If order of group $G $ is finite, say $|G|=n$, and if $exists; xin G$ such that $|x|=m$, then …?
$begingroup$
Claim: If order of group $G $ is finite, say $|G|=n$, and if $exists; xin G$ such that $|x|=m$, then $forall d>0; text{such that d divides m, we must have } y in G$ such that $|y|=d$
Proof: $|G|=n ; text{and} ; exists ;xin G, |x|=m$
Let $H=<x>Rightarrow |H|=m$
If $d>0$ divides $m$ $Rightarrow text{d divides order of H} $ and from Fundamental theorem on cyclic groups, there will exist a subgroup of $H$, say $N$, that is of order $d$. Since $N$ will be cyclic, it will be generated by an element of order $d$
Is this correct?
group-theory proof-verification
asked Jan 25 at 2:05
AbhayAbhay
3789
$endgroup$
add a comment |
$begingroup$
Claim: If order of group $G $ is finite, say $|G|=n$, and if $exists; xin G$ such that $|x|=m$, then $forall d>0; text{such that d divides m, we must have } y in G$ such that $|y|=d$
Proof: $|G|=n ; text{and} ; exists ;xin G, |x|=m$
Let $H=<x>Rightarrow |H|=m$
If $d>0$ divides $m$ $Rightarrow text{d divides order of H} $ and from Fundamental theorem on cyclic groups, there will exist a subgroup of $H$, say $N$, that is of order $d$. Since $N$ will be cyclic, it will be generated by an element of order $d$
Is this correct?
group-theory proof-verification
asked Jan 25 at 2:05
AbhayAbhay
3789
$endgroup$
add a comment |
$begingroup$
Claim: If order of group $G $ is finite, say $|G|=n$, and if $exists; xin G$ such that $|x|=m$, then $forall d>0; text{such that d divides m, we must have } y in G$ such that $|y|=d$
Proof: $|G|=n ; text{and} ; exists ;xin G, |x|=m$
Let $H=<x>Rightarrow |H|=m$
If $d>0$ divides $m$ $Rightarrow text{d divides order of H} $ and from Fundamental theorem on cyclic groups, there will exist a subgroup of $H$, say $N$, that is of order $d$. Since $N$ will be cyclic, it will be generated by an element of order $d$
Is this correct?
group-theory proof-verification
asked Jan 25 at 2:05
AbhayAbhay
3789
$endgroup$
Claim: If order of group $G $ is finite, say $|G|=n$, and if $exists; xin G$ such that $|x|=m$, then $forall d>0; text{such that d divides m, we must have } y in G$ such that $|y|=d$
Proof: $|G|=n ; text{and} ; exists ;xin G, |x|=m$
Let $H=<x>Rightarrow |H|=m$
If $d>0$ divides $m$ $Rightarrow text{d divides order of H} $ and from Fundamental theorem on cyclic groups, there will exist a subgroup of $H$, say $N$, that is of order $d$. Since $N$ will be cyclic, it will be generated by an element of order $d$
Is this correct?
group-theory proof-verification
group-theory proof-verification
asked Jan 25 at 2:05
AbhayAbhay
3789
asked Jan 25 at 2:05
AbhayAbhay
3789
asked Jan 25 at 2:05
AbhayAbhay
3789
asked Jan 25 at 2:05
AbhayAbhay
3789
asked Jan 25 at 2:05
AbhayAbhay
3789
3789
add a comment |
add a comment |
1 Answer
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$begingroup$
You are correct!
Another way of seeing this is that $x^{m/d}$ has order $d$.
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You are correct!
Another way of seeing this is that $x^{m/d}$ has order $d$.
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
$endgroup$
add a comment |
$begingroup$
You are correct!
Another way of seeing this is that $x^{m/d}$ has order $d$.
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
$endgroup$
add a comment |
$begingroup$
You are correct!
Another way of seeing this is that $x^{m/d}$ has order $d$.
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
$endgroup$
You are correct!
Another way of seeing this is that $x^{m/d}$ has order $d$.
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
answered Jan 25 at 2:35
Santana AftonSantana Afton
2,9132629
2,9132629
add a comment |
add a comment |
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