Existence of subgroup of order power of prime in a finite abelian group?
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Say I have a finite abelian group $G$ such that $left | G right |=p_1^n_1...p_m^n_m$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^n_1$? I'm thinking that I can use the structure theorem to write $G$ as $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1times...timesmathbbZ/p_m^n_m_1times ...times mathbbZ/p_m^n_m_s_m$ then just take $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1timesleft 1 right ...timesleft 1 right $ where $sum_i n_1_i=n_1$.
Is my reasoning correct? I would like to avoid using Sylow theorems.
Thanks in advance.
group-theory finite-groups abelian-groups
asked Dec 20 '18 at 21:38
John11
1,0071821
add a comment |
Say I have a finite abelian group $G$ such that $left | G right |=p_1^n_1...p_m^n_m$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^n_1$? I'm thinking that I can use the structure theorem to write $G$ as $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1times...timesmathbbZ/p_m^n_m_1times ...times mathbbZ/p_m^n_m_s_m$ then just take $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1timesleft 1 right ...timesleft 1 right $ where $sum_i n_1_i=n_1$.
Is my reasoning correct? I would like to avoid using Sylow theorems.
Thanks in advance.
group-theory finite-groups abelian-groups
asked Dec 20 '18 at 21:38
John11
1,0071821
1
Yes, you're on the right track (though those $;n_1_r;$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
– DonAntonio
Dec 20 '18 at 21:49
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
– John11
Dec 20 '18 at 21:56
add a comment |
Say I have a finite abelian group $G$ such that $left | G right |=p_1^n_1...p_m^n_m$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^n_1$? I'm thinking that I can use the structure theorem to write $G$ as $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1times...timesmathbbZ/p_m^n_m_1times ...times mathbbZ/p_m^n_m_s_m$ then just take $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1timesleft 1 right ...timesleft 1 right $ where $sum_i n_1_i=n_1$.
Is my reasoning correct? I would like to avoid using Sylow theorems.
Thanks in advance.
group-theory finite-groups abelian-groups
asked Dec 20 '18 at 21:38
John11
1,0071821
Say I have a finite abelian group $G$ such that $left | G right |=p_1^n_1...p_m^n_m$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^n_1$? I'm thinking that I can use the structure theorem to write $G$ as $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1times...timesmathbbZ/p_m^n_m_1times ...times mathbbZ/p_m^n_m_s_m$ then just take $mathbbZ/p_1^n_1_1times ...times mathbbZ/p_1^n_1_s_1timesleft 1 right ...timesleft 1 right $ where $sum_i n_1_i=n_1$.
Is my reasoning correct? I would like to avoid using Sylow theorems.
Thanks in advance.
group-theory finite-groups abelian-groups
group-theory finite-groups abelian-groups
asked Dec 20 '18 at 21:38
John11
1,0071821
asked Dec 20 '18 at 21:38
John11
1,0071821
edited Dec 20 '18 at 22:05
asked Dec 20 '18 at 21:38
John11
1,0071821
asked Dec 20 '18 at 21:38
John11
1,0071821
asked Dec 20 '18 at 21:38
John11
1,0071821
1,0071821
1
Yes, you're on the right track (though those $;n_1_r;$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
– DonAntonio
Dec 20 '18 at 21:49
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
– John11
Dec 20 '18 at 21:56
add a comment |
1
Yes, you're on the right track (though those $;n_1_r;$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
– DonAntonio
Dec 20 '18 at 21:49
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
– John11
Dec 20 '18 at 21:56
1
1
Yes, you're on the right track (though those $;n_1_r;$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
– DonAntonio
Dec 20 '18 at 21:49
Yes, you're on the right track (though those $;n_1_r;$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
– DonAntonio
Dec 20 '18 at 21:49
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
– John11
Dec 20 '18 at 21:56
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
– John11
Dec 20 '18 at 21:56
add a comment |
2 Answers
2
active
oldest
votes
Some highlights:
First, you can take $;K:=Bbb Z/p_1^n_1Bbb Ztimes1timesldotstimes{1|];$ to get a subgroup of order $;p_1^n_1;$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.
Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .
Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
add a comment |
Here is a roadmap.
Let $p$ be a prime dividing the order of $G$.
Let $P = g in G : ord(g) text is a power of $p$ $.
Then $P$ is a subgroup of $G$ (because $G$ is abelian).
The order of $P$ is a power of $p$ (by Cauchy's theorem).
The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).
answered Dec 21 '18 at 0:33
lhf
163k10167386
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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votes
Some highlights:
First, you can take $;K:=Bbb Z/p_1^n_1Bbb Ztimes1timesldotstimes{1|];$ to get a subgroup of order $;p_1^n_1;$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.
Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .
Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
add a comment |
Some highlights:
First, you can take $;K:=Bbb Z/p_1^n_1Bbb Ztimes1timesldotstimes{1|];$ to get a subgroup of order $;p_1^n_1;$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.
Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .
Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
add a comment |
Some highlights:
First, you can take $;K:=Bbb Z/p_1^n_1Bbb Ztimes1timesldotstimes{1|];$ to get a subgroup of order $;p_1^n_1;$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.
Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .
Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
Some highlights:
First, you can take $;K:=Bbb Z/p_1^n_1Bbb Ztimes1timesldotstimes{1|];$ to get a subgroup of order $;p_1^n_1;$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.
Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .
Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
answered Dec 20 '18 at 23:00
DonAntonio
177k1491225
177k1491225
add a comment |
add a comment |
Here is a roadmap.
Let $p$ be a prime dividing the order of $G$.
Let $P = g in G : ord(g) text is a power of $p$ $.
Then $P$ is a subgroup of $G$ (because $G$ is abelian).
The order of $P$ is a power of $p$ (by Cauchy's theorem).
The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).
answered Dec 21 '18 at 0:33
lhf
163k10167386
add a comment |
Here is a roadmap.
Let $p$ be a prime dividing the order of $G$.
Let $P = g in G : ord(g) text is a power of $p$ $.
Then $P$ is a subgroup of $G$ (because $G$ is abelian).
The order of $P$ is a power of $p$ (by Cauchy's theorem).
The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).
answered Dec 21 '18 at 0:33
lhf
163k10167386
add a comment |
Here is a roadmap.
Let $p$ be a prime dividing the order of $G$.
Let $P = g in G : ord(g) text is a power of $p$ $.
Then $P$ is a subgroup of $G$ (because $G$ is abelian).
The order of $P$ is a power of $p$ (by Cauchy's theorem).
The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).
answered Dec 21 '18 at 0:33
lhf
163k10167386
Here is a roadmap.
Let $p$ be a prime dividing the order of $G$.
Let $P = g in G : ord(g) text is a power of $p$ $.
Then $P$ is a subgroup of $G$ (because $G$ is abelian).
The order of $P$ is a power of $p$ (by Cauchy's theorem).
The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).
answered Dec 21 '18 at 0:33
lhf
163k10167386
answered Dec 21 '18 at 0:33
lhf
163k10167386
answered Dec 21 '18 at 0:33
lhf
163k10167386
answered Dec 21 '18 at 0:33
lhf
163k10167386
163k10167386
add a comment |
add a comment |
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Yes, you're on the right track (though those $;n_1_r;$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
– DonAntonio
Dec 20 '18 at 21:49
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
– John11
Dec 20 '18 at 21:56