Let G be a group and H a subgroup. Prove if [G:H]=2, then H is normal.

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SUMMARY

In group theory, if G is a group and H is a subgroup with the index [G:H] equal to 2, then H is a normal subgroup of G. This is established by noting that there are only two left cosets of H in G: H itself and G\H. Consequently, the left cosets gH and right cosets Hg must coincide, confirming that gH = Hg for all g in G. This property directly leads to the conclusion that H is normal in G.

PREREQUISITES
  • Understanding of group theory concepts, specifically subgroups and cosets.
  • Familiarity with the definition of index of a subgroup.
  • Knowledge of normal subgroups and their properties.
  • Basic comprehension of set-theoretic notation, particularly G\H.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Explore the implications of subgroup indices in finite groups.
  • Learn about the relationship between left and right cosets in groups.
  • Investigate examples of groups with normal subgroups and their indices.
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as educators and mathematicians seeking to deepen their understanding of subgroup properties and normality.

mathmajor2013
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Let G be a group and H be a subgroup of G. Prove that if [G:H]=2, then H is a normal subgroup of G.
 
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The same as in my other reply:

1) this should belong in the homework forums
2) what did you try?
 
I'm lost on this one. It doesn't make sense how the number of left cosets corresponds to the normality. #gH=#Hg doesn't seem like it necessarily means that gH=Hg.
 
That [G:H]=2 means that there are only two left cosets of H. Also, it means that there are only two right cosets of H: H and G\H.

Thus gH is H or G\H, and for Hg thesame thing. Does this help you?
 
I thought that G/H was the set of left or right cosets, not a coset itself? But yes that does help, thank you!
 
No, I mean G\H, not G/H. With G\H, I mean the set-theoretic difference, i.e. everything in G which is not in H.
 
Oh I see. Yes! Thank you.
 
(Thread moved and OP pinged)
 

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