Find all the Sylow 2-subgroups of S_3 and show that they are conjugate.
The correct answer and explanation is:
To find all the Sylow 2-subgroups of the symmetric group S3S_3 and show that they are conjugate, we proceed as follows:
Step 1: Order of the Group
The symmetric group S3S_3 has order ∣S3∣=3!=6|S_3| = 3! = 6
Step 2: Apply Sylow’s Theorems
We want to find the Sylow 2-subgroups. The highest power of 2 dividing 6 is 21=22^1 = 2, so any Sylow 2-subgroup of S3S_3 has order 2.
By Sylow’s theorems:
- The number n2n_2 of Sylow 2-subgroups divides 3 (since 6/2 = 3) and
- n2≡1mod 2n_2 \equiv 1 \mod 2
So n2n_2 divides 3 and is congruent to 1 mod 2. The possible values are 1 and 3.
We will find all subgroups of order 2 in S3S_3.
Step 3: List Elements of S3S_3
The elements of S3S_3 are: id, (12), (13), (23), (123), (132)\text{id},\ (12),\ (13),\ (23),\ (123),\ (132)
The three transpositions (12),(13),(23)(12), (13), (23) each generate a subgroup of order 2:
- H1={id,(12)}H_1 = \{ \text{id}, (12) \}
- H2={id,(13)}H_2 = \{ \text{id}, (13) \}
- H3={id,(23)}H_3 = \{ \text{id}, (23) \}
These are all the subgroups of order 2, so n2=3n_2 = 3. Thus, there are three Sylow 2-subgroups in S3S_3.
Step 4: Show They Are Conjugate
Two subgroups HH and KK are conjugate in a group GG if there exists g∈Gg \in G such that gHg−1=KgHg^{-1} = K.
Let us take H1={id,(12)}H_1 = \{ \text{id}, (12) \}. We can find elements of S3S_3 that conjugate it to the other two subgroups.
Let g=(123)∈S3g = (123) \in S_3. Then (123)(12)(132)=(23)(123)(12)(132) = (23)
So (123)H1(132)={id,(23)}=H3(123)H_1(132) = \{ \text{id}, (23) \} = H_3
Similarly, (132)(12)(123)=(13)(132)(12)(123) = (13), so (132)H1(123)={id,(13)}=H2(132)H_1(123) = \{ \text{id}, (13) \} = H_2
This shows that all three Sylow 2-subgroups are conjugate to each other in S3S_3.
Final Answer:
There are three Sylow 2-subgroups of S3S_3: {id,(12)}, {id,(13)}, {id,(23)}\{ \text{id}, (12) \},\ \{ \text{id}, (13) \},\ \{ \text{id}, (23) \}
They are all conjugate to each other under elements of S3S_3, which satisfies Sylow’s conjugacy condition.