Math Problem Statement

Show that the group An has no subgroups of order n!/4

Solution

To show that the alternating group AnA_n (the group of even permutations of the symmetric group SnS_n) has no subgroups of order n!4\frac{n!}{4}, we rely on fundamental properties of group theory and permutations. Here's the reasoning:


Step 1: Basic properties of AnA_n

  1. Order of AnA_n: The order of AnA_n is: An=n!2,|A_n| = \frac{n!}{2}, because AnA_n is the index-2 subgroup of SnS_n, consisting of all even permutations.

  2. Lagrange's theorem: For a subgroup HH of a finite group GG, the order of HH must divide the order of GG. Hence, any subgroup of AnA_n must have an order that divides n!2\frac{n!}{2}.


Step 2: Investigate subgroups of order n!4\frac{n!}{4}

If AnA_n has a subgroup HH of order n!4\frac{n!}{4}, then: H=n!4.|H| = \frac{n!}{4}. However, H|H| must divide An=n!2|A_n| = \frac{n!}{2}. This leads to: n!4n!2.\frac{n!}{4} \mid \frac{n!}{2}.

This is not possible because:

  • The division n!2÷n!4=2\frac{n!}{2} \div \frac{n!}{4} = 2 is not an integer (it is not a divisor relationship in the subgroup sense).
  • Thus, the existence of such a subgroup contradicts Lagrange’s theorem.

Step 3: Subgroup implications

  1. If n5n \geq 5, AnA_n is simple (it has no normal subgroups other than {e}\{e\} and itself). For AnA_n, this excludes the possibility of any proper subgroup of order n!4\frac{n!}{4} since such a subgroup would imply non-trivial factorizations of AnA_n.

  2. If n<5n < 5:

    • For n=3n = 3, A3A_3 has order 33, and 3!4=1.5\frac{3!}{4} = 1.5, which is not an integer.
    • For n=4n = 4, A4A_4 has order 1212, and 4!4=6\frac{4!}{4} = 6, but no subgroup of order 6 exists in A4A_4.

Conclusion:

No subgroup of AnA_n exists with order n!4\frac{n!}{4}. This follows from Lagrange’s theorem and, for n5n \geq 5, the simplicity of AnA_n.

Would you like further details or clarifications?


Relative Questions:

  1. Why is AnA_n simple for n5n \geq 5?
  2. What is the significance of Lagrange's theorem in subgroup analysis?
  3. How does the simplicity of AnA_n relate to its subgroups?
  4. Why does the order of HH dividing G|G| matter in group theory?
  5. Can you prove simplicity for any AnA_n with n5n \geq 5?

Tip: Always check divisors of the group order to analyze potential subgroup sizes.

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Math Problem Analysis

Mathematical Concepts

Group Theory
Permutations
Subgroups
Lagrange's Theorem
Simplicity of Groups

Formulas

|A_n| = n!/2
|H| = n!/4
|A_n| = n!/2

Theorems

Lagrange's Theorem
Simplicity of A_n for n >= 5

Suitable Grade Level

Undergraduate (Advanced Group Theory)