Question

Prove that 2-sylow subgroups of s4 are isomorphic to d4

Answer 1

Answer:

By the Sylow counting theorems, the number n2 of Sylow 2-subgroups is congruent to 1 mod 2 and divides 24/23=3. Thus, n2 equals 1 or 3. If n2 equals 1, then D4 is the unique Sylow-2 subgroup in S4, which means it is also normal in S4, but this is not the case. For it can be seen that D4 is not closed under conjugates: it contains (1234) but not (1324), for example, or it contains (13) but not (12). Thus n2=3. One of the subgroups in S4 of order 8 is D4=⟨(13),(1234)⟩. By the Sylow theorems, all subgroups of order 8 are conjugate. Hence the other two subgroups of order 8 can be obtained by relabeling the points in D4 (i.e. taking conjugates of D4). For example, conjugation by (23) gives the subgroup ⟨(12),(1324)⟩, and conjugation by (34) gives the subgroup ⟨(14),(1243)⟩.

Regarding your question on how to choose σ, it is known that the n-cycle (1,2,…,n) and the transposition (i,j) together generate all of Sn iff i−j and n are relatively prime.

Step-by-step explanation: