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

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SUMMARY

The action of conjugation by Sylow 2-subgroups of a group G of order 48 is onto, as demonstrated through the transitive nature of the action on the set of these subgroups. Given three Sylow 2-subgroups H, K, and M, the existence of elements x, y, and z in G allows for the mapping of these subgroups through conjugation, confirming that all permutations in S3 can be achieved. The analysis shows that the conjugation action corresponds to permutations, establishing that the action is indeed bijective and covers all elements of the set.

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  • Understanding of group theory concepts, particularly Sylow theorems.
  • Familiarity with group actions and transitivity.
  • Knowledge of permutation groups, specifically S3.
  • Basic understanding of conjugation in group theory.
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  • Study the properties of Sylow subgroups in finite group theory.
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  • Explore the structure and properties of permutation groups, focusing on S3.
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Mathematicians, particularly those specializing in group theory, algebraists, and students studying advanced algebra concepts, will benefit from this discussion.

Poirot1
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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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