Prove that if h is in H, then also h^-1

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Discussion Overview

The discussion revolves around proving that if an element \( h \) belongs to a subgroup \( H \) of a group \( G \), then its inverse \( h^{-1} \) also belongs to \( H \). The context involves group theory, specifically focusing on the properties of subgroups and the implications of closure under the group operation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how to select elements \( h \) and \( h' \) from \( H \) to demonstrate closure and the existence of inverses.
  • Another participant suggests considering the size of \( H \) given that \( G \) is finite, hinting at the implications for the subgroup's structure.
  • A participant proposes that since \( g^{m-n} = e \in H \), \( H \) contains all powers of \( g \) up to \( m-n \), leading to the conclusion that for any \( h = g^{s_2} \in H \), there exists an \( h' = g^{s_1} \in H \) such that \( hh' = e \in H \).
  • Another participant expresses confusion regarding the notation and the reasoning involving \( m \) and \( n \), suggesting a more straightforward approach to defining the relationship between \( s_1 \) and \( s_2 \).

Areas of Agreement / Disagreement

Participants express differing levels of understanding regarding the proof, with some agreeing on the closure property while others question the clarity of the reasoning and notation used. The discussion remains unresolved as participants have not reached a consensus on the proof's validity.

Contextual Notes

There are uncertainties regarding the definitions of \( m \) and \( n \), as well as the notation used in the proof. The discussion reflects varying interpretations of the subgroup properties and the implications of finite groups.

sutupidmath
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How would one prove that if h is in H, then also h^-1, that is its inverse of it is also in H, if Z is only positive integers.
Where [tex]H=\{a^n: n\in Z^+\}[/tex] a in G
I managed to show, as stated that [tex]e=a^{m-n}\in H[/tex]
since i supposed that the group G is finite, and i also know that now i have to take two elements in H, say h and h' and then use the property that H is closed, and [tex]hh'=...=e\in H[/tex]


, but i cannot figure out how to pick up h and h'?

Can u help me on this?
 
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Hi sutupidmath! :smile:

(I'm not sure I understand what G and H are, but in case I've got it right:)

Hint: if G is finite, and Hh = {hn : n ε Z+}, then how large is Hh? :wink:
 
I think i got it, let's see:

SInce [tex]g^{m-n}=e\in H[/tex] it means that H consists of m-n elements, that is all powers of g up to m-n. Now this means that there defenitely are:

[tex]s_1, s_2 \in Z^{+}[/tex] such that [tex]s_1+s_2 =m-n[/tex] while [/tex] g^{s_1},g^{s_2} \in H[/tex] this means that:

For every element in H, say [tex]h=g^{s_2}, s_2 \in Z^+, \exists h'=g^{s_1}, s_1 \in Z^+[/tex] such that

using closure property of H, we get:


[tex]hh'=g^{s_1}g^{s_2}=g^{s_1+s_2}=g^{m-n}=e \in H[/tex] which means that h and h' are inverses of each other, and still both h and h' are in H. Proof done. RIght?
 
sutupidmath said:
I think i got it, let's see:

SInce [tex]g^{m-n}=e\in H[/tex] it means that H consists of m-n elements, that is all powers of g up to m-n. Now this means that there defenitely are:

[tex]s_1, s_2 \in Z^{+}[/tex] such that [tex]s_1+s_2 =m-n[/tex] while [tex]g^{s_1},g^{s_2} \in H[/tex] this means that:

For every element in H, say [tex]h=g^{s_2}, s_2 \in Z^+, \exists h'=g^{s_1}, s_1 \in Z^+[/tex] such that

using closure property of H, we get:


[tex]hh'=g^{s_1}g^{s_2}=g^{s_1+s_2}=g^{m-n}=e \in H[/tex] which means that h and h' are inverses of each other, and still both h and h' are in H. Proof done. RIght?

Hi sutupidmath! :smile:

This is very confusing …

I take it you're using g instead of a? …

and I don't understand what m is, or why you're using m - n when n is just an index. :confused:

Can't you shorten it, by starting with something like "for any s2 ε Z+, define s1 = -s2 mod m - n …" ? :smile:
 

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