Exercise 2.1.5 ($G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G|2$)

Question

Prove that $G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G|>2$.

Richard Ganaye · Accepted Answer

\begin{proof} 
 Assume, for the sake of contradiction, that $|H| = n-1$.

By Lagrange's Theorem (see Ex. 1.7.22), 
$$ n- 1 = |H| 	ext{ divides } |G| = n.$$
Then $n-1 \mid n$, therefore $n-1 \mid n - (n-1) = 1$, where $n-1 = |H| \geq 0$, thus $n-1 \leq 1$, so $n \leq 2$. This contradicts the hypothesis.

A group $G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G|>2$.
\end{proof}

Exercise 2.1.5 ($G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G|>2$)

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Exercise 2.1.5 ($G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G|>2$)

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