exponent of a finite group divides order of the group
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Let $minBbb N$ be the exponent of a finite group $G $ ($|G|=n$). It' the smallest integer such that $g^m=e_G forall gin G$
Proving that $mmid |G|$ is the same as proving that the least common multiple of the orders of the elements of $G$ divides the order of the group once you realize the exponent is the LCM of $o(g)$ for $gin G$.
But how to conclude the proof?
I tried to reason with $q_icdot o(g_i)=|G| forall iin[n]$ but it doesn't seem to lead me enywhere
group-theory finite-groups
asked Nov 15 at 13:16
John Cataldo
8581216
add a comment |
up vote
2
down vote
favorite
Let $minBbb N$ be the exponent of a finite group $G $ ($|G|=n$). It' the smallest integer such that $g^m=e_G forall gin G$
Proving that $mmid |G|$ is the same as proving that the least common multiple of the orders of the elements of $G$ divides the order of the group once you realize the exponent is the LCM of $o(g)$ for $gin G$.
But how to conclude the proof?
I tried to reason with $q_icdot o(g_i)=|G| forall iin[n]$ but it doesn't seem to lead me enywhere
group-theory finite-groups
asked Nov 15 at 13:16
John Cataldo
8581216
Have you thought of showing that the prime-power divisors of the exponent divide the order of the group?
– ancientmathematician
Nov 15 at 14:36
@ancientmathematician no, I'll try that
– John Cataldo
Nov 15 at 14:39
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $minBbb N$ be the exponent of a finite group $G $ ($|G|=n$). It' the smallest integer such that $g^m=e_G forall gin G$
Proving that $mmid |G|$ is the same as proving that the least common multiple of the orders of the elements of $G$ divides the order of the group once you realize the exponent is the LCM of $o(g)$ for $gin G$.
But how to conclude the proof?
I tried to reason with $q_icdot o(g_i)=|G| forall iin[n]$ but it doesn't seem to lead me enywhere
group-theory finite-groups
asked Nov 15 at 13:16
John Cataldo
8581216
Let $minBbb N$ be the exponent of a finite group $G $ ($|G|=n$). It' the smallest integer such that $g^m=e_G forall gin G$
Proving that $mmid |G|$ is the same as proving that the least common multiple of the orders of the elements of $G$ divides the order of the group once you realize the exponent is the LCM of $o(g)$ for $gin G$.
But how to conclude the proof?
I tried to reason with $q_icdot o(g_i)=|G| forall iin[n]$ but it doesn't seem to lead me enywhere
group-theory finite-groups
group-theory finite-groups
asked Nov 15 at 13:16
John Cataldo
8581216
asked Nov 15 at 13:16
John Cataldo
8581216
asked Nov 15 at 13:16
John Cataldo
8581216
asked Nov 15 at 13:16
John Cataldo
8581216
asked Nov 15 at 13:16
John Cataldo
8581216
8581216
Have you thought of showing that the prime-power divisors of the exponent divide the order of the group?
– ancientmathematician
Nov 15 at 14:36
@ancientmathematician no, I'll try that
– John Cataldo
Nov 15 at 14:39
add a comment |
Have you thought of showing that the prime-power divisors of the exponent divide the order of the group?
– ancientmathematician
Nov 15 at 14:36
@ancientmathematician no, I'll try that
– John Cataldo
Nov 15 at 14:39
Have you thought of showing that the prime-power divisors of the exponent divide the order of the group?
– ancientmathematician
Nov 15 at 14:36
Have you thought of showing that the prime-power divisors of the exponent divide the order of the group?
– ancientmathematician
Nov 15 at 14:36
@ancientmathematician no, I'll try that
– John Cataldo
Nov 15 at 14:39
@ancientmathematician no, I'll try that
– John Cataldo
Nov 15 at 14:39
add a comment |
1 Answer
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Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $ain G$.
answered Nov 21 at 3:30
Alex Sanger
695
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $ain G$.
answered Nov 21 at 3:30
Alex Sanger
695
add a comment |
up vote
0
down vote
Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $ain G$.
answered Nov 21 at 3:30
Alex Sanger
695
add a comment |
up vote
0
down vote
up vote
0
down vote
Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $ain G$.
answered Nov 21 at 3:30
Alex Sanger
695
Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $ain G$.
answered Nov 21 at 3:30
Alex Sanger
695
answered Nov 21 at 3:30
Alex Sanger
695
answered Nov 21 at 3:30
Alex Sanger
695
answered Nov 21 at 3:30
Alex Sanger
695
695
add a comment |
add a comment |
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Have you thought of showing that the prime-power divisors of the exponent divide the order of the group?
– ancientmathematician
Nov 15 at 14:36
@ancientmathematician no, I'll try that
– John Cataldo
Nov 15 at 14:39