Conjugacy classes of non-Abelian group of order $p^3$
$begingroup$
Let $G$ be a non-abelian group of order $p^3$. How many are its
conjugacy classes?
The conjugacy classes are the orbits of $G$ under conjugation of
$G$ by itself. Since $G$ is non-abelian, its center has order $p$.
So the class equation yields
$p^3 = p + sum_{[x]} (G: G_x)$, where $G_x$ is the centralizer of $x$
and the sum is taken over disjoint orbits $[x]$. We can also see that
$(G:G_x)$ can only be $p$ or $p^2$. So we will have $p$ orbits of length
$1$ and then orbits of length $p$ and $p^2$. Any hints on determining the
number of the latter?
abstract-algebra group-theory finite-groups
edited Sep 8 '17 at 22:55
Ali Heydari
966
asked Oct 12 '11 at 15:57
ManosManos
14k33188
$endgroup$
add a comment |
$begingroup$
Let $G$ be a non-abelian group of order $p^3$. How many are its
conjugacy classes?
The conjugacy classes are the orbits of $G$ under conjugation of
$G$ by itself. Since $G$ is non-abelian, its center has order $p$.
So the class equation yields
$p^3 = p + sum_{[x]} (G: G_x)$, where $G_x$ is the centralizer of $x$
and the sum is taken over disjoint orbits $[x]$. We can also see that
$(G:G_x)$ can only be $p$ or $p^2$. So we will have $p$ orbits of length
$1$ and then orbits of length $p$ and $p^2$. Any hints on determining the
number of the latter?
abstract-algebra group-theory finite-groups
edited Sep 8 '17 at 22:55
Ali Heydari
966
asked Oct 12 '11 at 15:57
ManosManos
14k33188
$endgroup$
add a comment |
$begingroup$
Let $G$ be a non-abelian group of order $p^3$. How many are its
conjugacy classes?
The conjugacy classes are the orbits of $G$ under conjugation of
$G$ by itself. Since $G$ is non-abelian, its center has order $p$.
So the class equation yields
$p^3 = p + sum_{[x]} (G: G_x)$, where $G_x$ is the centralizer of $x$
and the sum is taken over disjoint orbits $[x]$. We can also see that
$(G:G_x)$ can only be $p$ or $p^2$. So we will have $p$ orbits of length
$1$ and then orbits of length $p$ and $p^2$. Any hints on determining the
number of the latter?
abstract-algebra group-theory finite-groups
edited Sep 8 '17 at 22:55
Ali Heydari
966
asked Oct 12 '11 at 15:57
ManosManos
14k33188
$endgroup$
Let $G$ be a non-abelian group of order $p^3$. How many are its
conjugacy classes?
The conjugacy classes are the orbits of $G$ under conjugation of
$G$ by itself. Since $G$ is non-abelian, its center has order $p$.
So the class equation yields
$p^3 = p + sum_{[x]} (G: G_x)$, where $G_x$ is the centralizer of $x$
and the sum is taken over disjoint orbits $[x]$. We can also see that
$(G:G_x)$ can only be $p$ or $p^2$. So we will have $p$ orbits of length
$1$ and then orbits of length $p$ and $p^2$. Any hints on determining the
number of the latter?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Sep 8 '17 at 22:55
Ali Heydari
966
asked Oct 12 '11 at 15:57
ManosManos
14k33188
edited Sep 8 '17 at 22:55
Ali Heydari
966
asked Oct 12 '11 at 15:57
ManosManos
14k33188
edited Sep 8 '17 at 22:55
Ali Heydari
966
edited Sep 8 '17 at 22:55
Ali Heydari
966
edited Sep 8 '17 at 22:55
Ali Heydari
966
966
asked Oct 12 '11 at 15:57
ManosManos
14k33188
asked Oct 12 '11 at 15:57
ManosManos
14k33188
asked Oct 12 '11 at 15:57
ManosManos
14k33188
14k33188
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
$endgroup$
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
add a comment |
$begingroup$
Suppose $g$ is a noncentral element. Then the centralizer of $g$ contains the centre of $G$, and also contains $g$ itself, giving it at least $p+1$ elements. Thus the centralizer has order $p^2$, so $g$ has $p$ conjugates. There are thus $p+frac{p^3-p}{p}=p^2+p-1$ conjugacy classes.
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
$endgroup$
add a comment |
$begingroup$
The centraliser of a non-central element contains at least $p+1$ elements, since it contains the centre, with order $p$. It cannot have order equal to $p^3$, or the element would be central. Thus, the order of the centraliser or a non-central element is equal to $p^2$, and hence the index is equal to $p$, which is to say that it has $p$ conjugates. Therefore, there are $frac{p^3 - p}{p} = p^2 - 1$ conjugacy classes of non-central elements, giving a total of $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
$endgroup$
add a comment |
$begingroup$
One can break it up into 3 steps. First, show that $|Z(G)|$ is equal to p. Second, show that if $a$ is not in $Z(G)$ then $|N_G(a)|=p^2$. Finally, the last step is to use the class equation to show that there are $p^2+p-1$ conjugacy classes.
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
$endgroup$
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
add a comment |
$begingroup$
We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
$endgroup$
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
add a comment |
$begingroup$
We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
$endgroup$
We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
edited Oct 12 '11 at 16:30
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
answered Oct 12 '11 at 16:17
Mikko KorhonenMikko Korhonen
18.6k34692
18.6k34692
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
add a comment |
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
$begingroup$
Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part.
$endgroup$
– awllower
Jun 3 '12 at 17:57
add a comment |
$begingroup$
Suppose $g$ is a noncentral element. Then the centralizer of $g$ contains the centre of $G$, and also contains $g$ itself, giving it at least $p+1$ elements. Thus the centralizer has order $p^2$, so $g$ has $p$ conjugates. There are thus $p+frac{p^3-p}{p}=p^2+p-1$ conjugacy classes.
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
$endgroup$
add a comment |
$begingroup$
Suppose $g$ is a noncentral element. Then the centralizer of $g$ contains the centre of $G$, and also contains $g$ itself, giving it at least $p+1$ elements. Thus the centralizer has order $p^2$, so $g$ has $p$ conjugates. There are thus $p+frac{p^3-p}{p}=p^2+p-1$ conjugacy classes.
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
$endgroup$
add a comment |
$begingroup$
Suppose $g$ is a noncentral element. Then the centralizer of $g$ contains the centre of $G$, and also contains $g$ itself, giving it at least $p+1$ elements. Thus the centralizer has order $p^2$, so $g$ has $p$ conjugates. There are thus $p+frac{p^3-p}{p}=p^2+p-1$ conjugacy classes.
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
$endgroup$
Suppose $g$ is a noncentral element. Then the centralizer of $g$ contains the centre of $G$, and also contains $g$ itself, giving it at least $p+1$ elements. Thus the centralizer has order $p^2$, so $g$ has $p$ conjugates. There are thus $p+frac{p^3-p}{p}=p^2+p-1$ conjugacy classes.
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
answered Oct 12 '11 at 16:08
Chris EagleChris Eagle
29.1k27199
29.1k27199
add a comment |
add a comment |
$begingroup$
The centraliser of a non-central element contains at least $p+1$ elements, since it contains the centre, with order $p$. It cannot have order equal to $p^3$, or the element would be central. Thus, the order of the centraliser or a non-central element is equal to $p^2$, and hence the index is equal to $p$, which is to say that it has $p$ conjugates. Therefore, there are $frac{p^3 - p}{p} = p^2 - 1$ conjugacy classes of non-central elements, giving a total of $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
$endgroup$
add a comment |
$begingroup$
The centraliser of a non-central element contains at least $p+1$ elements, since it contains the centre, with order $p$. It cannot have order equal to $p^3$, or the element would be central. Thus, the order of the centraliser or a non-central element is equal to $p^2$, and hence the index is equal to $p$, which is to say that it has $p$ conjugates. Therefore, there are $frac{p^3 - p}{p} = p^2 - 1$ conjugacy classes of non-central elements, giving a total of $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
$endgroup$
add a comment |
$begingroup$
The centraliser of a non-central element contains at least $p+1$ elements, since it contains the centre, with order $p$. It cannot have order equal to $p^3$, or the element would be central. Thus, the order of the centraliser or a non-central element is equal to $p^2$, and hence the index is equal to $p$, which is to say that it has $p$ conjugates. Therefore, there are $frac{p^3 - p}{p} = p^2 - 1$ conjugacy classes of non-central elements, giving a total of $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
$endgroup$
The centraliser of a non-central element contains at least $p+1$ elements, since it contains the centre, with order $p$. It cannot have order equal to $p^3$, or the element would be central. Thus, the order of the centraliser or a non-central element is equal to $p^2$, and hence the index is equal to $p$, which is to say that it has $p$ conjugates. Therefore, there are $frac{p^3 - p}{p} = p^2 - 1$ conjugacy classes of non-central elements, giving a total of $p^2 + p - 1$ conjugacy classes.
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
answered Oct 12 '11 at 16:36
JamesJames
7,09221527
7,09221527
add a comment |
add a comment |
$begingroup$
One can break it up into 3 steps. First, show that $|Z(G)|$ is equal to p. Second, show that if $a$ is not in $Z(G)$ then $|N_G(a)|=p^2$. Finally, the last step is to use the class equation to show that there are $p^2+p-1$ conjugacy classes.
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
$endgroup$
add a comment |
$begingroup$
One can break it up into 3 steps. First, show that $|Z(G)|$ is equal to p. Second, show that if $a$ is not in $Z(G)$ then $|N_G(a)|=p^2$. Finally, the last step is to use the class equation to show that there are $p^2+p-1$ conjugacy classes.
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
$endgroup$
add a comment |
$begingroup$
One can break it up into 3 steps. First, show that $|Z(G)|$ is equal to p. Second, show that if $a$ is not in $Z(G)$ then $|N_G(a)|=p^2$. Finally, the last step is to use the class equation to show that there are $p^2+p-1$ conjugacy classes.
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
$endgroup$
One can break it up into 3 steps. First, show that $|Z(G)|$ is equal to p. Second, show that if $a$ is not in $Z(G)$ then $|N_G(a)|=p^2$. Finally, the last step is to use the class equation to show that there are $p^2+p-1$ conjugacy classes.
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
answered Nov 25 '13 at 19:09
Israel HIsrael H
311
311
add a comment |
add a comment |
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