
     \draw (n5) edge (n1)  edge (n4) edge (n6) ;
   \end{tikzpicture}
\end{center}

By the second isomorphism theorem, since $H \unlhd G$, then  $HK \leq G,\ H \unlhd HK,\ K \cap H \unlhd K$ and
$$HK/ H \simeq K/K \cap H.$$
Therefore $|HK : H| >1$, where $|G : H | = p< \infty$, so
\begin{align}
p = |G : H | = |G : HK|\cdot  |HK : H|.
\end{align}
This shows that $| K : K \cap H| = |HK: H|$ divides $p$, where $|K : K \cap H | >1$. Since $p$ is prime,
$$|K : K \cap H | = p.$$
Then $|HK : H| = p$, thus (1) shows that $|G : HK| = 1$, so
$$G = HK.$$
In conclusion, if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K \leq G$ either
\begin{enumerate}
\item[{\bf(i)}] $K \leq H$ or
\item[{\bf(ii)}] $G = HK$ and $|K : K \cap H | = p$.
\end{enumerate}


\end{proof}

Exercise 3.3.3 (Application of the Second Isomorphism Theorem)

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Exercise 3.3.3 (Application of the Second Isomorphism Theorem)

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