- #1
mathmajor2013
- 26
- 0
Let G be a group and H be a subgroup of G. Prove that if [G
]=2, then H is a normal subgroup of G.
]=2, then H is a normal subgroup of G.Let G be a group and H a subgroup. Prove if [G:H]=2, then H is normal.
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Homework Help Overview
The discussion revolves around a group theory problem concerning the normality of a subgroup H within a group G, specifically when the index [G:H] equals 2.
Discussion Character
- Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants explore the implications of having only two left cosets and question how this relates to the normality of H. Some express confusion about the relationship between left and right cosets.
Discussion Status
Participants are actively engaging with the problem, clarifying terms and concepts related to cosets. Some guidance has been offered regarding the implications of the index being 2, and there is a recognition of the distinction between left cosets and the set-theoretic difference.
Contextual Notes
There is some confusion regarding terminology, particularly the distinction between G/H and G\H, which has been addressed in the discussion.
- #2
- 22,170
- 3,326
The same as in my other reply:
1) this should belong in the homework forums
2) what did you try?
1) this should belong in the homework forums
2) what did you try?
- #3
mathmajor2013
- 26
- 0
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.
- #4
- 22,170
- 3,326
That [G]=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?
]=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?
- #5
mathmajor2013
- 26
- 0
I thought that G/H was the set of left or right cosets, not a coset itself? But yes that does help, thank you!
- #6
- 22,170
- 3,326
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.
- #7
mathmajor2013
- 26
- 0
Oh I see. Yes! Thank you.
- #8
berkeman
Admin
- 69,481
- 25,011
(Thread moved and OP pinged)
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