The question asks about the order of a proper subgroup of the alternating group $A_4$. The multiple-choice options provided are:

• (a) 4
• (b) 5
• (c) 6
• (d) 8

Solution:

The group $A_4$ is the alternating group on 4 elements, consisting of all even permutations of the symmetric group $S_4$. The order of $A_4$ is given by:

$|A_4| = \frac{|S_4|}{2} = \frac{24}{2} = 12.$

A proper subgroup of $A_4$ must have an order that divides $12$ (by Lagrange's theorem). The divisors of $12$ are:

$1, 2, 3, 4, 6, 12.$

Since we are considering proper subgroups, we exclude $12$, which is the order of $A_4$ itself. The possible orders of proper subgroups are:

$1, 2, 3, 4, 6.$

Among these, we can identify:

1. Subgroups of order $4$: These correspond to Klein four-groups (e.g., $V_4$).
2. Subgroups of order $6$: These are isomorphic to cyclic groups or symmetric groups $S_3$.

Correct Answer: (c) 6

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Related Questions:

1. What is the structure of the group $A_4$?
2. Explain the significance of Lagrange's theorem in group theory.
3. What are the elements of $A_4$?
4. Why is $V_4$ (Klein four-group) an important subgroup of $A_4$?
5. Can $A_4$ have a subgroup of order $5$? Why or why not?

Tip:

When solving group theory problems, always check divisors of the group's order using Lagrange's theorem for valid subgroup sizes.