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

Proof by Counterexample:

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 $3$.

$\blacksquare$


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