Question

Which of the following statements is/are true?
(I) The maximum possible order of an element in the group \(S_5\) (with composition \(\circ\)) is 6.
(II) If \(f:(S_3,\circ)\to(\mathbb{Z}_6,+_6)\) is a group homomorphism, then the order of \(f(S_3)\) is 1, 2 or 3.

• A. Both (I) and (II)
• B. Only (I)
• C. Neither (I) nor (II)
• D. Only (II)

Hint:

Find the cycle type in \(S_5\) with the largest lcm, and note any homomorphism from a non-abelian group to an abelian group must kill the commutator subgroup.

Answer 1

The Correct Option is B

Step 1: Check statement (I).
Every element of \(S_5\) is a product of disjoint cycles, and its order equals the lcm of the cycle lengths. The possible cycle types (partitions of 5) and their orders are:\[5 \to 5,\quad 4+1\to 4,\quad 3+2\to 6,\quad 3+1+1\to 3,\quad 2+2+1\to 2,\quad 2+1+1+1\to 2,\quad 1+1+1+1+1\to 1\]The largest value here is \(\operatorname{lcm}(3,2)=6\), attained by a permutation that is a 3-cycle times a disjoint 2-cycle, for example \((123)(45)\). So the maximum possible order of an element of \(S_5\) is 6, and statement (I) is true.

Step 2: Check statement (II).
\(S_3\) is non-abelian while \(\mathbb{Z}_6\) is abelian, so for any homomorphism \(f:S_3\to\mathbb{Z}_6\), the kernel must be a normal subgroup of \(S_3\) containing the commutator subgroup \([S_3,S_3]=A_3\) (the quotient \(S_3/\ker f\) is abelian). The only normal subgroups of \(S_3\) are \(\{e\}\), \(A_3\), and \(S_3\), and only \(A_3\) and \(S_3\) contain \(A_3\).

Step 3: Compute the possible image orders.
By the first isomorphism theorem, \(|f(S_3)| = |S_3|/|\ker f|\).\[\ker f = A_3 \Rightarrow |f(S_3)| = 6/3 = 2, \qquad \ker f = S_3 \Rightarrow |f(S_3)| = 6/6 = 1\]So the order of \(f(S_3)\) is always 1 or 2, never 3. Statement (II) is false.

Step 4: Conclusion.
Only statement (I) is true.\[\boxed{\text{Only (I)}}\]