Mixing
A group of order 30 has at most 7 subgroups of order 5
$begingroup$
Show that a group of order 30 has at most 7 subgroups of order 5.
This should be a basic question (from an introductory algebra course), but I got no clue... Please help!
abstract-algebra group-theory
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
$endgroup$
add a comment |
$begingroup$
Show that a group of order 30 has at most 7 subgroups of order 5.
This should be a basic question (from an introductory algebra course), but I got no clue... Please help!
abstract-algebra group-theory
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
$endgroup$
1
$begingroup$
Each subgroup of order 5 will contain the identity element & 4 other elements. These four elements had better be different to four other elements in a different subgroup. Now count the elements.
$endgroup$
– Donald Splutterwit
Feb 20 '17 at 13:57
add a comment |
$begingroup$
Show that a group of order 30 has at most 7 subgroups of order 5.
This should be a basic question (from an introductory algebra course), but I got no clue... Please help!
abstract-algebra group-theory
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
$endgroup$
Show that a group of order 30 has at most 7 subgroups of order 5.
This should be a basic question (from an introductory algebra course), but I got no clue... Please help!
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
edited Jan 13 at 19:22
Matt Samuel
38.4k63768
38.4k63768
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
asked Feb 20 '17 at 13:54
A. ChuA. Chu
7,01593284
7,01593284
1
$begingroup$
Each subgroup of order 5 will contain the identity element & 4 other elements. These four elements had better be different to four other elements in a different subgroup. Now count the elements.
$endgroup$
– Donald Splutterwit
Feb 20 '17 at 13:57
add a comment |
1
$begingroup$
Each subgroup of order 5 will contain the identity element & 4 other elements. These four elements had better be different to four other elements in a different subgroup. Now count the elements.
$endgroup$
– Donald Splutterwit
Feb 20 '17 at 13:57
1
1
$begingroup$
Each subgroup of order 5 will contain the identity element & 4 other elements. These four elements had better be different to four other elements in a different subgroup. Now count the elements.
$endgroup$
– Donald Splutterwit
Feb 20 '17 at 13:57
$begingroup$
Each subgroup of order 5 will contain the identity element & 4 other elements. These four elements had better be different to four other elements in a different subgroup. Now count the elements.
$endgroup$
– Donald Splutterwit
Feb 20 '17 at 13:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It's going to have fewer than $7$, but proving that requires some tools it doesn't sound like you have.
Two distinct subgroups of order $5$ intersect only in the identity (by Lagrange's theorem). This means no two will share any nonidentity elements. Thus if there are $m$ subgroups of order $5$, you can calculate exactly how many elements are in their union. This can't be more than $30$, the order of the group.
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
$endgroup$
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
1
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
add a comment |
$begingroup$
The intersection of two subgroups is itself a subgroup. Lagrange's Theorem therefore implies that the intersection of two subgroups of order $5$ must have order $1$ or $5$. Therefore, two distinct subgroups of order $5$ must intersect only at the identity element.
If there are $k$ distinct subgroups of order $5$, then there are $1+4k$ distinct elements in those subgroups, which must be at most the order of the group.
Using this GAP code
for G in AllSmallGroups(30) do
S:=Set(List(G,g->Group(g)));
Print(StructureDescription(G)," ",Number(S,H->Size(H)=5),"n");
od;
We can check that every group of order $30$ actually has a unique subgroup of order $5$. These groups are $C_5 times S_3$, $C_3 times D_{10}$, $D_{30}$, and $C_{30}$. (See also Groups of order 30.)
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Apply Sylow theorem,
$$30 = 2 times 3 times 5.$$
No. Of Sylow 5 sub-groups $= 1 mod 5$ and divides 6, i.e. 1 or 6
So, max 6 subgroups of order 5 is possible.
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
$endgroup$
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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votes
$begingroup$
It's going to have fewer than $7$, but proving that requires some tools it doesn't sound like you have.
Two distinct subgroups of order $5$ intersect only in the identity (by Lagrange's theorem). This means no two will share any nonidentity elements. Thus if there are $m$ subgroups of order $5$, you can calculate exactly how many elements are in their union. This can't be more than $30$, the order of the group.
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
$endgroup$
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
1
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
add a comment |
$begingroup$
It's going to have fewer than $7$, but proving that requires some tools it doesn't sound like you have.
Two distinct subgroups of order $5$ intersect only in the identity (by Lagrange's theorem). This means no two will share any nonidentity elements. Thus if there are $m$ subgroups of order $5$, you can calculate exactly how many elements are in their union. This can't be more than $30$, the order of the group.
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
$endgroup$
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
1
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
add a comment |
$begingroup$
It's going to have fewer than $7$, but proving that requires some tools it doesn't sound like you have.
Two distinct subgroups of order $5$ intersect only in the identity (by Lagrange's theorem). This means no two will share any nonidentity elements. Thus if there are $m$ subgroups of order $5$, you can calculate exactly how many elements are in their union. This can't be more than $30$, the order of the group.
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
$endgroup$
It's going to have fewer than $7$, but proving that requires some tools it doesn't sound like you have.
Two distinct subgroups of order $5$ intersect only in the identity (by Lagrange's theorem). This means no two will share any nonidentity elements. Thus if there are $m$ subgroups of order $5$, you can calculate exactly how many elements are in their union. This can't be more than $30$, the order of the group.
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
answered Feb 20 '17 at 13:59
Matt SamuelMatt Samuel
38.4k63768
38.4k63768
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
1
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
add a comment |
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
1
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
$begingroup$
Thank you for your answer. Just curious, can you tell me more about "It's going to have fewer than 7"?
$endgroup$
– A. Chu
Feb 20 '17 at 14:46
1
1
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
$begingroup$
en.m.wikipedia.org/wiki/Sylow_theorems @JASON
$endgroup$
– Matt Samuel
Feb 20 '17 at 14:50
add a comment |
$begingroup$
The intersection of two subgroups is itself a subgroup. Lagrange's Theorem therefore implies that the intersection of two subgroups of order $5$ must have order $1$ or $5$. Therefore, two distinct subgroups of order $5$ must intersect only at the identity element.
If there are $k$ distinct subgroups of order $5$, then there are $1+4k$ distinct elements in those subgroups, which must be at most the order of the group.
Using this GAP code
for G in AllSmallGroups(30) do
S:=Set(List(G,g->Group(g)));
Print(StructureDescription(G)," ",Number(S,H->Size(H)=5),"n");
od;
We can check that every group of order $30$ actually has a unique subgroup of order $5$. These groups are $C_5 times S_3$, $C_3 times D_{10}$, $D_{30}$, and $C_{30}$. (See also Groups of order 30.)
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
The intersection of two subgroups is itself a subgroup. Lagrange's Theorem therefore implies that the intersection of two subgroups of order $5$ must have order $1$ or $5$. Therefore, two distinct subgroups of order $5$ must intersect only at the identity element.
If there are $k$ distinct subgroups of order $5$, then there are $1+4k$ distinct elements in those subgroups, which must be at most the order of the group.
Using this GAP code
for G in AllSmallGroups(30) do
S:=Set(List(G,g->Group(g)));
Print(StructureDescription(G)," ",Number(S,H->Size(H)=5),"n");
od;
We can check that every group of order $30$ actually has a unique subgroup of order $5$. These groups are $C_5 times S_3$, $C_3 times D_{10}$, $D_{30}$, and $C_{30}$. (See also Groups of order 30.)
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
The intersection of two subgroups is itself a subgroup. Lagrange's Theorem therefore implies that the intersection of two subgroups of order $5$ must have order $1$ or $5$. Therefore, two distinct subgroups of order $5$ must intersect only at the identity element.
If there are $k$ distinct subgroups of order $5$, then there are $1+4k$ distinct elements in those subgroups, which must be at most the order of the group.
Using this GAP code
for G in AllSmallGroups(30) do
S:=Set(List(G,g->Group(g)));
Print(StructureDescription(G)," ",Number(S,H->Size(H)=5),"n");
od;
We can check that every group of order $30$ actually has a unique subgroup of order $5$. These groups are $C_5 times S_3$, $C_3 times D_{10}$, $D_{30}$, and $C_{30}$. (See also Groups of order 30.)
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
The intersection of two subgroups is itself a subgroup. Lagrange's Theorem therefore implies that the intersection of two subgroups of order $5$ must have order $1$ or $5$. Therefore, two distinct subgroups of order $5$ must intersect only at the identity element.
If there are $k$ distinct subgroups of order $5$, then there are $1+4k$ distinct elements in those subgroups, which must be at most the order of the group.
Using this GAP code
for G in AllSmallGroups(30) do
S:=Set(List(G,g->Group(g)));
Print(StructureDescription(G)," ",Number(S,H->Size(H)=5),"n");
od;
We can check that every group of order $30$ actually has a unique subgroup of order $5$. These groups are $C_5 times S_3$, $C_3 times D_{10}$, $D_{30}$, and $C_{30}$. (See also Groups of order 30.)
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
edited Feb 20 '17 at 14:16
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
answered Feb 20 '17 at 13:59
Rebecca J. StonesRebecca J. Stones
21k22781
21k22781
add a comment |
add a comment |
$begingroup$
Apply Sylow theorem,
$$30 = 2 times 3 times 5.$$
No. Of Sylow 5 sub-groups $= 1 mod 5$ and divides 6, i.e. 1 or 6
So, max 6 subgroups of order 5 is possible.
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
$endgroup$
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
add a comment |
$begingroup$
Apply Sylow theorem,
$$30 = 2 times 3 times 5.$$
No. Of Sylow 5 sub-groups $= 1 mod 5$ and divides 6, i.e. 1 or 6
So, max 6 subgroups of order 5 is possible.
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
$endgroup$
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
add a comment |
$begingroup$
Apply Sylow theorem,
$$30 = 2 times 3 times 5.$$
No. Of Sylow 5 sub-groups $= 1 mod 5$ and divides 6, i.e. 1 or 6
So, max 6 subgroups of order 5 is possible.
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
$endgroup$
Apply Sylow theorem,
$$30 = 2 times 3 times 5.$$
No. Of Sylow 5 sub-groups $= 1 mod 5$ and divides 6, i.e. 1 or 6
So, max 6 subgroups of order 5 is possible.
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
edited Aug 13 '18 at 19:58
Xander Henderson
14.4k103554
14.4k103554
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
answered Aug 13 '18 at 19:27
uttam sumanuttam suman
1
1
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
add a comment |
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
$begingroup$
Your answer is quite hard to understand. I've edited it to include some MathJax, but it could use some additional work. Writing in complete sentences would be a good start.
$endgroup$
– Xander Henderson
Aug 13 '18 at 19:58
add a comment |
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$begingroup$
Each subgroup of order 5 will contain the identity element & 4 other elements. These four elements had better be different to four other elements in a different subgroup. Now count the elements.
$endgroup$
– Donald Splutterwit
Feb 20 '17 at 13:57