Why Does \( g^{m!} \in H \) for All \( g \in G \)?

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Discussion Overview

The discussion revolves around proving that \( g^{m!} \in H \) for all \( g \in G \) where \( G \) is a finite group and \( H \) is a subgroup of \( G \) with the relationship \( |G| = m|H| \). The approach suggested involves using Lagrange's theorem and the concept of cosets.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant introduces the problem and suggests using Lagrange's theorem to prove that \( g^{m!} \in H \) for all \( g \in G \).
  • Another participant applies Lagrange's theorem to establish that there are \( m \) cosets of \( H \) in \( G \) and discusses the implications of these cosets potentially being different.
  • A participant questions the reasoning behind the conclusion that if the cosets cannot all be the same, there must exist an integer \( n \) such that \( g^n \in H \).
  • Further clarification is provided by another participant, explaining that if two cosets are equal, it leads to the conclusion that \( g^{b-a} \in H \), where \( b \) and \( a \) are indices of the cosets.

Areas of Agreement / Disagreement

Participants express uncertainty about the implications of the cosets being different and the existence of the integer \( n \). The discussion does not reach a consensus on the clarity of the reasoning involved.

Contextual Notes

The discussion includes assumptions about the properties of groups and cosets, and the reasoning relies on the structure of finite groups as defined by Lagrange's theorem. There are unresolved steps in the argument regarding the implications of coset equality.

mathmadx
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Dear all,
The question I've been struggling with is supposed to be solved using the way Lagrange's thm was proven( with number of cosets and stuff). However, it remains a mystery how to do it:

Let G be a finite group and H<G with |G|=m|H|. Proof that

[tex]g^{m!} \in H, \forall g \in G[/tex]
 
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Using Lagrange's theorem on your expression we get [itex][G : H] = |G|/|H| = m[/itex] so there are m cosets of H in G. Consider:
[tex]1H,gH,g^2H,\ldots,g^mH[/tex]
Can they all be different? If not then we have an integer [itex]n \in \{1,2,\ldots,m\}[/itex] such that [itex]g^n \in H[/itex]. Now since H is a group and [itex]m!/n[/itex] is an integer, then H contains:
[tex]\left(g^n\right)^{m!/n} = g^{m!}[/tex]
 
Ok, thanks, however, it's still a bit blurry to me why the fact that they can't be all the same implies that there is such an integer n..
 
mathmadx said:
Ok, thanks, however, it's still a bit blurry to me why the fact that they can't be all the same implies that there is such an integer n..

Since they are not all the same we can find two that are equal. Let these be [itex]g^a H[/itex] and [itex]g^b H[/itex] with [itex]a,b \in \{0,1,\ldots,m\}[/itex]. Assume without loss of generality that a < b, then we have:
[tex]g^b H = g^a H[/tex]
which implies
[tex]g^{b-a} H = g^{a-a}H = H[/tex]
which is equivalent to [itex]g^{b-a} \in H[/itex].

Since [itex]b \leq m[/itex] and a is non-negative [itex]b-a\leq m[/itex] and since b>a we have 0<b-a which together gives us [itex]0 < b-a \leq m[/itex] so n=b-a is the number we wanted to prove existed.
 

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