Mixing
Let $G$ be finite and every nonidentity element have prime order. If $Z(G)neq{e}$, prove that every...
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This question appears to be new here.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.51.$^dagger$
Suppose that $G$ is a finite group with the property that every nonidentity element has prime order (e.g., $D_3$). If $Z(G)$ is not trivial, prove that every nonidentity element of $G$ has the same order.
Thoughts:
Lemma: If $G$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.
Proof: If $G$ is abelian (i.e., $G=Z(G)$), then consider $g,hin G$ such that $lvert grvert=p$ and $lvert hrvert=q$ for distinct primes $p$ and $q$. We have
begin{align}
(gh)^{pq}&=(g^p)^q(h^q)^p\
&=e,
end{align}
so that $lvert ghrvert$ divides $pq$.
If $lvert ghrvert=pq$, then it is composite, a contradiction; thus without loss of generality $lvert ghrvert=p$. Now we have
begin{align}
e&=(gh)^p\
&=g^ph^p\
&=eh^p \
&=h^p,
end{align}
but now $qmid p$, which is a contradiction since $pneq q$ and $p$ is prime.
Thus all nonidentity elements of $G$ have the same prime order.$square$
That's all I have so far.
I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.
Edit:
This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $G$ comes into play.
Please help :)
$dagger$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.
group-theory finite-groups
asked Jan 16 at 11:51
ShaunShaun
9,268113684
$endgroup$
|
show 1 more comment
$begingroup$
This question appears to be new here.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.51.$^dagger$
Suppose that $G$ is a finite group with the property that every nonidentity element has prime order (e.g., $D_3$). If $Z(G)$ is not trivial, prove that every nonidentity element of $G$ has the same order.
Thoughts:
Lemma: If $G$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.
Proof: If $G$ is abelian (i.e., $G=Z(G)$), then consider $g,hin G$ such that $lvert grvert=p$ and $lvert hrvert=q$ for distinct primes $p$ and $q$. We have
begin{align}
(gh)^{pq}&=(g^p)^q(h^q)^p\
&=e,
end{align}
so that $lvert ghrvert$ divides $pq$.
If $lvert ghrvert=pq$, then it is composite, a contradiction; thus without loss of generality $lvert ghrvert=p$. Now we have
begin{align}
e&=(gh)^p\
&=g^ph^p\
&=eh^p \
&=h^p,
end{align}
but now $qmid p$, which is a contradiction since $pneq q$ and $p$ is prime.
Thus all nonidentity elements of $G$ have the same prime order.$square$
That's all I have so far.
I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.
Edit:
This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $G$ comes into play.
Please help :)
$dagger$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.
group-theory finite-groups
asked Jan 16 at 11:51
ShaunShaun
9,268113684
$endgroup$
2
$begingroup$
How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center?
$endgroup$
– Mindlack
Jan 16 at 11:55
$begingroup$
I see what you mean, @Mindlack! That's a fun little trick! Thank you.
$endgroup$
– Shaun
Jan 16 at 11:57
$begingroup$
But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack.
$endgroup$
– Shaun
Jan 16 at 11:58
3
$begingroup$
@Shaun They might just be trying to avoid awkward questions like "is infinity prime".
$endgroup$
– user3482749
Jan 16 at 12:17
1
$begingroup$
@Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it.
$endgroup$
– user3482749
Jan 16 at 12:24
|
show 1 more comment
$begingroup$
This question appears to be new here.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.51.$^dagger$
Suppose that $G$ is a finite group with the property that every nonidentity element has prime order (e.g., $D_3$). If $Z(G)$ is not trivial, prove that every nonidentity element of $G$ has the same order.
Thoughts:
Lemma: If $G$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.
Proof: If $G$ is abelian (i.e., $G=Z(G)$), then consider $g,hin G$ such that $lvert grvert=p$ and $lvert hrvert=q$ for distinct primes $p$ and $q$. We have
begin{align}
(gh)^{pq}&=(g^p)^q(h^q)^p\
&=e,
end{align}
so that $lvert ghrvert$ divides $pq$.
If $lvert ghrvert=pq$, then it is composite, a contradiction; thus without loss of generality $lvert ghrvert=p$. Now we have
begin{align}
e&=(gh)^p\
&=g^ph^p\
&=eh^p \
&=h^p,
end{align}
but now $qmid p$, which is a contradiction since $pneq q$ and $p$ is prime.
Thus all nonidentity elements of $G$ have the same prime order.$square$
That's all I have so far.
I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.
Edit:
This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $G$ comes into play.
Please help :)
$dagger$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.
group-theory finite-groups
asked Jan 16 at 11:51
ShaunShaun
9,268113684
$endgroup$
This question appears to be new here.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.51.$^dagger$
Suppose that $G$ is a finite group with the property that every nonidentity element has prime order (e.g., $D_3$). If $Z(G)$ is not trivial, prove that every nonidentity element of $G$ has the same order.
Thoughts:
Lemma: If $G$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.
Proof: If $G$ is abelian (i.e., $G=Z(G)$), then consider $g,hin G$ such that $lvert grvert=p$ and $lvert hrvert=q$ for distinct primes $p$ and $q$. We have
begin{align}
(gh)^{pq}&=(g^p)^q(h^q)^p\
&=e,
end{align}
so that $lvert ghrvert$ divides $pq$.
If $lvert ghrvert=pq$, then it is composite, a contradiction; thus without loss of generality $lvert ghrvert=p$. Now we have
begin{align}
e&=(gh)^p\
&=g^ph^p\
&=eh^p \
&=h^p,
end{align}
but now $qmid p$, which is a contradiction since $pneq q$ and $p$ is prime.
Thus all nonidentity elements of $G$ have the same prime order.$square$
That's all I have so far.
I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.
Edit:
This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $G$ comes into play.
Please help :)
$dagger$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.
group-theory finite-groups
group-theory finite-groups
asked Jan 16 at 11:51
ShaunShaun
9,268113684
asked Jan 16 at 11:51
ShaunShaun
9,268113684
edited Jan 16 at 15:52
Shaun
asked Jan 16 at 11:51
ShaunShaun
9,268113684
asked Jan 16 at 11:51
ShaunShaun
9,268113684
asked Jan 16 at 11:51
ShaunShaun
9,268113684
9,268113684
2
$begingroup$
How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center?
$endgroup$
– Mindlack
Jan 16 at 11:55
$begingroup$
I see what you mean, @Mindlack! That's a fun little trick! Thank you.
$endgroup$
– Shaun
Jan 16 at 11:57
$begingroup$
But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack.
$endgroup$
– Shaun
Jan 16 at 11:58
3
$begingroup$
@Shaun They might just be trying to avoid awkward questions like "is infinity prime".
$endgroup$
– user3482749
Jan 16 at 12:17
1
$begingroup$
@Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it.
$endgroup$
– user3482749
Jan 16 at 12:24
|
show 1 more comment
2
$begingroup$
How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center?
$endgroup$
– Mindlack
Jan 16 at 11:55
$begingroup$
I see what you mean, @Mindlack! That's a fun little trick! Thank you.
$endgroup$
– Shaun
Jan 16 at 11:57
$begingroup$
But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack.
$endgroup$
– Shaun
Jan 16 at 11:58
3
$begingroup$
@Shaun They might just be trying to avoid awkward questions like "is infinity prime".
$endgroup$
– user3482749
Jan 16 at 12:17
1
$begingroup$
@Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it.
$endgroup$
– user3482749
Jan 16 at 12:24
2
2
$begingroup$
How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center?
$endgroup$
– Mindlack
Jan 16 at 11:55
$begingroup$
How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center?
$endgroup$
– Mindlack
Jan 16 at 11:55
$begingroup$
I see what you mean, @Mindlack! That's a fun little trick! Thank you.
$endgroup$
– Shaun
Jan 16 at 11:57
$begingroup$
I see what you mean, @Mindlack! That's a fun little trick! Thank you.
$endgroup$
– Shaun
Jan 16 at 11:57
$begingroup$
But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack.
$endgroup$
– Shaun
Jan 16 at 11:58
$begingroup$
But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack.
$endgroup$
– Shaun
Jan 16 at 11:58
3
3
$begingroup$
@Shaun They might just be trying to avoid awkward questions like "is infinity prime".
$endgroup$
– user3482749
Jan 16 at 12:17
$begingroup$
@Shaun They might just be trying to avoid awkward questions like "is infinity prime".
$endgroup$
– user3482749
Jan 16 at 12:17
1
1
$begingroup$
@Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it.
$endgroup$
– user3482749
Jan 16 at 12:24
$begingroup$
@Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it.
$endgroup$
– user3482749
Jan 16 at 12:24
|
show 1 more comment
1 Answer
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oldest
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The following paraphrases the proof in the solutions section of the book the exercise is from.
Let $zin Z(G)$ such that $lvert zrvert=p$ for a prime $p$. Consider $gin G$. We have $lvert grvert=q$ is prime. Then
begin{align}
(zg)^{pq}&=(z^p)^q(g^q)^p \
&=e^qe^p \
&=e
end{align}
and thus $lvert zgrvertin{p, q}$ (since it has to be prime). Assume without loss of generality that $lvert zgrvert=p$. Then
begin{align}
e&=(zg)^p \
&=z^pg^p \
&=eg^p \
&=g^p,
end{align}
so $qmid p$. Hence $p=q$.
answered Jan 16 at 15:35
ShaunShaun
9,268113684
$endgroup$
add a comment |
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$begingroup$
The following paraphrases the proof in the solutions section of the book the exercise is from.
Let $zin Z(G)$ such that $lvert zrvert=p$ for a prime $p$. Consider $gin G$. We have $lvert grvert=q$ is prime. Then
begin{align}
(zg)^{pq}&=(z^p)^q(g^q)^p \
&=e^qe^p \
&=e
end{align}
and thus $lvert zgrvertin{p, q}$ (since it has to be prime). Assume without loss of generality that $lvert zgrvert=p$. Then
begin{align}
e&=(zg)^p \
&=z^pg^p \
&=eg^p \
&=g^p,
end{align}
so $qmid p$. Hence $p=q$.
answered Jan 16 at 15:35
ShaunShaun
9,268113684
$endgroup$
add a comment |
$begingroup$
The following paraphrases the proof in the solutions section of the book the exercise is from.
Let $zin Z(G)$ such that $lvert zrvert=p$ for a prime $p$. Consider $gin G$. We have $lvert grvert=q$ is prime. Then
begin{align}
(zg)^{pq}&=(z^p)^q(g^q)^p \
&=e^qe^p \
&=e
end{align}
and thus $lvert zgrvertin{p, q}$ (since it has to be prime). Assume without loss of generality that $lvert zgrvert=p$. Then
begin{align}
e&=(zg)^p \
&=z^pg^p \
&=eg^p \
&=g^p,
end{align}
so $qmid p$. Hence $p=q$.
answered Jan 16 at 15:35
ShaunShaun
9,268113684
$endgroup$
add a comment |
$begingroup$
The following paraphrases the proof in the solutions section of the book the exercise is from.
Let $zin Z(G)$ such that $lvert zrvert=p$ for a prime $p$. Consider $gin G$. We have $lvert grvert=q$ is prime. Then
begin{align}
(zg)^{pq}&=(z^p)^q(g^q)^p \
&=e^qe^p \
&=e
end{align}
and thus $lvert zgrvertin{p, q}$ (since it has to be prime). Assume without loss of generality that $lvert zgrvert=p$. Then
begin{align}
e&=(zg)^p \
&=z^pg^p \
&=eg^p \
&=g^p,
end{align}
so $qmid p$. Hence $p=q$.
answered Jan 16 at 15:35
ShaunShaun
9,268113684
$endgroup$
The following paraphrases the proof in the solutions section of the book the exercise is from.
Let $zin Z(G)$ such that $lvert zrvert=p$ for a prime $p$. Consider $gin G$. We have $lvert grvert=q$ is prime. Then
begin{align}
(zg)^{pq}&=(z^p)^q(g^q)^p \
&=e^qe^p \
&=e
end{align}
and thus $lvert zgrvertin{p, q}$ (since it has to be prime). Assume without loss of generality that $lvert zgrvert=p$. Then
begin{align}
e&=(zg)^p \
&=z^pg^p \
&=eg^p \
&=g^p,
end{align}
so $qmid p$. Hence $p=q$.
answered Jan 16 at 15:35
ShaunShaun
9,268113684
answered Jan 16 at 15:35
ShaunShaun
9,268113684
answered Jan 16 at 15:35
ShaunShaun
9,268113684
answered Jan 16 at 15:35
ShaunShaun
9,268113684
9,268113684
add a comment |
add a comment |
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2
$begingroup$
How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center?
$endgroup$
– Mindlack
Jan 16 at 11:55
$begingroup$
I see what you mean, @Mindlack! That's a fun little trick! Thank you.
$endgroup$
– Shaun
Jan 16 at 11:57
$begingroup$
But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack.
$endgroup$
– Shaun
Jan 16 at 11:58
3
$begingroup$
@Shaun They might just be trying to avoid awkward questions like "is infinity prime".
$endgroup$
– user3482749
Jan 16 at 12:17
1
$begingroup$
@Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it.
$endgroup$
– user3482749
Jan 16 at 12:24