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If there is an $x_0$ of order two and $operatornameordf(x_0)^k=2$, $k=0,…,p-1$, then $x^2=e$.

Order of $b$ when $ba=ab^3$ and $operatornameord(a)=4$.If $G$ is a finite group, then $operatornameordS(a)=operatornameord(G)/operatornameOrd(C(a))$Relationship between $operatornameord(ab), operatornameord(a)$, and $operatornameord(b)$Natural action of $operatornameAut(G)$ on sets of subgroups of $G$ of same order is transitive.$operatornameord(a^k)$ is a divisor of $operatornameord(a)$Proof of $prod_limitsi=1^nG_i$ is cyclic iff $gcd(operatornameord G_i,operatornameord G_j)=1$$gcd(operatornameord G_i,operatornameord G_j)=1$ if $ineq j$ and cyclic groups ex$DeclareMathOperatorordordgcd(ord(a),ord(b)) = 1$ then $ord(ab)=ord(a)cdot ord(b)$ where $a,b$ do not commuteThe order of $(g,h)$ equals the least common multiple of ord(g) and ord(h)?Automorphisms of Quaternion Group $Q$. Prove that $operatornameAut(Q)$ contains no element of order $6$, and so $operatornameAut(Q)cong S_4 $

0

$begingroup$

I have a group $G=e,f(x),dots ,f(x)^p-1$ and $fin operatornameAut(G)$ and as $|G|$ is even, I read the following property:

If there is an $x_0$ of order two and $ operatornameordf(x_0)^k=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.

I am not sure I understand it as I should.

abstract-algebra group-theory finite-groups

share|cite|improve this question

edited Mar 13 at 22:00

Shaun

9,569113684

asked Mar 13 at 8:41

user651754

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you know about generators?
  $endgroup$
  – Shaun
  Mar 13 at 22:01
  

  
  
  
  

add a comment | 

0

$begingroup$

I have a group $G=e,f(x),dots ,f(x)^p-1$ and $fin operatornameAut(G)$ and as $|G|$ is even, I read the following property:

If there is an $x_0$ of order two and $ operatornameordf(x_0)^k=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.

I am not sure I understand it as I should.

abstract-algebra group-theory finite-groups

share|cite|improve this question

edited Mar 13 at 22:00

Shaun

9,569113684

asked Mar 13 at 8:41

user651754

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you know about generators?
  $endgroup$
  – Shaun
  Mar 13 at 22:01
  

  
  
  
  

add a comment | 

$begingroup$

I have a group $G=e,f(x),dots ,f(x)^p-1$ and $fin operatornameAut(G)$ and as $|G|$ is even, I read the following property:

If there is an $x_0$ of order two and $ operatornameordf(x_0)^k=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.

I am not sure I understand it as I should.

abstract-algebra group-theory finite-groups

share|cite|improve this question

edited Mar 13 at 22:00

Shaun

9,569113684

asked Mar 13 at 8:41

user651754

$endgroup$

share|cite|improve this question

edited Mar 13 at 22:00

Shaun

9,569113684

asked Mar 13 at 8:41

user651754

share|cite|improve this question

edited Mar 13 at 22:00

Shaun

9,569113684

asked Mar 13 at 8:41

user651754

edited Mar 13 at 22:00

Shaun

9,569113684

edited Mar 13 at 22:00

Shaun

9,569113684