Exponent of a finite group $G$
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Let $G$ be a finite group. The exponent of $G$, $exp(G)$ is defined as the minimal positive integer $m$ such that $x^m=1$ for all $x in G$.
prove:
$a)$ if $G$ is abelian then $exp(G)= max{text{ord}(x):x in G}$
$b)$ if $G$ is not abelian the previous statement may fail.
finite-groups
edited Nov 18 at 2:25
hanszimmer
195
asked Nov 18 at 2:10
mathnoob
1,133116
add a comment |
up vote
1
down vote
favorite
Let $G$ be a finite group. The exponent of $G$, $exp(G)$ is defined as the minimal positive integer $m$ such that $x^m=1$ for all $x in G$.
prove:
$a)$ if $G$ is abelian then $exp(G)= max{text{ord}(x):x in G}$
$b)$ if $G$ is not abelian the previous statement may fail.
finite-groups
edited Nov 18 at 2:25
hanszimmer
195
asked Nov 18 at 2:10
mathnoob
1,133116
Is my solution correct?
– mathnoob
Nov 18 at 2:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G$ be a finite group. The exponent of $G$, $exp(G)$ is defined as the minimal positive integer $m$ such that $x^m=1$ for all $x in G$.
prove:
$a)$ if $G$ is abelian then $exp(G)= max{text{ord}(x):x in G}$
$b)$ if $G$ is not abelian the previous statement may fail.
finite-groups
edited Nov 18 at 2:25
hanszimmer
195
asked Nov 18 at 2:10
mathnoob
1,133116
Let $G$ be a finite group. The exponent of $G$, $exp(G)$ is defined as the minimal positive integer $m$ such that $x^m=1$ for all $x in G$.
prove:
$a)$ if $G$ is abelian then $exp(G)= max{text{ord}(x):x in G}$
$b)$ if $G$ is not abelian the previous statement may fail.
finite-groups
finite-groups
edited Nov 18 at 2:25
hanszimmer
195
asked Nov 18 at 2:10
mathnoob
1,133116
edited Nov 18 at 2:25
hanszimmer
195
asked Nov 18 at 2:10
mathnoob
1,133116
edited Nov 18 at 2:25
hanszimmer
195
edited Nov 18 at 2:25
hanszimmer
195
edited Nov 18 at 2:25
hanszimmer
195
195
asked Nov 18 at 2:10
mathnoob
1,133116
asked Nov 18 at 2:10
mathnoob
1,133116
asked Nov 18 at 2:10
mathnoob
1,133116
1,133116
Is my solution correct?
– mathnoob
Nov 18 at 2:30
add a comment |
Is my solution correct?
– mathnoob
Nov 18 at 2:30
Is my solution correct?
– mathnoob
Nov 18 at 2:30
Is my solution correct?
– mathnoob
Nov 18 at 2:30
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
Here is my attempt for
$a)$
$G$ is finite abelian, so $G$ is direct sums of $k$ $mathbb{Z}/p_i mathbb{Z}$'s for some $k in mathbb{N}$ and $p_i|p_{i+1}$ for all $i$. Hence element with highest order is $(1,...,1)$ with order $lcm(p_1,...,p_k)=p_k$ which is the highest order of any element in $G$ which is $exp(G)$.
$b)$ $S_3$ is a counterexample.
answered Nov 18 at 2:10
mathnoob
1,133116
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here is my attempt for
$a)$
$G$ is finite abelian, so $G$ is direct sums of $k$ $mathbb{Z}/p_i mathbb{Z}$'s for some $k in mathbb{N}$ and $p_i|p_{i+1}$ for all $i$. Hence element with highest order is $(1,...,1)$ with order $lcm(p_1,...,p_k)=p_k$ which is the highest order of any element in $G$ which is $exp(G)$.
$b)$ $S_3$ is a counterexample.
answered Nov 18 at 2:10
mathnoob
1,133116
add a comment |
up vote
0
down vote
Here is my attempt for
$a)$
$G$ is finite abelian, so $G$ is direct sums of $k$ $mathbb{Z}/p_i mathbb{Z}$'s for some $k in mathbb{N}$ and $p_i|p_{i+1}$ for all $i$. Hence element with highest order is $(1,...,1)$ with order $lcm(p_1,...,p_k)=p_k$ which is the highest order of any element in $G$ which is $exp(G)$.
$b)$ $S_3$ is a counterexample.
answered Nov 18 at 2:10
mathnoob
1,133116
add a comment |
up vote
0
down vote
up vote
0
down vote
Here is my attempt for
$a)$
$G$ is finite abelian, so $G$ is direct sums of $k$ $mathbb{Z}/p_i mathbb{Z}$'s for some $k in mathbb{N}$ and $p_i|p_{i+1}$ for all $i$. Hence element with highest order is $(1,...,1)$ with order $lcm(p_1,...,p_k)=p_k$ which is the highest order of any element in $G$ which is $exp(G)$.
$b)$ $S_3$ is a counterexample.
answered Nov 18 at 2:10
mathnoob
1,133116
Here is my attempt for
$a)$
$G$ is finite abelian, so $G$ is direct sums of $k$ $mathbb{Z}/p_i mathbb{Z}$'s for some $k in mathbb{N}$ and $p_i|p_{i+1}$ for all $i$. Hence element with highest order is $(1,...,1)$ with order $lcm(p_1,...,p_k)=p_k$ which is the highest order of any element in $G$ which is $exp(G)$.
$b)$ $S_3$ is a counterexample.
answered Nov 18 at 2:10
mathnoob
1,133116
answered Nov 18 at 2:10
mathnoob
1,133116
answered Nov 18 at 2:10
mathnoob
1,133116
answered Nov 18 at 2:10
mathnoob
1,133116
1,133116
add a comment |
add a comment |
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Is my solution correct?
– mathnoob
Nov 18 at 2:30