Exercise 3.2.19 (If $N \unlhd G$ and ${(|N|, |G:N|) = 1}$ then $N$ is the unique subgroup of order $|N|$)

Question

Prove that if $N$ is a normal subgroup of the finite group $G$ and ${(|N|, |G:N|) = 1}$ then $N$ is the unique subgroup of $G$ of order $|N|$.

Richard Ganaye · Accepted Answer

\begin{proof} We suppose that $N \unlhd G$ and ${|N| \wedge |G:N| = 1}$ .

Let $H$ be a subgroup of $G$ of order $|H| = |N|$. Then
$$|H| \wedge |G:N| =1.$$
By Exercise 18, $H \leq N$.
Since $|H| = |N|$, this gives
$$H = N.$$
So $N$ is the unique subgroup of $G$ of order $|N|$.
\end{proof}

Exercise 3.2.19 (If $N \unlhd G$ and ${(|N|, |G:N|) = 1}$ then $N$ is the unique subgroup of order $|N|$)

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Exercise 3.2.19 (If $N \unlhd G$ and ${(|N|, |G:N|) = 1}$ then $N$ is the unique subgroup of order $|N|$)

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