Groups of order $p^2$ abelian
$begingroup$
I know that $|Z(G)|= p or p^2$ In the second case there's is nothing to do. And if $|Z(G)|=p$ then $G/Z(G)$is cyclic hence G abelian.
But my book has a different proof that I really want to understand.
Here it is:
If $|G|=p^2$ and $Z(G) neq G$, then let $a in G - Z (G)$. Then $C(a)$
is a subgroup containing both $a$ and $Z(G)$, with $| Z(G)| = p$. This shows that $C(a) = G$, a contradiction. Thus $Z(G) = G$, and so G is abelian.
I don't see why $Z(G) subseteq C(a)$
and I guess $C(a) = G$ implies $ain Z(G)$, a contradiction.
abstract-algebra group-theory
asked Apr 4 '14 at 21:23
abeabe
594318
$endgroup$
add a comment |
$begingroup$
I know that $|Z(G)|= p or p^2$ In the second case there's is nothing to do. And if $|Z(G)|=p$ then $G/Z(G)$is cyclic hence G abelian.
But my book has a different proof that I really want to understand.
Here it is:
If $|G|=p^2$ and $Z(G) neq G$, then let $a in G - Z (G)$. Then $C(a)$
is a subgroup containing both $a$ and $Z(G)$, with $| Z(G)| = p$. This shows that $C(a) = G$, a contradiction. Thus $Z(G) = G$, and so G is abelian.
I don't see why $Z(G) subseteq C(a)$
and I guess $C(a) = G$ implies $ain Z(G)$, a contradiction.
abstract-algebra group-theory
asked Apr 4 '14 at 21:23
abeabe
594318
$endgroup$
add a comment |
$begingroup$
I know that $|Z(G)|= p or p^2$ In the second case there's is nothing to do. And if $|Z(G)|=p$ then $G/Z(G)$is cyclic hence G abelian.
But my book has a different proof that I really want to understand.
Here it is:
If $|G|=p^2$ and $Z(G) neq G$, then let $a in G - Z (G)$. Then $C(a)$
is a subgroup containing both $a$ and $Z(G)$, with $| Z(G)| = p$. This shows that $C(a) = G$, a contradiction. Thus $Z(G) = G$, and so G is abelian.
I don't see why $Z(G) subseteq C(a)$
and I guess $C(a) = G$ implies $ain Z(G)$, a contradiction.
abstract-algebra group-theory
asked Apr 4 '14 at 21:23
abeabe
594318
$endgroup$
I know that $|Z(G)|= p or p^2$ In the second case there's is nothing to do. And if $|Z(G)|=p$ then $G/Z(G)$is cyclic hence G abelian.
But my book has a different proof that I really want to understand.
Here it is:
If $|G|=p^2$ and $Z(G) neq G$, then let $a in G - Z (G)$. Then $C(a)$
is a subgroup containing both $a$ and $Z(G)$, with $| Z(G)| = p$. This shows that $C(a) = G$, a contradiction. Thus $Z(G) = G$, and so G is abelian.
I don't see why $Z(G) subseteq C(a)$
and I guess $C(a) = G$ implies $ain Z(G)$, a contradiction.
abstract-algebra group-theory
abstract-algebra group-theory
asked Apr 4 '14 at 21:23
abeabe
594318
asked Apr 4 '14 at 21:23
abeabe
594318
edited Apr 4 '14 at 21:33
abe
asked Apr 4 '14 at 21:23
abeabe
594318
asked Apr 4 '14 at 21:23
abeabe
594318
asked Apr 4 '14 at 21:23
abeabe
594318
594318
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Well I'm assuming that $C(a)$ is the centralizer of the set ${a}$ i.e. the group consisting of all elements of $G$ commuting with $a$. Since elements of $Z(G)$ commute with $pmb{all}$ elements of $G$ they commute with $a$ in particular. Thus $Z(G)subset C(a)$. Also you state $|Z(G)|> p$, in the fourth paragraph, but do you mean $|Z(G)| = p$?
Anyway, is it now clear how the result follows (given that the order of a subgroup must divide the order of the group?
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
$endgroup$
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
1
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well I'm assuming that $C(a)$ is the centralizer of the set ${a}$ i.e. the group consisting of all elements of $G$ commuting with $a$. Since elements of $Z(G)$ commute with $pmb{all}$ elements of $G$ they commute with $a$ in particular. Thus $Z(G)subset C(a)$. Also you state $|Z(G)|> p$, in the fourth paragraph, but do you mean $|Z(G)| = p$?
Anyway, is it now clear how the result follows (given that the order of a subgroup must divide the order of the group?
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
$endgroup$
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
1
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
add a comment |
$begingroup$
Well I'm assuming that $C(a)$ is the centralizer of the set ${a}$ i.e. the group consisting of all elements of $G$ commuting with $a$. Since elements of $Z(G)$ commute with $pmb{all}$ elements of $G$ they commute with $a$ in particular. Thus $Z(G)subset C(a)$. Also you state $|Z(G)|> p$, in the fourth paragraph, but do you mean $|Z(G)| = p$?
Anyway, is it now clear how the result follows (given that the order of a subgroup must divide the order of the group?
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
$endgroup$
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
1
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
add a comment |
$begingroup$
Well I'm assuming that $C(a)$ is the centralizer of the set ${a}$ i.e. the group consisting of all elements of $G$ commuting with $a$. Since elements of $Z(G)$ commute with $pmb{all}$ elements of $G$ they commute with $a$ in particular. Thus $Z(G)subset C(a)$. Also you state $|Z(G)|> p$, in the fourth paragraph, but do you mean $|Z(G)| = p$?
Anyway, is it now clear how the result follows (given that the order of a subgroup must divide the order of the group?
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
$endgroup$
Well I'm assuming that $C(a)$ is the centralizer of the set ${a}$ i.e. the group consisting of all elements of $G$ commuting with $a$. Since elements of $Z(G)$ commute with $pmb{all}$ elements of $G$ they commute with $a$ in particular. Thus $Z(G)subset C(a)$. Also you state $|Z(G)|> p$, in the fourth paragraph, but do you mean $|Z(G)| = p$?
Anyway, is it now clear how the result follows (given that the order of a subgroup must divide the order of the group?
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
edited Jan 17 at 0:25
J. W. Tanner
4,7721420
4,7721420
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
answered Apr 4 '14 at 21:31
Daniel MckenzieDaniel Mckenzie
617512
617512
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
1
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
add a comment |
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
1
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Thanks :D, your wording made it easier to understand.
$endgroup$
– abe
Apr 4 '14 at 21:49
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Since $Z(G)=p$ and $a,Z(G)subset C(a) rightarrow C(a) geq p+1$ but since $|G|$divides$ | C(a)|$, $|C(a)|=p^2$, So from the class equation $p^2=|G|neq p +p^2 = |Z(G)|+ sum_{forall xin G-Z(G)}[G:C(x)]$ Contradiction
$endgroup$
– abe
Apr 4 '14 at 21:53
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
$begingroup$
Does it look good?
$endgroup$
– abe
Apr 4 '14 at 21:55
1
1
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
$begingroup$
Looks good, but you could make this shorter. By Lagrange's theorem $|C(a)| | |G| = p^{2}$ so $|C(a)| = 1,p$ or $p^{2}$ ($p$ is prime so these are the only factors of $p^{2}$). Since $|Z(G)| = p$, $anotin Z(G)$ and ${a}, Z(G)subset C(a)$ we must have $|C(a)| geq p+1$. But then by the first line $|C(a)| = p^{2}$, thus $C(a) = G$ implying that $a$ commutes with everything in $G$, contradicting the assumption that $anotin Z(G)$
$endgroup$
– Daniel Mckenzie
Apr 5 '14 at 20:21
add a comment |
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