Let n > 1 and i ∈ ℕ with 1 ≤ i ≤ n. Show that the set S(i) = {σ ∈ S_n | σ(i) = i} is a subgroup of S_n.
For 1 ≤ i, j ≤ n, show
that the groups S(i) and S(j) are isomorphic. If i ≠ j, determine if
S(i, j) = {σ ∈ S_n | σ(i) = j} is a subgroup of S_n.
I have no idea where to start
any ideas would be appreciated !!
Group theory isomorphism.?
2015-11-09 2:34 am
回答 (1)
2015-11-09 3:17 am
None of this is hard. I think you just need more practice working with groups and doing proves.
For the first part, you always start with showing the same 3 things about S(i):
1. It contains the identity.
2. The product of two elements in S(i) is still in S(i)
3. The inverse of any element in S(i) is in S(i).
Those are all very easy to show, and the conclusion is that S(i) is a subgroup.
S(i) is the set of permutations that leaves i unchanged.
S(j) is the set of permutations that leaves j unchanged.
It's pretty obvious that those should look the same.
Note that t = (ij)^-1 = (ij).
Show that conjugation of a subgroup is an isomorphism, and then show that tS(i)t^-1 = S(j) -- if s leaves i fixed, then the 3 elements making up tst^-1 do what to j?
For the first part, you always start with showing the same 3 things about S(i):
1. It contains the identity.
2. The product of two elements in S(i) is still in S(i)
3. The inverse of any element in S(i) is in S(i).
Those are all very easy to show, and the conclusion is that S(i) is a subgroup.
S(i) is the set of permutations that leaves i unchanged.
S(j) is the set of permutations that leaves j unchanged.
It's pretty obvious that those should look the same.
Note that t = (ij)^-1 = (ij).
Show that conjugation of a subgroup is an isomorphism, and then show that tS(i)t^-1 = S(j) -- if s leaves i fixed, then the 3 elements making up tst^-1 do what to j?
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