Is it true all centralizer of G are abelian?
$begingroup$
Suppose $G$ is a finite group such that $frac{G}{Z(G)}cong Z_ptimes Z_ptimes Z_p$.
Is it true all centralizers of G are abelian?
finite-groups
asked Dec 9 '18 at 7:21
m a sm a s
333
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a finite group such that $frac{G}{Z(G)}cong Z_ptimes Z_ptimes Z_p$.
Is it true all centralizers of G are abelian?
finite-groups
asked Dec 9 '18 at 7:21
m a sm a s
333
$endgroup$
$begingroup$
I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$.
$endgroup$
– Alex Vong
Dec 9 '18 at 11:20
add a comment |
$begingroup$
Suppose $G$ is a finite group such that $frac{G}{Z(G)}cong Z_ptimes Z_ptimes Z_p$.
Is it true all centralizers of G are abelian?
finite-groups
asked Dec 9 '18 at 7:21
m a sm a s
333
$endgroup$
Suppose $G$ is a finite group such that $frac{G}{Z(G)}cong Z_ptimes Z_ptimes Z_p$.
Is it true all centralizers of G are abelian?
finite-groups
finite-groups
asked Dec 9 '18 at 7:21
m a sm a s
333
asked Dec 9 '18 at 7:21
m a sm a s
333
asked Dec 9 '18 at 7:21
m a sm a s
333
asked Dec 9 '18 at 7:21
m a sm a s
333
asked Dec 9 '18 at 7:21
m a sm a s
333
333
$begingroup$
I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$.
$endgroup$
– Alex Vong
Dec 9 '18 at 11:20
add a comment |
$begingroup$
I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$.
$endgroup$
– Alex Vong
Dec 9 '18 at 11:20
$begingroup$
I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$.
$endgroup$
– Alex Vong
Dec 9 '18 at 11:20
$begingroup$
I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$.
$endgroup$
– Alex Vong
Dec 9 '18 at 11:20
add a comment |
1 Answer
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$begingroup$
As mentioned in the comments, if $G$ was non-abelian, then for $zin Z(G)$, $C(z)=G$ would be a contradiction. But of course, if $G$ is abelian, then $G/Z(G)$ is trivial. So your claim is equivalent to: there exists no finite group $G$ with $G/Z(G)=Z_p^3$.
That's probably a hint this isn't true. Since $G$ is nilpotent, we can restrict attention to finite $p$-groups. Then we can get a counterexample for every $p$:
Let $H$ be the elementary abelian group of order $p^3$, generated by $x,y,z$. Let $alpha,betain Aut(H)$ be two order $p$ elements:
begin{align*}
alpha(x) = xy && alpha(y)=y && alpha(z)=z\
beta(x)=xz && beta(y)=y && beta(z)=z
end{align*}
Then if $G$ is the group of order $p^5$, given by the semidirect product $Hrtimeslanglealpha,betarangle$, we have $Z(G)=langle y,zrangle$ and $G/Z(G)$ is the elementary abelian group generated by the images of $x$, $alpha$, and $beta$.
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
$endgroup$
add a comment |
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$begingroup$
As mentioned in the comments, if $G$ was non-abelian, then for $zin Z(G)$, $C(z)=G$ would be a contradiction. But of course, if $G$ is abelian, then $G/Z(G)$ is trivial. So your claim is equivalent to: there exists no finite group $G$ with $G/Z(G)=Z_p^3$.
That's probably a hint this isn't true. Since $G$ is nilpotent, we can restrict attention to finite $p$-groups. Then we can get a counterexample for every $p$:
Let $H$ be the elementary abelian group of order $p^3$, generated by $x,y,z$. Let $alpha,betain Aut(H)$ be two order $p$ elements:
begin{align*}
alpha(x) = xy && alpha(y)=y && alpha(z)=z\
beta(x)=xz && beta(y)=y && beta(z)=z
end{align*}
Then if $G$ is the group of order $p^5$, given by the semidirect product $Hrtimeslanglealpha,betarangle$, we have $Z(G)=langle y,zrangle$ and $G/Z(G)$ is the elementary abelian group generated by the images of $x$, $alpha$, and $beta$.
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
$endgroup$
add a comment |
$begingroup$
As mentioned in the comments, if $G$ was non-abelian, then for $zin Z(G)$, $C(z)=G$ would be a contradiction. But of course, if $G$ is abelian, then $G/Z(G)$ is trivial. So your claim is equivalent to: there exists no finite group $G$ with $G/Z(G)=Z_p^3$.
That's probably a hint this isn't true. Since $G$ is nilpotent, we can restrict attention to finite $p$-groups. Then we can get a counterexample for every $p$:
Let $H$ be the elementary abelian group of order $p^3$, generated by $x,y,z$. Let $alpha,betain Aut(H)$ be two order $p$ elements:
begin{align*}
alpha(x) = xy && alpha(y)=y && alpha(z)=z\
beta(x)=xz && beta(y)=y && beta(z)=z
end{align*}
Then if $G$ is the group of order $p^5$, given by the semidirect product $Hrtimeslanglealpha,betarangle$, we have $Z(G)=langle y,zrangle$ and $G/Z(G)$ is the elementary abelian group generated by the images of $x$, $alpha$, and $beta$.
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
$endgroup$
add a comment |
$begingroup$
As mentioned in the comments, if $G$ was non-abelian, then for $zin Z(G)$, $C(z)=G$ would be a contradiction. But of course, if $G$ is abelian, then $G/Z(G)$ is trivial. So your claim is equivalent to: there exists no finite group $G$ with $G/Z(G)=Z_p^3$.
That's probably a hint this isn't true. Since $G$ is nilpotent, we can restrict attention to finite $p$-groups. Then we can get a counterexample for every $p$:
Let $H$ be the elementary abelian group of order $p^3$, generated by $x,y,z$. Let $alpha,betain Aut(H)$ be two order $p$ elements:
begin{align*}
alpha(x) = xy && alpha(y)=y && alpha(z)=z\
beta(x)=xz && beta(y)=y && beta(z)=z
end{align*}
Then if $G$ is the group of order $p^5$, given by the semidirect product $Hrtimeslanglealpha,betarangle$, we have $Z(G)=langle y,zrangle$ and $G/Z(G)$ is the elementary abelian group generated by the images of $x$, $alpha$, and $beta$.
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
$endgroup$
As mentioned in the comments, if $G$ was non-abelian, then for $zin Z(G)$, $C(z)=G$ would be a contradiction. But of course, if $G$ is abelian, then $G/Z(G)$ is trivial. So your claim is equivalent to: there exists no finite group $G$ with $G/Z(G)=Z_p^3$.
That's probably a hint this isn't true. Since $G$ is nilpotent, we can restrict attention to finite $p$-groups. Then we can get a counterexample for every $p$:
Let $H$ be the elementary abelian group of order $p^3$, generated by $x,y,z$. Let $alpha,betain Aut(H)$ be two order $p$ elements:
begin{align*}
alpha(x) = xy && alpha(y)=y && alpha(z)=z\
beta(x)=xz && beta(y)=y && beta(z)=z
end{align*}
Then if $G$ is the group of order $p^5$, given by the semidirect product $Hrtimeslanglealpha,betarangle$, we have $Z(G)=langle y,zrangle$ and $G/Z(G)$ is the elementary abelian group generated by the images of $x$, $alpha$, and $beta$.
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
answered Dec 10 '18 at 5:44
HempeliciousHempelicious
147110
147110
add a comment |
add a comment |
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$begingroup$
I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$.
$endgroup$
– Alex Vong
Dec 9 '18 at 11:20