# The Image of a Group Action

#### Poirot

##### Banned
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.

#### Poirot

##### Banned
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

#### Deveno

##### Well-known member
MHB Math Scholar
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).