Exercise 3.2.18 (If $(|H|,|G : N |) = 1$ then $H \leq N$)

Question

Let $G$ be a finite group, let $H$ be a subgroup of $G$ and let $N \unlhd G$. Prove that if $|H|$ and $|G : N |$ are relatively prime then $H \leq N$.

Richard Ganaye · Accepted Answer

\begin{proof} 

Since $N \unlhd G$, then $NH$ is a subgroup of $G$. By the Second Isomorphism Theorem (Theorem 18 p. 97), $N \unlhd NH$, $H \cap N \unlhd H$, and 
\begin{align}
NH/N \simeq H/H\cap N.
\end{align}

\bigskip
\begin{center}
\begin{tikzpicture}
    
ode (n) at (3,-1) {$\{1\}$};
    
ode (n0) at (3,0) {$H\cap N$};
    
ode (n1) at (1.5,1.5) {$N$};
    
ode (n4) at (4.5,1) {$H$};
    
ode (n5) at (3,2.5) {$NH$};
    
ode (n6) at (3, 3.5) {$G$};
    \draw (n) edge (n0);
     \draw (n0) edge (n1)  edge (n4)  ;
     \draw (n5) edge (n1)  edge (n4) edge (n6) ;
   \end{tikzpicture}
\end{center}

Put $n = |H : H \cap N| = |H/H\cap N|$. Then $n$ divides $|H|$.

Moreover the isomorphism (1) shows that $n = |NH : N|$.

 Since $|G:N| = |G : NH| \cdot |NH : N| = n |G:NH|$, we know that $n$ divides $|G/N|$. So
$$n \mid |H| \qquad 	ext{and} \qquad n \mid |G:N|.$$
Since $|H|$ and $|G : N |$ are relatively prime, this shows that $n \mid 1$, thus $n = 1$.

So $|H /H \cap N| = 1$, therefore $H = H \cap N$. Hence 
$$H \leq N.$$
\end{proof}

Exercise 3.2.18 (If $(|H|,|G : N |) = 1$ then $H \leq N$)

Answers

Comments

Exercise 3.2.18 (If $(|H|,|G : N |) = 1$ then $H \leq N$)

Answers

Comments

Add answer