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$$gcd(operatornameord G_i,operatornameord G_j)=1$ if $ineq j$ and cyclic groups ex$DeclareMathOperatorordordgcd(ord(a),ord(b)) = 1$ then $ord(ab)=ord(a)cdot ord(b)$ where $a,b$ do not commuteThe order of $(g,h)$ equals the least common multiple of ord(g) and ord(h)?Automorphisms of Quaternion Group $Q$. Prove that $operatornameAut(Q)$ contains no element of order $6$, and so $operatornameAut(Q)cong S_4 $
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If there is an $x_0$ of order two and $operatornameordf(x_0)^k=2$, $k=0,…,p-1$, then $x^2=e$.
Order of $b$ when $ba=ab^3$ and $operatornameord(a)=4$.If $G$ is a finite group, then $operatornameordS(a)=operatornameord(G)/operatornameOrd(C(a))$Relationship between $operatornameord(ab), operatornameord(a)$, and $operatornameord(b)$Natural action of $operatornameAut(G)$ on sets of subgroups of $G$ of same order is transitive.$operatornameord(a^k)$ is a divisor of $operatornameord(a)$Proof of $prod_limitsi=1^nG_i$ is cyclic iff $gcd(operatornameord G_i,operatornameord G_j)=1$$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 $
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
abstract-algebra group-theory finite-groups
edited Mar 13 at 22:00
Shaun
9,569113684
9,569113684
asked Mar 13 at 8:41
user651754
$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01
add a comment |
$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01
$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01
$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01
add a comment |
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$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01