Similarly (14), (24) ∈ NG(H).
(Clearly, their products (1 2 4), (1 4 2) ∈ NG(H) as well.)
So far, we have {(1), (12), (14), (24), (124), (142)} is contained in NG(H).
This may very well be it for NG(H).
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However, for (13):
(13) * (1 2 4) * (13)⁻¹ = (2 4 3) ∉ H.
So, (13) ∉ NG(H).
Through quick proofs by contradiction, we see that (23), (34) ∉ NG(H).
(For instance, if (23) ∈ NG(H), then (12)(23) = (13) ∈ NG(H) by closure.
By the previous remarks, this is a contradiction.)
In a similar manner, can you rule out the remaining permutations from S₄ being in NG(H)?
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(c) Let N denote NG(H) as shorthand.
Define F : X --> Y via F(xN) = xHx⁻¹
Show that F is a well-defined map which yields a bijection.
(i) F is well-defined.
[Show: If xN = yN, then F(xN) = F(yN).]
Since xN = yN, we have y⁻¹x ∈ N.
==> (y⁻¹x) H (y⁻¹x)⁻¹ = H
==> (y⁻¹x) H (x⁻¹y) = H
==> xHx⁻¹ = yHy⁻¹
==> F(xN) = F(yN).
(ii) F is onto.
Obvious; given a subgroup K of G which is conjugate to H, we have K = gHg⁻¹ for some g ∈ G.
So, F(gN) = K, as required.
(iii) F is 1-1.
Suppose there are cosets xN, yN such that F(xN) = F(yN).
==> xHx⁻¹ = yHy⁻¹
==> (y⁻¹x) H (x⁻¹y) = H
==> (y⁻¹x) H (y⁻¹x)⁻¹ = H
==> y⁻¹x ∈ N
==> xN = yN.
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I hope this helps!