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 September 28th, 2007, 09:25 AM #1 Newbie   Joined: Sep 2007 Posts: 10 Thanks: 0 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.
 September 28th, 2007, 04:39 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,622 Thanks: 2075 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?
 October 1st, 2007, 07:31 AM #3 Newbie   Joined: Sep 2007 Posts: 10 Thanks: 0 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.
 October 1st, 2007, 04:24 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,622 Thanks: 2075 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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