Question

Let \( G \) be a group of order 20 in which the conjugacy classes have sizes 1, 4, 5, 5, 5. Then which of the following is/are TRUE?

• A. \( G \) contains a normal subgroup of order 5
• B. \( G \) contains a non-normal subgroup of order 5
• C. \( G \) contains a subgroup of order 10
• D. \( G \) contains a normal subgroup of order 4

Hint:

Sylow’s theorems help us determine the number and normality of Sylow subgroups based on the group’s order.

Answer 1

The Correct Option is A, C

Step 1: Applying Sylow's theorems.
By Sylow's theorems, since the order of \( G \) is 20, the number of Sylow \( 5 \)-subgroups \( n_5 \) must divide 20 and satisfy \( n_5 \equiv 1 \mod 5 \). The possible values of \( n_5 \) are 1 and 4. If \( n_5 = 1 \), there is a unique Sylow \( 5 \)-subgroup, and this subgroup is normal. Thus, \( G \) contains a normal subgroup of order 5.
Step 2: Checking the other options.
- (B) A non-normal subgroup of order 5 does not necessarily exist, as the Sylow \( 5 \)-subgroup can be unique and normal. - (C) A subgroup of order 10 exists because \( 20 = 2 \times 10 \), and such a subgroup is normal. - (D) A normal subgroup of order 4 does not necessarily exist because there is no direct guarantee from Sylow's theorems for the existence of such a subgroup.
Step 3: Conclusion.
Thus, the correct answers are (A) and (C).