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September 28th, 2007, 10:25 AM   #1
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normal subgroup

Let G be a group. Let N be a normal subgroup, H an arbitrary subgroup. Let HN be the subset consisting of all products hn with: h contained in H, and n contained in N. Prove that HN is a subgroup of G.
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September 28th, 2007, 05:39 PM   #2
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Can you see why HN = NH?

Also h ∈ H, n ∈ N implies both hn ∈ HN and (n^-1)h^-1) ∈ NH.

Combining the above, you can deduce that the inverse of any element of HN is also an element of HN.

What remains to be proved? What progress can you now make?
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October 1st, 2007, 08:31 AM   #3
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i am still really confused about this problem. I am not really sure how to show the inverses are also in HN, or what remains to be proved.
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October 1st, 2007, 05:24 PM   #4
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I assume you know the definition of a normal subgroup. It should be easy to see from that definition that HN = NH. Give that step some more thought.
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