Centralizers of non-central elements of a special group
$begingroup$
Let $G$ be a finite group so that $frac{G}{Z(G)}cong mathbb{Z}_p times mathbb{Z}_p$, where $p$ is a prime. Then $frac{G}{Z(G)}$ has the presentation $langle aZ(G),bZ(G)|a^p ,b^p , aba^{-1}b^{-1}in Z(G)rangle$.
I want to determine the centralizer $C_G (g)$ of an arbitrary element $gin Gsetminus Z(G)$ in term of cosets of $Z(G)$ in $C_G (g)$, exactly.
For doing this, I take $gin Gsetminus Z(G)$. Then $Z(G)neq gZ(G)in frac{G}{Z(G)}$. Every element of $frac{G}{Z(G)}$ has the form $a^i b^j Z(G)$ for some $1leq ileq p$ and $1leq jleq p$. Especially, I assume that $gZ(G)=a^i Z(G)$ for some $1leq ileq p$. So $g=a^i z$ for some $zin Z(G)$. Then $C_G (g)=C_G (a^i z)=C_G (a^i)$. Now it is enough to compute $C_{G}(a^i )$ in term of cosets of $Z(G)$ in $C_G (a^i )$. I know that $C_G (a^i )=Z(G)sqcup Big(bigsqcup_{tin C_G (a^i)setminus Z(G)}tZ(G)Big)$. But I don't know for which $t$ the equation holds.
Thanks in advance.
group-theory finite-groups group-presentation quotient-group
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
$endgroup$
|
show 6 more comments
$begingroup$
Let $G$ be a finite group so that $frac{G}{Z(G)}cong mathbb{Z}_p times mathbb{Z}_p$, where $p$ is a prime. Then $frac{G}{Z(G)}$ has the presentation $langle aZ(G),bZ(G)|a^p ,b^p , aba^{-1}b^{-1}in Z(G)rangle$.
I want to determine the centralizer $C_G (g)$ of an arbitrary element $gin Gsetminus Z(G)$ in term of cosets of $Z(G)$ in $C_G (g)$, exactly.
For doing this, I take $gin Gsetminus Z(G)$. Then $Z(G)neq gZ(G)in frac{G}{Z(G)}$. Every element of $frac{G}{Z(G)}$ has the form $a^i b^j Z(G)$ for some $1leq ileq p$ and $1leq jleq p$. Especially, I assume that $gZ(G)=a^i Z(G)$ for some $1leq ileq p$. So $g=a^i z$ for some $zin Z(G)$. Then $C_G (g)=C_G (a^i z)=C_G (a^i)$. Now it is enough to compute $C_{G}(a^i )$ in term of cosets of $Z(G)$ in $C_G (a^i )$. I know that $C_G (a^i )=Z(G)sqcup Big(bigsqcup_{tin C_G (a^i)setminus Z(G)}tZ(G)Big)$. But I don't know for which $t$ the equation holds.
Thanks in advance.
group-theory finite-groups group-presentation quotient-group
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
$endgroup$
$begingroup$
You already have $t=a$ obviously (and all powers of $a$ as well), and you don't have $t=b$ otherwise $gin Z(G)$. You could have powers of $b$ though, I can't see why not. Perhaps $p$ being prime prevents that.
$endgroup$
– Arnaud Mortier
Feb 15 '18 at 13:35
1
$begingroup$
For $1 le i < p$, we have $langle a,Z(G) rangle le C_G(a^iz) < G$, and since $|G:langle a,Z(G) rangle|= p$ is prime, we must have $langle a,Z(G) rangle = C_G(a^iz) $.
$endgroup$
– Derek Holt
Feb 15 '18 at 14:11
$begingroup$
@ArnaudMortier I understood the case $t=b$. But I couldn't understand why $b^j not in C_G (a^i )$!
$endgroup$
– user481657
Feb 16 '18 at 14:47
$begingroup$
@DerekHolt Why $|G:langle a,Z(G)rangle |=p$?
$endgroup$
– user481657
Feb 16 '18 at 14:59
$begingroup$
That follows from $|G/Z(G)| = p^2$ and $Z(G) = langle aZ(G), bZ(G) rangle$.
$endgroup$
– Derek Holt
Feb 16 '18 at 15:18
|
show 6 more comments
$begingroup$
Let $G$ be a finite group so that $frac{G}{Z(G)}cong mathbb{Z}_p times mathbb{Z}_p$, where $p$ is a prime. Then $frac{G}{Z(G)}$ has the presentation $langle aZ(G),bZ(G)|a^p ,b^p , aba^{-1}b^{-1}in Z(G)rangle$.
I want to determine the centralizer $C_G (g)$ of an arbitrary element $gin Gsetminus Z(G)$ in term of cosets of $Z(G)$ in $C_G (g)$, exactly.
For doing this, I take $gin Gsetminus Z(G)$. Then $Z(G)neq gZ(G)in frac{G}{Z(G)}$. Every element of $frac{G}{Z(G)}$ has the form $a^i b^j Z(G)$ for some $1leq ileq p$ and $1leq jleq p$. Especially, I assume that $gZ(G)=a^i Z(G)$ for some $1leq ileq p$. So $g=a^i z$ for some $zin Z(G)$. Then $C_G (g)=C_G (a^i z)=C_G (a^i)$. Now it is enough to compute $C_{G}(a^i )$ in term of cosets of $Z(G)$ in $C_G (a^i )$. I know that $C_G (a^i )=Z(G)sqcup Big(bigsqcup_{tin C_G (a^i)setminus Z(G)}tZ(G)Big)$. But I don't know for which $t$ the equation holds.
Thanks in advance.
group-theory finite-groups group-presentation quotient-group
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
$endgroup$
Let $G$ be a finite group so that $frac{G}{Z(G)}cong mathbb{Z}_p times mathbb{Z}_p$, where $p$ is a prime. Then $frac{G}{Z(G)}$ has the presentation $langle aZ(G),bZ(G)|a^p ,b^p , aba^{-1}b^{-1}in Z(G)rangle$.
I want to determine the centralizer $C_G (g)$ of an arbitrary element $gin Gsetminus Z(G)$ in term of cosets of $Z(G)$ in $C_G (g)$, exactly.
For doing this, I take $gin Gsetminus Z(G)$. Then $Z(G)neq gZ(G)in frac{G}{Z(G)}$. Every element of $frac{G}{Z(G)}$ has the form $a^i b^j Z(G)$ for some $1leq ileq p$ and $1leq jleq p$. Especially, I assume that $gZ(G)=a^i Z(G)$ for some $1leq ileq p$. So $g=a^i z$ for some $zin Z(G)$. Then $C_G (g)=C_G (a^i z)=C_G (a^i)$. Now it is enough to compute $C_{G}(a^i )$ in term of cosets of $Z(G)$ in $C_G (a^i )$. I know that $C_G (a^i )=Z(G)sqcup Big(bigsqcup_{tin C_G (a^i)setminus Z(G)}tZ(G)Big)$. But I don't know for which $t$ the equation holds.
Thanks in advance.
group-theory finite-groups group-presentation quotient-group
group-theory finite-groups group-presentation quotient-group
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
edited Dec 14 '18 at 4:17
Shaun
10.4k113686
10.4k113686
asked Feb 15 '18 at 13:21
user481657
$begingroup$
You already have $t=a$ obviously (and all powers of $a$ as well), and you don't have $t=b$ otherwise $gin Z(G)$. You could have powers of $b$ though, I can't see why not. Perhaps $p$ being prime prevents that.
$endgroup$
– Arnaud Mortier
Feb 15 '18 at 13:35
1
$begingroup$
For $1 le i < p$, we have $langle a,Z(G) rangle le C_G(a^iz) < G$, and since $|G:langle a,Z(G) rangle|= p$ is prime, we must have $langle a,Z(G) rangle = C_G(a^iz) $.
$endgroup$
– Derek Holt
Feb 15 '18 at 14:11
$begingroup$
@ArnaudMortier I understood the case $t=b$. But I couldn't understand why $b^j not in C_G (a^i )$!
$endgroup$
– user481657
Feb 16 '18 at 14:47
$begingroup$
@DerekHolt Why $|G:langle a,Z(G)rangle |=p$?
$endgroup$
– user481657
Feb 16 '18 at 14:59
$begingroup$
That follows from $|G/Z(G)| = p^2$ and $Z(G) = langle aZ(G), bZ(G) rangle$.
$endgroup$
– Derek Holt
Feb 16 '18 at 15:18
|
show 6 more comments
$begingroup$
You already have $t=a$ obviously (and all powers of $a$ as well), and you don't have $t=b$ otherwise $gin Z(G)$. You could have powers of $b$ though, I can't see why not. Perhaps $p$ being prime prevents that.
$endgroup$
– Arnaud Mortier
Feb 15 '18 at 13:35
1
$begingroup$
For $1 le i < p$, we have $langle a,Z(G) rangle le C_G(a^iz) < G$, and since $|G:langle a,Z(G) rangle|= p$ is prime, we must have $langle a,Z(G) rangle = C_G(a^iz) $.
$endgroup$
– Derek Holt
Feb 15 '18 at 14:11
$begingroup$
@ArnaudMortier I understood the case $t=b$. But I couldn't understand why $b^j not in C_G (a^i )$!
$endgroup$
– user481657
Feb 16 '18 at 14:47
$begingroup$
@DerekHolt Why $|G:langle a,Z(G)rangle |=p$?
$endgroup$
– user481657
Feb 16 '18 at 14:59
$begingroup$
That follows from $|G/Z(G)| = p^2$ and $Z(G) = langle aZ(G), bZ(G) rangle$.
$endgroup$
– Derek Holt
Feb 16 '18 at 15:18
$begingroup$
You already have $t=a$ obviously (and all powers of $a$ as well), and you don't have $t=b$ otherwise $gin Z(G)$. You could have powers of $b$ though, I can't see why not. Perhaps $p$ being prime prevents that.
$endgroup$
– Arnaud Mortier
Feb 15 '18 at 13:35
$begingroup$
You already have $t=a$ obviously (and all powers of $a$ as well), and you don't have $t=b$ otherwise $gin Z(G)$. You could have powers of $b$ though, I can't see why not. Perhaps $p$ being prime prevents that.
$endgroup$
– Arnaud Mortier
Feb 15 '18 at 13:35
1
1
$begingroup$
For $1 le i < p$, we have $langle a,Z(G) rangle le C_G(a^iz) < G$, and since $|G:langle a,Z(G) rangle|= p$ is prime, we must have $langle a,Z(G) rangle = C_G(a^iz) $.
$endgroup$
– Derek Holt
Feb 15 '18 at 14:11
$begingroup$
For $1 le i < p$, we have $langle a,Z(G) rangle le C_G(a^iz) < G$, and since $|G:langle a,Z(G) rangle|= p$ is prime, we must have $langle a,Z(G) rangle = C_G(a^iz) $.
$endgroup$
– Derek Holt
Feb 15 '18 at 14:11
$begingroup$
@ArnaudMortier I understood the case $t=b$. But I couldn't understand why $b^j not in C_G (a^i )$!
$endgroup$
– user481657
Feb 16 '18 at 14:47
$begingroup$
@ArnaudMortier I understood the case $t=b$. But I couldn't understand why $b^j not in C_G (a^i )$!
$endgroup$
– user481657
Feb 16 '18 at 14:47
$begingroup$
@DerekHolt Why $|G:langle a,Z(G)rangle |=p$?
$endgroup$
– user481657
Feb 16 '18 at 14:59
$begingroup$
@DerekHolt Why $|G:langle a,Z(G)rangle |=p$?
$endgroup$
– user481657
Feb 16 '18 at 14:59
$begingroup$
That follows from $|G/Z(G)| = p^2$ and $Z(G) = langle aZ(G), bZ(G) rangle$.
$endgroup$
– Derek Holt
Feb 16 '18 at 15:18
$begingroup$
That follows from $|G/Z(G)| = p^2$ and $Z(G) = langle aZ(G), bZ(G) rangle$.
$endgroup$
– Derek Holt
Feb 16 '18 at 15:18
|
show 6 more comments
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$begingroup$
You already have $t=a$ obviously (and all powers of $a$ as well), and you don't have $t=b$ otherwise $gin Z(G)$. You could have powers of $b$ though, I can't see why not. Perhaps $p$ being prime prevents that.
$endgroup$
– Arnaud Mortier
Feb 15 '18 at 13:35
1
$begingroup$
For $1 le i < p$, we have $langle a,Z(G) rangle le C_G(a^iz) < G$, and since $|G:langle a,Z(G) rangle|= p$ is prime, we must have $langle a,Z(G) rangle = C_G(a^iz) $.
$endgroup$
– Derek Holt
Feb 15 '18 at 14:11
$begingroup$
@ArnaudMortier I understood the case $t=b$. But I couldn't understand why $b^j not in C_G (a^i )$!
$endgroup$
– user481657
Feb 16 '18 at 14:47
$begingroup$
@DerekHolt Why $|G:langle a,Z(G)rangle |=p$?
$endgroup$
– user481657
Feb 16 '18 at 14:59
$begingroup$
That follows from $|G/Z(G)| = p^2$ and $Z(G) = langle aZ(G), bZ(G) rangle$.
$endgroup$
– Derek Holt
Feb 16 '18 at 15:18