Existence of elements of order $p^k$ in a finite group
By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.
My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?
Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists.
group-theory finite-groups
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
add a comment |
By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.
My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?
Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists.
group-theory finite-groups
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
2
It's good to keep small examples of finite groups in mind. Here, the non-cyclic group of order $4$ is already a counterexample.
– lulu
Dec 13 '18 at 11:01
add a comment |
By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.
My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?
Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists.
group-theory finite-groups
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.
My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?
Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists.
group-theory finite-groups
group-theory finite-groups
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
edited Dec 13 '18 at 20:04
Arnaud Casteigts
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
asked Dec 13 '18 at 10:23
Arnaud CasteigtsArnaud Casteigts
83
83
2
It's good to keep small examples of finite groups in mind. Here, the non-cyclic group of order $4$ is already a counterexample.
– lulu
Dec 13 '18 at 11:01
add a comment |
2
It's good to keep small examples of finite groups in mind. Here, the non-cyclic group of order $4$ is already a counterexample.
– lulu
Dec 13 '18 at 11:01
2
2
It's good to keep small examples of finite groups in mind. Here, the non-cyclic group of order $4$ is already a counterexample.
– lulu
Dec 13 '18 at 11:01
It's good to keep small examples of finite groups in mind. Here, the non-cyclic group of order $4$ is already a counterexample.
– lulu
Dec 13 '18 at 11:01
add a comment |
3 Answers
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No, because that would imply that $G$ has a cyclic Sylow subgroup, which is not always true. A simple example: $G=mathbb{F}_p^2$ for addition.
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
add a comment |
No, $G= C_p times cdots times C_p$ ($k$ factors) has order $p^k$ but all nontrivial elements have order $p$.
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
add a comment |
No. $mathbb{Z}_2 oplus mathbb{Z}_2$ has order $2^2$ but no elements of order $2^2$.
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
add a comment |
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3 Answers
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3 Answers
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No, because that would imply that $G$ has a cyclic Sylow subgroup, which is not always true. A simple example: $G=mathbb{F}_p^2$ for addition.
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
add a comment |
No, because that would imply that $G$ has a cyclic Sylow subgroup, which is not always true. A simple example: $G=mathbb{F}_p^2$ for addition.
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
add a comment |
No, because that would imply that $G$ has a cyclic Sylow subgroup, which is not always true. A simple example: $G=mathbb{F}_p^2$ for addition.
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
No, because that would imply that $G$ has a cyclic Sylow subgroup, which is not always true. A simple example: $G=mathbb{F}_p^2$ for addition.
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
answered Dec 13 '18 at 10:24
MindlackMindlack
2,59217
2,59217
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
add a comment |
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
Your answer seem to suggest that such an element could however exist in some cases. If we restrict G to be the symmetric group (of any order), would your answer change? (If so, I'll update the question accordingly.) Thanks.
– Arnaud Casteigts
Dec 13 '18 at 11:00
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
if $n$ is large enough and $p$ is any prime factor of $n!$ (say, $p < n/2$), the corresponding $p$-Sylow has cardinality $p^k$, where $k>n/p -1$. To have an element of corresponding order, you need that $p^k leq n$. If this holds, then $k geq 2$ thus $n geq p^2$ hence by Legendre $k geq n/p$ and $n geq p^k > kp geq n$. Thus, in a symmetric group $S_n$, there is a cyclic $p$-Sylow iff $p > n/2$, when, say, $n geq 5$.
– Mindlack
Dec 13 '18 at 11:33
add a comment |
No, $G= C_p times cdots times C_p$ ($k$ factors) has order $p^k$ but all nontrivial elements have order $p$.
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
add a comment |
No, $G= C_p times cdots times C_p$ ($k$ factors) has order $p^k$ but all nontrivial elements have order $p$.
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
add a comment |
No, $G= C_p times cdots times C_p$ ($k$ factors) has order $p^k$ but all nontrivial elements have order $p$.
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
No, $G= C_p times cdots times C_p$ ($k$ factors) has order $p^k$ but all nontrivial elements have order $p$.
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
answered Dec 13 '18 at 10:42
lhflhf
163k10168388
163k10168388
add a comment |
add a comment |
No. $mathbb{Z}_2 oplus mathbb{Z}_2$ has order $2^2$ but no elements of order $2^2$.
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
add a comment |
No. $mathbb{Z}_2 oplus mathbb{Z}_2$ has order $2^2$ but no elements of order $2^2$.
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
add a comment |
No. $mathbb{Z}_2 oplus mathbb{Z}_2$ has order $2^2$ but no elements of order $2^2$.
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
No. $mathbb{Z}_2 oplus mathbb{Z}_2$ has order $2^2$ but no elements of order $2^2$.
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
answered Dec 13 '18 at 10:25
MathematicsStudent1122MathematicsStudent1122
8,55122466
8,55122466
add a comment |
add a comment |
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It's good to keep small examples of finite groups in mind. Here, the non-cyclic group of order $4$ is already a counterexample.
– lulu
Dec 13 '18 at 11:01