Does this group theory question require an additional hypothesis?
The problem is to
Show that if $G$ is a finite group and for all nontrivial elements $a, b$ there exists an automorphism taking $a$ to $b$, then $G$ is a $C_p$ vector space, where $C_p$ is the group of prime order $p$.
My question is if an additional hypothesis that $G$ is abelian is needed.
I cannot seem to prove that $G$ is abelian from the hypotheses.
Of course, the point is that all the elements of $G$ have the same prime order $p$.
But I cannot seem to get the result without showing $G$ is abelian, in which case the normality of all subgroups gives what I want.
linear-algebra finite-groups abelian-groups group-homomorphism
edited Nov 28 '18 at 23:51
Shaun
8,805113680
asked Oct 28 '14 at 1:39
user187877
162
|
show 1 more comment
The problem is to
Show that if $G$ is a finite group and for all nontrivial elements $a, b$ there exists an automorphism taking $a$ to $b$, then $G$ is a $C_p$ vector space, where $C_p$ is the group of prime order $p$.
My question is if an additional hypothesis that $G$ is abelian is needed.
I cannot seem to prove that $G$ is abelian from the hypotheses.
Of course, the point is that all the elements of $G$ have the same prime order $p$.
But I cannot seem to get the result without showing $G$ is abelian, in which case the normality of all subgroups gives what I want.
linear-algebra finite-groups abelian-groups group-homomorphism
edited Nov 28 '18 at 23:51
Shaun
8,805113680
asked Oct 28 '14 at 1:39
user187877
162
If the additional hypothesis is not necessary, then please don't answer the question, by the way.
– user187877
Oct 28 '14 at 1:43
Are you sure you can't show that the group is already abelian by your hypothesis?
– Jonny
Oct 28 '14 at 1:46
Can you? I gave it a try but not too hard of a try.
– user187877
Oct 28 '14 at 1:48
I just saw this result online in the case of abelian groups. But my professor phrased it this way, and I thought he was just trying to be sneaky about saying every element has the same order. But now I'm not so sure.
– user187877
Oct 28 '14 at 1:49
2
These hypotheses can show that $G$ is abelian, so this solves the problem.
– user187877
Oct 28 '14 at 2:24
|
show 1 more comment
The problem is to
Show that if $G$ is a finite group and for all nontrivial elements $a, b$ there exists an automorphism taking $a$ to $b$, then $G$ is a $C_p$ vector space, where $C_p$ is the group of prime order $p$.
My question is if an additional hypothesis that $G$ is abelian is needed.
I cannot seem to prove that $G$ is abelian from the hypotheses.
Of course, the point is that all the elements of $G$ have the same prime order $p$.
But I cannot seem to get the result without showing $G$ is abelian, in which case the normality of all subgroups gives what I want.
linear-algebra finite-groups abelian-groups group-homomorphism
edited Nov 28 '18 at 23:51
Shaun
8,805113680
asked Oct 28 '14 at 1:39
user187877
162
The problem is to
Show that if $G$ is a finite group and for all nontrivial elements $a, b$ there exists an automorphism taking $a$ to $b$, then $G$ is a $C_p$ vector space, where $C_p$ is the group of prime order $p$.
My question is if an additional hypothesis that $G$ is abelian is needed.
I cannot seem to prove that $G$ is abelian from the hypotheses.
Of course, the point is that all the elements of $G$ have the same prime order $p$.
But I cannot seem to get the result without showing $G$ is abelian, in which case the normality of all subgroups gives what I want.
linear-algebra finite-groups abelian-groups group-homomorphism
linear-algebra finite-groups abelian-groups group-homomorphism
edited Nov 28 '18 at 23:51
Shaun
8,805113680
asked Oct 28 '14 at 1:39
user187877
162
edited Nov 28 '18 at 23:51
Shaun
8,805113680
asked Oct 28 '14 at 1:39
user187877
162
edited Nov 28 '18 at 23:51
Shaun
8,805113680
edited Nov 28 '18 at 23:51
Shaun
8,805113680
edited Nov 28 '18 at 23:51
Shaun
8,805113680
8,805113680
asked Oct 28 '14 at 1:39
user187877
162
asked Oct 28 '14 at 1:39
user187877
162
asked Oct 28 '14 at 1:39
user187877
162
162
If the additional hypothesis is not necessary, then please don't answer the question, by the way.
– user187877
Oct 28 '14 at 1:43
Are you sure you can't show that the group is already abelian by your hypothesis?
– Jonny
Oct 28 '14 at 1:46
Can you? I gave it a try but not too hard of a try.
– user187877
Oct 28 '14 at 1:48
I just saw this result online in the case of abelian groups. But my professor phrased it this way, and I thought he was just trying to be sneaky about saying every element has the same order. But now I'm not so sure.
– user187877
Oct 28 '14 at 1:49
2
These hypotheses can show that $G$ is abelian, so this solves the problem.
– user187877
Oct 28 '14 at 2:24
|
show 1 more comment
If the additional hypothesis is not necessary, then please don't answer the question, by the way.
– user187877
Oct 28 '14 at 1:43
Are you sure you can't show that the group is already abelian by your hypothesis?
– Jonny
Oct 28 '14 at 1:46
Can you? I gave it a try but not too hard of a try.
– user187877
Oct 28 '14 at 1:48
I just saw this result online in the case of abelian groups. But my professor phrased it this way, and I thought he was just trying to be sneaky about saying every element has the same order. But now I'm not so sure.
– user187877
Oct 28 '14 at 1:49
2
These hypotheses can show that $G$ is abelian, so this solves the problem.
– user187877
Oct 28 '14 at 2:24
If the additional hypothesis is not necessary, then please don't answer the question, by the way.
– user187877
Oct 28 '14 at 1:43
If the additional hypothesis is not necessary, then please don't answer the question, by the way.
– user187877
Oct 28 '14 at 1:43
Are you sure you can't show that the group is already abelian by your hypothesis?
– Jonny
Oct 28 '14 at 1:46
Are you sure you can't show that the group is already abelian by your hypothesis?
– Jonny
Oct 28 '14 at 1:46
Can you? I gave it a try but not too hard of a try.
– user187877
Oct 28 '14 at 1:48
Can you? I gave it a try but not too hard of a try.
– user187877
Oct 28 '14 at 1:48
I just saw this result online in the case of abelian groups. But my professor phrased it this way, and I thought he was just trying to be sneaky about saying every element has the same order. But now I'm not so sure.
– user187877
Oct 28 '14 at 1:49
I just saw this result online in the case of abelian groups. But my professor phrased it this way, and I thought he was just trying to be sneaky about saying every element has the same order. But now I'm not so sure.
– user187877
Oct 28 '14 at 1:49
2
2
These hypotheses can show that $G$ is abelian, so this solves the problem.
– user187877
Oct 28 '14 at 2:24
These hypotheses can show that $G$ is abelian, so this solves the problem.
– user187877
Oct 28 '14 at 2:24
|
show 1 more comment
1 Answer
1
active
oldest
votes
We may assume $G$ is not trivial. As mentioned in the statement of the problem, it is quick to go from the hypotheses to the fact that $G$ is a $p$-group. Now, every nontrivial $p$-group has a nontrivial center. Let $z$ be some nonzero element of the center. Given $g ,h in G$, by hypothesis we may choose an automorphism $phi$ carrying $g$ to $z$. Then, $phi(gh) = phi(g)phi(h) = phi(h)phi(g) = phi(hg)$. Since $phi$ is injective, we see that $gh = hg$. So, $G$ is abelian.
answered Oct 29 '14 at 2:59
user187877
162
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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oldest
votes
We may assume $G$ is not trivial. As mentioned in the statement of the problem, it is quick to go from the hypotheses to the fact that $G$ is a $p$-group. Now, every nontrivial $p$-group has a nontrivial center. Let $z$ be some nonzero element of the center. Given $g ,h in G$, by hypothesis we may choose an automorphism $phi$ carrying $g$ to $z$. Then, $phi(gh) = phi(g)phi(h) = phi(h)phi(g) = phi(hg)$. Since $phi$ is injective, we see that $gh = hg$. So, $G$ is abelian.
answered Oct 29 '14 at 2:59
user187877
162
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
add a comment |
We may assume $G$ is not trivial. As mentioned in the statement of the problem, it is quick to go from the hypotheses to the fact that $G$ is a $p$-group. Now, every nontrivial $p$-group has a nontrivial center. Let $z$ be some nonzero element of the center. Given $g ,h in G$, by hypothesis we may choose an automorphism $phi$ carrying $g$ to $z$. Then, $phi(gh) = phi(g)phi(h) = phi(h)phi(g) = phi(hg)$. Since $phi$ is injective, we see that $gh = hg$. So, $G$ is abelian.
answered Oct 29 '14 at 2:59
user187877
162
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
add a comment |
We may assume $G$ is not trivial. As mentioned in the statement of the problem, it is quick to go from the hypotheses to the fact that $G$ is a $p$-group. Now, every nontrivial $p$-group has a nontrivial center. Let $z$ be some nonzero element of the center. Given $g ,h in G$, by hypothesis we may choose an automorphism $phi$ carrying $g$ to $z$. Then, $phi(gh) = phi(g)phi(h) = phi(h)phi(g) = phi(hg)$. Since $phi$ is injective, we see that $gh = hg$. So, $G$ is abelian.
answered Oct 29 '14 at 2:59
user187877
162
We may assume $G$ is not trivial. As mentioned in the statement of the problem, it is quick to go from the hypotheses to the fact that $G$ is a $p$-group. Now, every nontrivial $p$-group has a nontrivial center. Let $z$ be some nonzero element of the center. Given $g ,h in G$, by hypothesis we may choose an automorphism $phi$ carrying $g$ to $z$. Then, $phi(gh) = phi(g)phi(h) = phi(h)phi(g) = phi(hg)$. Since $phi$ is injective, we see that $gh = hg$. So, $G$ is abelian.
answered Oct 29 '14 at 2:59
user187877
162
answered Oct 29 '14 at 2:59
user187877
162
answered Oct 29 '14 at 2:59
user187877
162
answered Oct 29 '14 at 2:59
user187877
162
162
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
add a comment |
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
Note that for a group automorphism $phi$, the order of $g$ = order of $phi(g)$. So what can be said about the order of G?
– Joel Pereira
Nov 29 '18 at 3:49
add a comment |
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If the additional hypothesis is not necessary, then please don't answer the question, by the way.
– user187877
Oct 28 '14 at 1:43
Are you sure you can't show that the group is already abelian by your hypothesis?
– Jonny
Oct 28 '14 at 1:46
Can you? I gave it a try but not too hard of a try.
– user187877
Oct 28 '14 at 1:48
I just saw this result online in the case of abelian groups. But my professor phrased it this way, and I thought he was just trying to be sneaky about saying every element has the same order. But now I'm not so sure.
– user187877
Oct 28 '14 at 1:49
2
These hypotheses can show that $G$ is abelian, so this solves the problem.
– user187877
Oct 28 '14 at 2:24