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Abstract Algebra?
Let H and K be normal subgroups of G such that the intersection of H and K = {e}. Show that hk = kh for all h in H and k in K.
I have so far that since gH = Hg and gK = Kg for g in G, gh = hg and gk = kg for all h in H and all k in K. Then, g = hgh^-1 and g = kgk^-1. So, hgh^-1 = kgk^-1. I've had no success in canceling the g, and I don't even know if what I have is correct or on the right track.
Any help would be much appreciated. Thanks.
1 Answer
- JCSLv 51 decade agoFavorite Answer
Start with:
(hk)(kh)^(-1) = (hk)(h^(-1))(k^(-1)) =
= (hkh^(-1))(k^(-1))
But hkh^(-1) ∈ K, so (hkh^(-1))(k^(-1)) ∈ K (note that it doesn't have to be equal to e). On the other hand:
(hk)(kh)^(-1) = (hk)(h^(-1))(k^(-1)) =
= h(kh^(-1)k^(-1))
And, as kh^(-1)k^(-1) ∈ H, so does (hk)(kh)^(-1).
Therefore, (hk)(kh)^(-1) ∈ H ∩ K = {e}, so:
(hk)(kh)^(-1) = e ⇒ hk = kh
Your mistake was to assume that, from gH = Hg, you have gh = hg, for the same h ∈ H.
