"Does Finite Group Contain Subgroup of Index 2 if Element has Order 2?

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Discussion Overview

The discussion revolves around whether a finite group containing a subgroup of index 2 necessarily has an element of order 2. Participants explore the implications of subgroup properties and group order in relation to this question.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions if the existence of a subgroup of index 2 implies the presence of an element of order 2.
  • Another participant suggests that since a subgroup of index 2 is normal, it leads to the conclusion that elements in the group must satisfy certain properties, potentially indicating they are their own inverses.
  • A different participant asserts that the statement is true based on Lagrange's theorem, which implies that the group must have even order, suggesting the existence of an element that is its own inverse.
  • Concerns are raised about the validity of the proof regarding the multiplication of cosets and the assumptions made about elements being in the subgroup.
  • One participant notes that for elements not in the subgroup, certain conditions must hold for the group structure, indicating a complexity in proving the initial claim.

Areas of Agreement / Disagreement

Participants express differing views on the implications of having a subgroup of index 2 and whether it guarantees an element of order 2. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations regarding the assumptions made about group operations and the definitions of cosets, which are not fully explored or resolved in the discussion.

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Is it true that if a finite group G contains a subgroup of index 2, then there is an element of G with order 2?
 
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I'm thinking that if H is a subgroup of index 2, then H is normal in G. Then for all x in G, xH=Hx, so (xH)(Hx)=H and xHx=H. So x=x^-1 and x has order 2. Is this correct?
 
The statement is true. If G has finite order and a subgroup of index two, G must have even order by Lagrange's theorem. Then, you can easily prove there must exist an element which is its own inverse. In fact, this is a standard exercise in any text.

Regarding your proof: Is multiplication of a left coset by a right coset well defined? And why do you end up with H on the right side? Doing this assumes xx=e.
 
You're right.

But you have to have x^2 in H for all x not in H. (xH)(xH)=H because otherwise, it would be an identity: (xH)(xH)=(xH) and cancellation gives that x is in H.

I don't see any way to prove this.
 

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