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- anonymous

Anyone on who knows some group theory?
I'm stuck:
Prove that G cannot have a subgroup H with
|H| = n-1, where n = |G| > 2

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- anonymous

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- anonymous

the order of the subgroup must divide the order of the group. so if the order of the group is n, there is no way that n- 1 | n unless n = 2

- anonymous

Is there a reason why the order of the subgroup must divide the order of the group?

- anonymous

btw this theorem actually has a name: lagrange's theorem

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- anonymous

okay, that's what I needed to be able to read more...

- anonymous

as i recall the proof is pretty straight forward although i cannot imagine that you are supposed to prove it. the gist of the proof is that the left (right) cosets of the subgroup are equivalence classes, and since no two elements can be in distinct equivalence classes they form a partition of G

- anonymous

It's a self study in prep for applying to grad school, so I am asking, not a class

- anonymous

aah. i was going to say that whatever book you are reading must have a proof of langrange

- anonymous

It probably will, subgroups come pretty early, Chapter 2, no lagrange yet.

- anonymous

well not sure how to do it without lagrange, but no reason to tie your hands. i just looked the proof up and it is very short. just google 'lagrange theorem' and you will find lots.
good luck getting in to grad school.

- anonymous

I just read the wikipedia on lagrange and the book I am using is one of four cited in the end notes! (Dummit and Foote) Thanks for the help.

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