Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements
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Theorem
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $3$.
Then it is not necessarily the case that:
- $\forall x \in G: x^3 \in H$
Proof
Consider $S_3$, the symmetric group on $3$ letters.
From Subgroups of Symmetric Group on 3 Letters, the subgroups of $S_3$ are:
subsets of $S_3$ which form subgroups of $S_3$ are:
\(\ds \) | \(\) | \(\ds S_3\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set e\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {123}, \tuple {132} }\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {12} }\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {13} }\) | ||||||||||||
\(\ds \) | \(\) | \(\ds \set {e, \tuple {23} }\) |
One such subgroup of $G$ whose index is $3$ is $\set {e, \tuple {12} }$
But $\set {e, \tuple {12} }$ does not contain $\tuple {123}$ or $\tuple {132}$, both of which are of order $3$.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $18 \ \text {(ii)}$