Question

Which of the following statements are false?

• A. $S_3$ is a subgroup of $S_4$
• B. $\mathbb{Z}_3$ is a subgroup of $S_4$
• C. $S_3$ is a quotient group of $S_4$
• D. $\mathbb{Z}_6$ is a quotient group of $S_4$

Hint:

Always check quotient groups using normal subgroups. $S_4$ has a normal subgroup $V_4$, and $S_4/V_4 \cong S_3$, not $\mathbb{Z}_6$.

Answer 1

The Correct Option is D

Step 1: Compute order of $S_4$.
\[ |S_4| = 4! = 24 \]
Step 2: Analyze each statement.
(A) $S_3$ is a subgroup of $S_4$.
By fixing one element in $\{1,2,3,4\}$ and permuting the remaining three, we obtain a subgroup isomorphic to $S_3$.
Hence, (A) is true.
(B) $\mathbb{Z}_3$ is a subgroup of $S_4$.
$S_4$ contains 3-cycles such as $(123)$.
A 3-cycle generates a cyclic subgroup of order 3.
Hence, (B) is true.
(C) $S_3$ is a quotient group of $S_4$.
For $S_4/N \cong S_3$, we require a normal subgroup $N$ of order: \[ \frac{24}{6} = 4 \] However, $S_4$ does not have a normal subgroup of order 4 that produces quotient isomorphic to $S_3$.
Hence, this statement is not valid in this context.
(D) $\mathbb{Z}_6$ is a quotient group of $S_4$.
For this to happen, we need a normal subgroup of order: \[ \frac{24}{6} = 4 \] Although $S_4$ has the Klein four subgroup $V_4$ of order 4, the quotient group $S_4/V_4$ is isomorphic to $S_3$, not $\mathbb{Z}_6$.
Since $\mathbb{Z}_6$ is cyclic but $S_4/V_4$ is non-cyclic, the statement claiming $\mathbb{Z}_6$ as a quotient is false.
Step 3: Conclusion.
The false statement is (D).