Intersection of abelian subgroups is a normal subgroup

"Prove that if H and K are abelian subgroups of a group G, then the intersection of H and K is a normal subgroup of <H,K>"

Just to be clear a normal subgroup of G is a subgroup where its elements multiplied by an element of G is an element in the normal subgroup, and <H,K> is the group where its elements are linear combinations of the elements of H and the elements of K, correct? I ask this just to be sure I understand the question, but I still don't know how to do this proof?

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Re: Intersection of abelian subgroups is a normal subgroup

Hi,

Your description of normal subgroups is faulty. A subgroup H of G is a normal subgroup iff for any x in G x^{-1}Hx = H. Also, <H,K> is not "linear combinations" of elements of H and K. Here's a solution with a few details left for you:

Attachment 27822

Re: Intersection of abelian subgroups is a normal subgroup

Ok it looks like my definition of a normal subgroup was too much like the definition of an ideal under a ring. So far I've only worked with cyclic groups with only one generating element like <H> or <K>, but never <H,K> so that's how I got confused. I don't have a very firm grasp of group theory, but to make sure I understand this, "if x_1.x_2.x_3....x_n where each x_i is in H or K" means that we are working with multiplicative groups and NOT additive groups, correct? And N_G means "normal group of"?

Re: Intersection of abelian subgroups is a normal subgroup

Yes, the convention is that the group operation is "multiplication" unless specified otherwise. Now for G a group and H a subgroup of G, N_{G}(H) is called the normalizer (in G) of H. $\displaystyle N_G(H)=\{x\in G\,:\,x^{-1}Hx=x\}$. This is easily shown to be a subgroup of G that contains H. H is a normal subgroup of G iff N_{G}(H) = G.