Step 1: Understanding the Topic
This question asks to identify a false statement about the structure of the symmetric group $S_4$ (the group of all permutations of 4 elements). We need to analyze its subgroups and quotient groups. The order of $S_4$ is $|S_4| = 4! = 24$.
Step 2: Key Approach - Analyzing Each Statement
We will evaluate the truthfulness of each statement using fundamental concepts of group theory, including Lagrange's theorem, normal subgroups, and quotient groups.
Step 3: Detailed Explanation
(A) $S_3$ is a subgroup of $S_4$:
We can consider the set of all permutations in $S_4$ that fix one element (e.g., the element '4'). These permutations only act on the set $\{1, 2, 3\}$, and this set of permutations forms a group that is isomorphic to $S_3$. Thus, $S_4$ contains a subgroup isomorphic to $S_3$. Statement (A) is true.
(B) $\mathbb{Z_3$ is a subgroup of $S_4$:}
$S_4$ contains elements of order 3, such as the 3-cycle permutation $(1\ 2\ 3)$. This element generates a cyclic subgroup of order 3: $\{e, (1\ 2\ 3), (1\ 3\ 2)\}$. This subgroup is isomorphic to $\mathbb{Z}_3$. Statement (B) is true.
(C) $S_3$ is a quotient group of $S_4$:
A quotient group $S_4/N$ is isomorphic to $S_3$ if there exists a normal subgroup $N$ of $S_4$ such that $|S_4/N| = |S_3|=6$. This would require $|N| = |S_4|/6 = 24/6 = 4$. The group $S_4$ has a unique normal subgroup of order 4, the Klein four-group $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$. The quotient group $S_4/V_4$ has order 6 and is indeed isomorphic to $S_3$. Statement (C) is true.
(D) $\mathbb{Z_6$ is a quotient group of $S_4$:}
For $S_4/N$ to be isomorphic to $\mathbb{Z}_6$, we would again need a normal subgroup $N$ of order 4, which must be $V_4$. However, as established above, $S_4/V_4 \cong S_3$. The group $S_3$ is non-abelian, while the group $\mathbb{Z}_6$ is abelian (and cyclic). Since non-abelian and abelian groups cannot be isomorphic, $S_4/V_4$ is not isomorphic to $\mathbb{Z}_6$. Statement (D) is false.
Step 4: Final Answer
The only false statement among the options is (D).