MHB Is the Action of Conjugation by Sylow 2-Subgroups Onto?

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The discussion centers on the action of conjugation by Sylow 2-subgroups of a group G of order 48, which has three such subgroups. The participants explore whether this action is onto, noting that all three subgroups are conjugate, indicating a transitive action. They demonstrate that for any two subgroups, there exists an element in G that conjugates one to the other, leading to a correspondence with permutations in S3. The analysis shows that the action is indeed onto, as the permutations cover all possible mappings between the subgroups. Ultimately, the conclusion is reached that the conjugation action is transitive and onto.
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Assume that G of order 48 has 3 sylow 2-subgroups. Let G act on the set of such subgroups by conjugation. How do I know that this action is onto? I know that all 3 subgroups are conjugate but I'm not sure this is enough.
 
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I can't imagine this is too difficultto find that an action is onto i.e. can I map to every permutation in S3. My problem is (1 2 3) is in S3 so is this telling me I need to find a g such that gHg^-1=M, gMg^-1=K and gkg^-1=H, for the slylow subgroups H,k,M
 
ok, suppose the three subgroups are H,K,M.

we know there is x,y,z in G with:

xHx-1 = K

yKy-1= M

zMz-1= H.

so the action is transitive. because we have an action (these are bijective maps), if:

x:H-->K

then xKx-1 can only be H or M. if xKx-1 = H, then x corresponds to the permutation (1 2).

otherwise xKx-1 = M, and x corresponds to (1 2 3).

now if xKx-1 corresponds to (1 2), we can look at y.

since y takes K to M, either y takes H to K, in which case y corresponds to (1 2 3)

or y takes H to H, in which case y corresponds to (2 3).

in this last case, we have xy corresponds to (1 2)(2 3) = (1 2 3):

xyH(xy)-1 = x(yHy-1)x-1 = xHx-1 = K

xyK(xy)-1 = x(yKy-1)x-1 = xMx-1 = M

xyM(xy)-1 = x(yMy-1)x-1 = xKx-1 = H.

(and we didn't even need to use z).
 

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