Prove Normal Subgroup When G Order is Twice H Order

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The discussion centers on proving that a subgroup H of a finite group G is a normal subgroup when the order of G is twice the order of H. The proof relies on the definition of normal subgroups, specifically that for all g in G, the left cosets gH and right cosets Hg must agree. Given that H has index two, it follows that there are exactly two cosets, leading to the conclusion that H is indeed a normal subgroup of G.

PREREQUISITES
  • Understanding of group theory concepts, particularly normal subgroups.
  • Familiarity with the definitions of cosets and their properties.
  • Knowledge of finite groups and subgroup orders.
  • Basic proof techniques in abstract algebra.
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  • Study the properties of normal subgroups in group theory.
  • Learn about cosets and their implications in finite groups.
  • Explore the concept of group indices and their significance.
  • Investigate examples of finite groups to solidify understanding of subgroup structures.
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Students and enthusiasts of abstract algebra, particularly those studying group theory and normal subgroups, as well as educators seeking to clarify these concepts.

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I'm working on introductory group theory and am stuck on this proof. I don't even know where to start, so I'd appreciate any help at all!

"A subgroup H of a finite group G is said to be a normal subgroup if, for each element h ∈ H and each element g ∈ G, the element g^1hg ∈ H

Prove that if the order of G is twice the order of H, then H is a normal subgroup of G."
 
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Take the definition of normal, and play with it. What it says is that for all g in G, the set gHg^{-1}=H, or that gH=Hg, ie left and right cosets agree. Suppose H has index two (i.e. |G|=2|H|). How many left cosets are there? How many right cosets are there. What properties of cosets do you know? How about cosets are disjoint or equal? Can you see how this implies the result?
 

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